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<item xml:lang="es">
		<title>[P(8031)] EBAU Andaluc&#237;a: qu&#237;mica (junio 2023) - ejercicio B.4 (8032)</title>
		<link>https://ejercicios-fyq.com/P-8031-EBAU-Andalucia-quimica-junio-2023-ejercicio-B-4-8032</link>
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		<dc:date>2023-08-26T06:20:45Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>pH</dc:subject>
		<dc:subject>Constante basicidad</dc:subject>
		<dc:subject>&#193;cidos y bases</dc:subject>
		<dc:subject>EBAU</dc:subject>
		<dc:subject>Selectividad</dc:subject>
		<dc:subject>EvAU</dc:subject>

		<description>
&lt;p&gt;Para ver el enunciado y las respuestas a los apartado del ejercicio que se resuelve en el v&#237;deo, solo tienes que hacer clic aqu&#237;.&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/11-PAU-Ejercicios-y-problemas-de-EBAU-y-PAU" rel="directory"&gt;11 - (PAU) Ejercicios y problemas de EBAU y PAU&lt;/a&gt;

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&lt;a href="https://ejercicios-fyq.com/mot47" rel="tag"&gt;pH&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Constante-basicidad" rel="tag"&gt;Constante basicidad&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Acidos-y-bases" rel="tag"&gt;&#193;cidos y bases&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/EBAU-329" rel="tag"&gt;EBAU&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Selectividad" rel="tag"&gt;Selectividad&lt;/a&gt;, 
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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Para ver el enunciado y las respuestas a los apartado del ejercicio que se resuelve en el v&#237;deo, &lt;b&gt;&lt;a href='https://ejercicios-fyq.com/EBAU-Andalucia-quimica-junio-2023-ejercicio-B-4-8031' class=&#034;spip_in&#034;&gt;solo tienes que hacer clic aqu&#237;&lt;/a&gt;&lt;/b&gt;.&lt;/p&gt;
&lt;p&gt; &lt;br/&gt;&lt;/p&gt;
&lt;iframe width=&#034;560&#034; height=&#034;315&#034; src=&#034;https://www.youtube.com/embed/D8eypZbBtXI&#034; title=&#034;YouTube video player&#034; frameborder=&#034;0&#034; allow=&#034;accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture&#034; allowfullscreen&gt;&lt;/iframe&gt;&lt;/div&gt;
		
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	</item>
<item xml:lang="es">
		<title>EBAU Andaluc&#237;a: qu&#237;mica (junio 2023) - ejercicio B.4 (8031)</title>
		<link>https://ejercicios-fyq.com/EBAU-Andalucia-quimica-junio-2023-ejercicio-B-4-8031</link>
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		<dc:date>2023-08-25T05:35:18Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>pH</dc:subject>
		<dc:subject>Constante basicidad</dc:subject>
		<dc:subject>&#193;cidos y bases</dc:subject>
		<dc:subject>EBAU</dc:subject>
		<dc:subject>Selectividad</dc:subject>
		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>EvAU</dc:subject>

		<description>
&lt;p&gt;La metilamina, , es una base d&#233;bil de acuerdo con la teor&#237;a de Br&#246;nsted-Lowry. &lt;br class='autobr' /&gt;
a) Escribe su equilibrio de disoluci&#243;n acuosa. &lt;br class='autobr' /&gt;
b) Escribe la expresi&#243;n de su constante de basicidad, . &lt;br class='autobr' /&gt;
c) &#191;Podr&#237;a un disoluci&#243;n acuosa de metilamina tener un valor de pH = 5? Razona la respuesta.&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Reacciones-de-Transferencia-de-Protones" rel="directory"&gt;Reacciones de Transferencia de Protones&lt;/a&gt;

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&lt;a href="https://ejercicios-fyq.com/mot47" rel="tag"&gt;pH&lt;/a&gt;, 
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&lt;a href="https://ejercicios-fyq.com/Selectividad" rel="tag"&gt;Selectividad&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/EvAU" rel="tag"&gt;EvAU&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;La metilamina, &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L60xH16/1b971ea7f63fd65b664809651ee8ff2f-cbcbf.png?1733055399' style='vertical-align:middle;' width='60' height='16' alt=&#034;\ce{CH3NH2}&#034; title=&#034;\ce{CH3NH2}&#034; /&gt;, es una base d&#233;bil de acuerdo con la teor&#237;a de Br&#246;nsted-Lowry.&lt;/p&gt;
&lt;p&gt;a) Escribe su equilibrio de disoluci&#243;n acuosa.&lt;/p&gt;
&lt;p&gt;b) Escribe la expresi&#243;n de su constante de basicidad, &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L18xH15/42729cb9c390ae177647363e3449bc68-7f283.png?1733055399' style='vertical-align:middle;' width='18' height='15' alt=&#034;\ce{K_b}&#034; title=&#034;\ce{K_b}&#034; /&gt;.&lt;/p&gt;
&lt;p&gt;c) &#191;Podr&#237;a un disoluci&#243;n acuosa de metilamina tener un valor de pH = 5? Razona la respuesta.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;a) &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/5cab0b597f8f138dcd07976ce389a913.png' style=&#034;vertical-align:middle;&#034; width=&#034;331&#034; height=&#034;27&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bf \ce{CH3NH2 + H2O &lt;&lt;=&gt; CH3NH3+ + OH-}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bf \ce{CH3NH2 + H2O &lt;&lt;=&gt; CH3NH3+ + OH-}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; b) &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a47710e5707d9ecd144c4439ef03327d.png' style=&#034;vertical-align:middle;&#034; width=&#034;169&#034; height=&#034;37&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{K_b = \frac{[\ce{CH3NH3+}][\ce{OH-}]}{[\ce{CH3NH2}]}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{K_b = \frac{[\ce{CH3NH3+}][\ce{OH-}]}{[\ce{CH3NH2}]}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; c) &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/3c2d8f6be25fde60bcdcd1674536aa2a.png' style=&#034;vertical-align:middle;&#034; width=&#034;119&#034; height=&#034;25&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\textbf{No es posible}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\textbf{No es posible}}}&#034; /&gt;&lt;/math&gt;&lt;/p&gt;
&lt;p&gt; &lt;br/&gt;&lt;/p&gt;
&lt;p&gt;&lt;u&gt;RESOLUCI&#211;N DEL EJERCICIO EN V&#205;DEO&lt;/u&gt;.&lt;/p&gt;
&lt;iframe width=&#034;560&#034; height=&#034;315&#034; src=&#034;https://www.youtube.com/embed/D8eypZbBtXI&#034; title=&#034;YouTube video player&#034; frameborder=&#034;0&#034; allow=&#034;accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture&#034; allowfullscreen&gt;&lt;/iframe&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>[P(1463)] EBAU Andaluc&#237;a: qu&#237;mica (junio 2011) - ejercicio A.5 (7820)</title>
		<link>https://ejercicios-fyq.com/P-1463-EBAU-Andalucia-quimica-junio-2011-ejercicio-A-5-7820</link>
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		<dc:date>2023-01-02T05:58:54Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Constante basicidad</dc:subject>
		<dc:subject>Constante equilibrio</dc:subject>
		<dc:subject>Concentraciones</dc:subject>
		<dc:subject>EBAU</dc:subject>
		<dc:subject>Selectividad</dc:subject>
		<dc:subject>EvAU</dc:subject>

		<description>
&lt;p&gt;Si quieres ver el enunciado y las respuestas al problema que se resuelve en el v&#237;deo clica AQU&#205;.&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/11-PAU-Ejercicios-y-problemas-de-EBAU-y-PAU" rel="directory"&gt;11 - (PAU) Ejercicios y problemas de EBAU y PAU&lt;/a&gt;

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&lt;a href="https://ejercicios-fyq.com/Constante-equilibrio" rel="tag"&gt;Constante equilibrio&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Concentraciones" rel="tag"&gt;Concentraciones&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/EBAU-329" rel="tag"&gt;EBAU&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Selectividad" rel="tag"&gt;Selectividad&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/EvAU" rel="tag"&gt;EvAU&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Si quieres ver el enunciado y las respuestas al problema que se resuelve en el v&#237;deo clica &lt;b&gt;&lt;a href='https://ejercicios-fyq.com/EBAU-Andalucia-quimica-junio-2011-ejercicio-A-5-1463' class=&#034;spip_in&#034;&gt;AQU&#205;&lt;/a&gt;&lt;/b&gt;.&lt;/p&gt;
&lt;p&gt; &lt;br/&gt;&lt;/p&gt;
&lt;iframe width=&#034;560&#034; height=&#034;315&#034; src=&#034;https://www.youtube.com/embed/UUIP24RpDWw&#034; title=&#034;YouTube video player&#034; frameborder=&#034;0&#034; allow=&#034;accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture&#034; allowfullscreen&gt;&lt;/iframe&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>pH de un b&#250;fer de amoniaco y amonio (6608)</title>
		<link>https://ejercicios-fyq.com/pH-de-un-bufer-de-amoniaco-y-amonio-6608</link>
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		<dc:date>2020-05-27T16:18:20Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Constante basicidad</dc:subject>
		<dc:subject>Par conjugado</dc:subject>
		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Disoluci&#243;n tamp&#243;n</dc:subject>

		<description>
&lt;p&gt;Se prepar&#243; una soluci&#243;n reguladora agregando a 1.0 L de soluci&#243;n de amoniaco, de concentraci&#243;n 3.4 g/L, cloruro de amonio en cantidad suficiente para que su concentraci&#243;n fuera 0.15 M. Calcula el pH de dicho amortiguador. &lt;br class='autobr' /&gt;
Dato: .&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Reacciones-de-Transferencia-de-Protones" rel="directory"&gt;Reacciones de Transferencia de Protones&lt;/a&gt;

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&lt;a href="https://ejercicios-fyq.com/Constante-basicidad" rel="tag"&gt;Constante basicidad&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Par-conjugado" rel="tag"&gt;Par conjugado&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Disolucion-tampon" rel="tag"&gt;Disoluci&#243;n tamp&#243;n&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Se prepar&#243; una soluci&#243;n reguladora agregando a 1.0 L de soluci&#243;n de amoniaco, de concentraci&#243;n 3.4 g/L, cloruro de amonio en cantidad suficiente para que su concentraci&#243;n fuera 0.15 M. Calcula el pH de dicho amortiguador.&lt;/p&gt;
&lt;p&gt;Dato: &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L148xH20/7ce6d063b7ea6ea7d12534da7be62222-b0b3a.png?1732973732' style='vertical-align:middle;' width='148' height='20' alt=&#034;\ce{K_b(NH3)} = 1.8\cdot 10^{-5}&#034; title=&#034;\ce{K_b(NH3)} = 1.8\cdot 10^{-5}&#034; /&gt; .&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Este problema puedes resolverlo de dos modos distintos y voy a ilustrar ambos modos. &lt;br/&gt; &lt;br/&gt; PRIMER MODO: &lt;u&gt;Aplicando la ecuaci&#243;n de Henderson&lt;/u&gt;. &lt;br/&gt; &lt;br/&gt; Lo primero que debes hacer el calcular la molaridad del amoniaco en el amortiguador: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/7f6a22d4be00dea598d429f3ecab0f3a.png' style=&#034;vertical-align:middle;&#034; width=&#034;183&#034; height=&#034;42&#034; alt=&#034;3.4\ \frac{\cancel{g}}{L}\cdot \frac{1\ mol}{17\ \cancel{g}} = \color[RGB]{0,112,192}{\bm{0.2\ \frac{mol}{L}}}&#034; title=&#034;3.4\ \frac{\cancel{g}}{L}\cdot \frac{1\ mol}{17\ \cancel{g}} = \color[RGB]{0,112,192}{\bm{0.2\ \frac{mol}{L}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Esta ecuaci&#243;n relaciona el valor del &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/888f8245209273b8d6ee38683da95513.png' style=&#034;vertical-align:middle;&#034; width=&#034;34&#034; height=&#034;16&#034; alt=&#034;\text{pOH}&#034; title=&#034;\text{pOH}&#034; /&gt; con el valor de la constante de basicidad del amoniaco y las concentraciones iniciales de base y &#225;cido conjugado: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/b5f9ed2d267a46c0d82d36c9c5d6b375.png' style=&#034;vertical-align:middle;&#034; width=&#034;181&#034; height=&#034;42&#034; alt=&#034;\text{pOH} = \ce{pK_b} + log\ \frac{[\ce{NH4+}]}{[\ce{NH3}]}&#034; title=&#034;\text{pOH} = \ce{pK_b} + log\ \frac{[\ce{NH4+}]}{[\ce{NH3}]}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Sustituyes los datos del enunciado y calculas: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/74c98ced24f50539992ec5ede1efcb41.png' style=&#034;vertical-align:middle;&#034; width=&#034;445&#034; height=&#034;51&#034; alt=&#034;\text{pOH} = - log (1.8\cdot 10^{-5}) + log\ \Big(\frac{0.15\ \cancel{M}}{0.2\ \cancel{M}}\Big) = \color[RGB]{2,112,20}{\bf 4.62}&#034; title=&#034;\text{pOH} = - log (1.8\cdot 10^{-5}) + log\ \Big(\frac{0.15\ \cancel{M}}{0.2\ \cancel{M}}\Big) = \color[RGB]{2,112,20}{\bf 4.62}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; Como te piden que calcules el &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/76ae34dbf07ae7b53692f602b8c6943b.png' style=&#034;vertical-align:middle;&#034; width=&#034;20&#034; height=&#034;16&#034; alt=&#034;\ce{pH}&#034; title=&#034;\ce{pH}&#034; /&gt; debes a&#250;n aplicar que la suma de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/76ae34dbf07ae7b53692f602b8c6943b.png' style=&#034;vertical-align:middle;&#034; width=&#034;20&#034; height=&#034;16&#034; alt=&#034;\ce{pH}&#034; title=&#034;\ce{pH}&#034; /&gt; y &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/05953b017f72aec14b2259def83de32a.png' style=&#034;vertical-align:middle;&#034; width=&#034;34&#034; height=&#034;16&#034; alt=&#034;\ce{pOH}&#034; title=&#034;\ce{pOH}&#034; /&gt; es igual a 14: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/090e4e69f9aee331487c1cfc9083924c.png' style=&#034;vertical-align:middle;&#034; width=&#034;483&#034; height=&#034;25&#034; alt=&#034;\text{pH} + \text{pOH} = 14\ \to\ \text{pH} = 14 - \text{pOH} = 14 - 4.62\ \to\ \fbox{\color[RGB]{192,0,0}{\bf \text{pH} = 9.38}}&#034; title=&#034;\text{pH} + \text{pOH} = 14\ \to\ \text{pH} = 14 - \text{pOH} = 14 - 4.62\ \to\ \fbox{\color[RGB]{192,0,0}{\bf \text{pH} = 9.38}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; SEGUNDO MODO: &lt;u&gt;A partir de las concentraciones en el equilibrio y la constante de basicidad&lt;/u&gt;. &lt;br/&gt; &lt;br/&gt; Dado que el amoniaco es un base d&#233;bil reaccionar&#225; poco. Si llamas &lt;i&gt;x&lt;/i&gt; a la parte del amoniaco que reacciona, las concentraciones en el equilibrio, escritas en la constante de basicidad, quedar&#225;n como: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6105cfbc657b0a680aa5de634760813d.png' style=&#034;vertical-align:middle;&#034; width=&#034;248&#034; height=&#034;18&#034; alt=&#034;\color[RGB]{2,112,20}{\textbf{\ce{NH3 + H2O &lt;=&gt; NH4+ + OH-}}}&#034; title=&#034;\color[RGB]{2,112,20}{\textbf{\ce{NH3 + H2O &lt;=&gt; NH4+ + OH-}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/603b2c83474522a5a10230ed6f6d7b23.png' style=&#034;vertical-align:middle;&#034; width=&#034;253&#034; height=&#034;42&#034; alt=&#034;\ce{K_b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = \frac{(0.15 + x)\cdot x}{(0.2 - x)}&#034; title=&#034;\ce{K_b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = \frac{(0.15 + x)\cdot x}{(0.2 - x)}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La ecuaci&#243;n de segundo grado que obtienes al sustituir es &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/8899b658fa92808d7d96cc57d397ec1b.png' style=&#034;vertical-align:middle;&#034; width=&#034;188&#034; height=&#034;17&#034; alt=&#034;x^2 + 0.15x - 3.6\cdot 10^{-6} = 0&#034; title=&#034;x^2 + 0.15x - 3.6\cdot 10^{-6} = 0&#034; /&gt; . El valor de &lt;i&gt;x&lt;/i&gt; al resolver la ecuaci&#243;n es muy peque&#241;o, como era de esperar: &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/eb98c50830812b7ba86cc89fa2dc5aee.png' style=&#034;vertical-align:middle;&#034; width=&#034;106&#034; height=&#034;16&#034; alt=&#034;\color[RGB]{0,112,192}{\bm{x = 2.4\cdot 10^{-5}}}&#034; title=&#034;\color[RGB]{0,112,192}{\bm{x = 2.4\cdot 10^{-5}}}&#034; /&gt;. &lt;br/&gt; &lt;br/&gt; Solo tienes que calcular el &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/888f8245209273b8d6ee38683da95513.png' style=&#034;vertical-align:middle;&#034; width=&#034;34&#034; height=&#034;16&#034; alt=&#034;\text{pOH}&#034; title=&#034;\text{pOH}&#034; /&gt; y luego el &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/eed14945050753792ee3c0f19a8c6d8d.png' style=&#034;vertical-align:middle;&#034; width=&#034;20&#034; height=&#034;16&#034; alt=&#034;\text{pH}&#034; title=&#034;\text{pH}&#034; /&gt; de manera similar al m&#233;todo anterior: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/60497fe9a576ab25e2b63be4e865fa92.png' style=&#034;vertical-align:middle;&#034; width=&#034;274&#034; height=&#034;24&#034; alt=&#034;\text{pOH} = -log\ 2.4\cdot 10^{-5} = 4.62&#034; title=&#034;\text{pOH} = -log\ 2.4\cdot 10^{-5} = 4.62&#034; /&gt; &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/090e4e69f9aee331487c1cfc9083924c.png' style=&#034;vertical-align:middle;&#034; width=&#034;483&#034; height=&#034;25&#034; alt=&#034;\text{pH} + \text{pOH} = 14\ \to\ \text{pH} = 14 - \text{pOH} = 14 - 4.62\ \to\ \fbox{\color[RGB]{192,0,0}{\bf \text{pH} = 9.38}}&#034; title=&#034;\text{pH} + \text{pOH} = 14\ \to\ \text{pH} = 14 - \text{pOH} = 14 - 4.62\ \to\ \fbox{\color[RGB]{192,0,0}{\bf \text{pH} = 9.38}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>pH de una disoluci&#243;n de metilamina y pH del b&#250;fer formado al a&#241;adir cloruro de metilamonio (5518)</title>
		<link>https://ejercicios-fyq.com/pH-de-una-disolucion-de-metilamina-y-pH-del-bufer-formado-al-anadir-cloruro-de</link>
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		<dc:date>2019-08-05T06:30:43Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>pH</dc:subject>
		<dc:subject>Constante basicidad</dc:subject>
		<dc:subject>pOH</dc:subject>
		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Disoluci&#243;n tamp&#243;n</dc:subject>

		<description>
&lt;p&gt;a) Se dispone de una disoluci&#243;n acuosa de metilamina 0.25 M. Sabiendo que su , &#191;cu&#225;l es el pH de la disoluci&#243;n? Escribe la ecuaci&#243;n de disociaci&#243;n. &lt;br class='autobr' /&gt;
b) A de la disoluci&#243;n anterior se le a&#241;aden de una disoluci&#243;n 0.40 M de cloruro de metilamonio. Calcula el pH de la nueva disoluci&#243;n. Escribe el equilibrio que se establece y c&#243;mo act&#250;a la mezcla b&#250;fer si se agrega NaOH.&lt;/p&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;a) Se dispone de una disoluci&#243;n acuosa de metilamina 0.25 M. Sabiendo que su &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L78xH16/7125a1836c9fd10dd8f6e48f6c08ec3f-a1288.png?1733048880' style='vertical-align:middle;' width='78' height='16' alt=&#034;\ce{pK_b} = 3.33&#034; title=&#034;\ce{pK_b} = 3.33&#034; /&gt;, &#191;cu&#225;l es el pH de la disoluci&#243;n? Escribe la ecuaci&#243;n de disociaci&#243;n.&lt;/p&gt;
&lt;p&gt;b) A &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L49xH16/b7f77b39218ff155cb6f97ee0f23e194-7054e.png?1732997747' style='vertical-align:middle;' width='49' height='16' alt=&#034;50 \ cm^3&#034; title=&#034;50 \ cm^3&#034; /&gt; de la disoluci&#243;n anterior se le a&#241;aden &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L49xH16/b7f77b39218ff155cb6f97ee0f23e194-7054e.png?1732997747' style='vertical-align:middle;' width='49' height='16' alt=&#034;50 \ cm^3&#034; title=&#034;50 \ cm^3&#034; /&gt; de una disoluci&#243;n 0.40 M de cloruro de metilamonio. Calcula el pH de la nueva disoluci&#243;n. Escribe el equilibrio que se establece y c&#243;mo act&#250;a la mezcla b&#250;fer si se agrega NaOH.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;a) A partir del dato del &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6d0efe9636bcf057367a43af36f58163.png' style=&#034;vertical-align:middle;&#034; width=&#034;27&#034; height=&#034;16&#034; alt=&#034;\ce{pK_b}&#034; title=&#034;\ce{pK_b}&#034; /&gt; podemos obtener el valor de la constante de disociaci&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/942f2bea4c077b48f27908346cba3f3f.png' style=&#034;vertical-align:middle;&#034; width=&#034;305&#034; height=&#034;17&#034; alt=&#034;\ce{K_b} = 10^{-\ce{pK_b}} = 10^{-3.33}\ \to\ \ce{K_b} = 4.68\cdot 10^{-4}&#034; title=&#034;\ce{K_b} = 10^{-\ce{pK_b}} = 10^{-3.33}\ \to\ \ce{K_b} = 4.68\cdot 10^{-4}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El equilibrio de disociaci&#243;n es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/7f1d1903b01686d9f7e90846f237f4da.png' style=&#034;vertical-align:middle;&#034; width=&#034;362&#034; height=&#034;23&#034; alt=&#034;\ce{CH3NH2 + H2O &lt;=&gt; CH3NH3+ + OH-}&#034; title=&#034;\ce{CH3NH2 + H2O &lt;=&gt; CH3NH3+ + OH-}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; Las concentraciones en el equilbrio son: &lt;br&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/240ed70ade12c62a2832aba98cb078c9.png' style=&#034;vertical-align:middle;&#034; width=&#034;165&#034; height=&#034;18&#034; alt=&#034;[\ce{CH3NH2}]_{eq} = c_0(1 - \alpha)&#034; title=&#034;[\ce{CH3NH2}]_{eq} = c_0(1 - \alpha)&#034; /&gt; &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/01d77d37be7648183497354179570e81.png' style=&#034;vertical-align:middle;&#034; width=&#034;278&#034; height=&#034;25&#034; alt=&#034;[\ce{CH3NH3^+}]_{eq} = [\ce{OH-}]_{eq} = c_0\alpha&#034; title=&#034;[\ce{CH3NH3^+}]_{eq} = [\ce{OH-}]_{eq} = c_0\alpha&#034; /&gt; &lt;br/&gt; &lt;br/&gt; A partir de la ecuaci&#243;n de la constante de ionizaci&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/c91128d4465138d76b130e94316b0874.png' style=&#034;vertical-align:middle;&#034; width=&#034;388&#034; height=&#034;43&#034; alt=&#034;\ce{K_b} = \frac{[\ce{CH3NH3+}]_{eq}\cdot [\ce{OH-}]_{eq}}{[\ce{CH3NH2}]_{eq}} = \frac{c_0\cancel{^2}\cdot \alpha^2}{\cancel{c_0}(1 - \alpha)} = 4.68\cdot 10^{-4}&#034; title=&#034;\ce{K_b} = \frac{[\ce{CH3NH3+}]_{eq}\cdot [\ce{OH-}]_{eq}}{[\ce{CH3NH2}]_{eq}} = \frac{c_0\cancel{^2}\cdot \alpha^2}{\cancel{c_0}(1 - \alpha)} = 4.68\cdot 10^{-4}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Sustituimos y despejamos para obtener una ecuaci&#243;n de segundo grado que hay que resolver: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/9f828da745bbfb9347892fbfa1db1b46.png' style=&#034;vertical-align:middle;&#034; width=&#034;414&#034; height=&#034;17&#034; alt=&#034;0.25\alpha^2 + 4.68\cdot 10^{-4}\alpha - 4.68\cdot 10^{-4} = 0\ \to\ \alpha = 4.23\cdot 10^{-2}&#034; title=&#034;0.25\alpha^2 + 4.68\cdot 10^{-4}\alpha - 4.68\cdot 10^{-4} = 0\ \to\ \alpha = 4.23\cdot 10^{-2}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La concentraci&#243;n de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/949ec6bcf1e80e902a4fb107ebfd4b00.png' style=&#034;vertical-align:middle;&#034; width=&#034;43&#034; height=&#034;17&#034; alt=&#034;\ce{OH^-}&#034; title=&#034;\ce{OH^-}&#034; /&gt; en el equilibrio nos da el valor del pOH: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/5d1cb19feec74793cb9f268052152e94.png' style=&#034;vertical-align:middle;&#034; width=&#034;318&#034; height=&#034;20&#034; alt=&#034;[\ce{OH-}] = 0.25\ M\cdot 4.23\cdot01^{-2} = 1.06\cdot 10^{-2}\ M&#034; title=&#034;[\ce{OH-}] = 0.25\ M\cdot 4.23\cdot01^{-2} = 1.06\cdot 10^{-2}\ M&#034; /&gt; &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/f259e44ae4bbc368199ab94af4104935.png' style=&#034;vertical-align:middle;&#034; width=&#034;324&#034; height=&#034;20&#034; alt=&#034;\text{pOH} = -log\ [\ce{OH-}] = -log\ 1.06\cdot 10^{-2} = 1.98&#034; title=&#034;\text{pOH} = -log\ [\ce{OH-}] = -log\ 1.06\cdot 10^{-2} = 1.98&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El valor del pH es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/44c10e8de9300c44d273e032094fe35d.png' style=&#034;vertical-align:middle;&#034; width=&#034;333&#034; height=&#034;21&#034; alt=&#034;\text{pH} + \text{pOH} = 14\ \to\ \text{pH} = 14 - 1.98 = \fbox{\color[RGB]{192,0,0}{\bf 12.02}}&#034; title=&#034;\text{pH} + \text{pOH} = 14\ \to\ \text{pH} = 14 - 1.98 = \fbox{\color[RGB]{192,0,0}{\bf 12.02}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) Ahora a&#241;adimos la sal al sistema, por lo que a&#241;adimos cati&#243;n &lt;i&gt;metilamonio&lt;/i&gt; (&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/b0664dc4650cc6c3a30efff5a191dedc.png' style=&#034;vertical-align:middle;&#034; width=&#034;70&#034; height=&#034;18&#034; alt=&#034;\ce{CH3NH3^+}&#034; title=&#034;\ce{CH3NH3^+}&#034; /&gt;) al sistema, provocando el &lt;u&gt;efecto ion com&#250;n&lt;/u&gt;. Como la sal se disocia completamente, podemos considerar que la concentraci&#243;n del cati&#243;n &lt;i&gt;metilamonio&lt;/i&gt; es la misma que la de la sal, con lo que el equilibrio del apartado anterior se desplaza hacia la izquierda, form&#225;ndose m&#225;s especie &lt;i&gt;metilamina&lt;/i&gt; y descenciendo la concentraci&#243;n de i&#243;n &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/949ec6bcf1e80e902a4fb107ebfd4b00.png' style=&#034;vertical-align:middle;&#034; width=&#034;43&#034; height=&#034;17&#034; alt=&#034;\ce{OH^-}&#034; title=&#034;\ce{OH^-}&#034; /&gt;. &lt;br/&gt; &lt;br/&gt; Las concentraciones de cada especie, al sumar los vol&#250;menes, se reducen a la mitad por lo que podemos considerar que son: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a1ad34944a189f297f965bbf40cec9f0.png' style=&#034;vertical-align:middle;&#034; width=&#034;270&#034; height=&#034;42&#034; alt=&#034;[\ce{CH3NH3+}] = \frac{0.4\frac{mol}{\cancel{L}}\cdot 0.05\ \cancel{L}}{0.1\ L} = 0.2\ M&#034; title=&#034;[\ce{CH3NH3+}] = \frac{0.4\frac{mol}{\cancel{L}}\cdot 0.05\ \cancel{L}}{0.1\ L} = 0.2\ M&#034; /&gt; &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/df72c9add2bcec8176a2ea75ba8e0304.png' style=&#034;vertical-align:middle;&#034; width=&#034;284&#034; height=&#034;42&#034; alt=&#034;[\ce{CH3NH2}] = \frac{0.25\frac{mol}{\cancel{L}}\cdot 0.05\ \cancel{L}}{0.1\ L} = 0.125\ M&#034; title=&#034;[\ce{CH3NH2}] = \frac{0.25\frac{mol}{\cancel{L}}\cdot 0.05\ \cancel{L}}{0.1\ L} = 0.125\ M&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Calculamos el pOH de la disoluci&#243;n b&#250;fer formada a partir de la expresi&#243;n que lo relaciona con la &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6d0efe9636bcf057367a43af36f58163.png' style=&#034;vertical-align:middle;&#034; width=&#034;27&#034; height=&#034;16&#034; alt=&#034;\ce{pK_b}&#034; title=&#034;\ce{pK_b}&#034; /&gt; y el cociente entre las especies &#225;cida y b&#225;sica: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/2b45d39d9e87d4693deb132b9245e8ad.png' style=&#034;vertical-align:middle;&#034; width=&#034;480&#034; height=&#034;42&#034; alt=&#034;\text{pOH} = \ce{pK_b} + log\ \left(\frac{[\ce{CH3NH3+}]}{[\ce{CH3NH2}]}\right) = 3.33 + log\ \left(\frac{0.2\ \cancel{M}}{0.125\ \cancel{M}}\right) = 3.53&#034; title=&#034;\text{pOH} = \ce{pK_b} + log\ \left(\frac{[\ce{CH3NH3+}]}{[\ce{CH3NH2}]}\right) = 3.33 + log\ \left(\frac{0.2\ \cancel{M}}{0.125\ \cancel{M}}\right) = 3.53&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El pH ser&#225;, por lo tanto: &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a30cb4f69b396ec885d9942bdc7f016b.png' style=&#034;vertical-align:middle;&#034; width=&#034;333&#034; height=&#034;21&#034; alt=&#034;\text{pH} + \text{pOH} = 14\ \to\ \text{pH} = 14 - 3.53 = \fbox{\color[RGB]{192,0,0}{\bf 10.47}}&#034; title=&#034;\text{pH} + \text{pOH} = 14\ \to\ \text{pH} = 14 - 3.53 = \fbox{\color[RGB]{192,0,0}{\bf 10.47}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>pH final de una mezcla de disoluciones de HCl y NH3 (5471)</title>
		<link>https://ejercicios-fyq.com/pH-final-de-una-mezcla-de-disoluciones-de-HCl-y-NH3-5471</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/pH-final-de-una-mezcla-de-disoluciones-de-HCl-y-NH3-5471</guid>
		<dc:date>2019-07-23T09:19:05Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>pH</dc:subject>
		<dc:subject>Constante basicidad</dc:subject>
		<dc:subject>Ionizaci&#243;n</dc:subject>
		<dc:subject>Neutralizaci&#243;n</dc:subject>
		<dc:subject>RESUELTO</dc:subject>

		<description>
&lt;p&gt;&#191;Qu&#233; pH tendr&#225; una disoluci&#243;n de 20.0 mL de HCl de concentraci&#243;n 0.5 M a la que se le a&#241;aden 60.0 mL de de 0.5 M de concentraci&#243;n? &lt;br class='autobr' /&gt;
Dato:&lt;/p&gt;


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		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;&#191;Qu&#233; pH tendr&#225; una disoluci&#243;n de 20.0 mL de HCl de concentraci&#243;n 0.5 M a la que se le a&#241;aden 60.0 mL de &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L57xH18/48f9d52492394fa3ac7b19a0441a3982-2db8d.png?1733007664' style='vertical-align:middle;' width='57' height='18' alt=&#034;\ce{NH_3(ac)}&#034; title=&#034;\ce{NH_3(ac)}&#034; /&gt; de 0.5 M de concentraci&#243;n?&lt;/math&gt;&lt;/p&gt;
&lt;p&gt;Dato: &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L190xH25/1d092df7f238e7b41fbba3b5d0be61b0-6e275.png?1733007664' style='vertical-align:middle;' width='190' height='25' alt=&#034;\ce{K_b(NH3)}=1.8\cdot 10^{-5}&#034; title=&#034;\ce{K_b(NH3)}=1.8\cdot 10^{-5}&#034; /&gt;&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Dado que el amoniaco es una base d&#233;bil, es necesario conocer el grado de disociaci&#243;n de la base para poder calcular los moles de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/d3ae778e752a042743bcfcf19b7a416d.png' style=&#034;vertical-align:middle;&#034; width=&#034;53&#034; height=&#034;23&#034; alt=&#034;[\ce{OH-}]&#034; title=&#034;[\ce{OH-}]&#034; /&gt; que aporta a la neutralizaci&#243;n del &#225;cido. Los moles de protones que el &#225;cido aporta son: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/c2b85c3d2225a574e8db3180ebe02945.png' style=&#034;vertical-align:middle;&#034; width=&#034;312&#034; height=&#034;47&#034; alt=&#034;20\ \cancel{mL}\cdot \frac{0.5\ mol}{10^3\ \cancel{mL}} = \color[RGB]{0,112,192}{\bm{10^{-2}}}\ \color[RGB]{0,112,192}{\textbf{mol\ \ce{H+}}}&#034; title=&#034;20\ \cancel{mL}\cdot \frac{0.5\ mol}{10^3\ \cancel{mL}} = \color[RGB]{0,112,192}{\bm{10^{-2}}}\ \color[RGB]{0,112,192}{\textbf{mol\ \ce{H+}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Para el amoniaco tendr&#237;as que la constante de basicidad, escrita en funci&#243;n del grado de disociaci&#243;n es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/ace6df48709af4d4cd8afd1c05524822.png' style=&#034;vertical-align:middle;&#034; width=&#034;405&#034; height=&#034;57&#034; alt=&#034;K_b = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = \frac{c_0\cancel{^2}\cdot \alpha^2}{\cancel{c_0}(1-\alpha)} = \color[RGB]{2,112,20}{\bm{\frac{c_0\cdot \alpha^2}{(1-\alpha)}}}&#034; title=&#034;K_b = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = \frac{c_0\cancel{^2}\cdot \alpha^2}{\cancel{c_0}(1-\alpha)} = \color[RGB]{2,112,20}{\bm{\frac{c_0\cdot \alpha^2}{(1-\alpha)}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Como el valor de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6e78513fed604635a00deacc0008fa65.png' style=&#034;vertical-align:middle;&#034; width=&#034;27&#034; height=&#034;22&#034; alt=&#034;\ce{K _b}&#034; title=&#034;\ce{K _b}&#034; /&gt; es peque&#241;o, supones que &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/7b7f9dbfea05c83784f8b85149852f08.png' style=&#034;vertical-align:middle;&#034; width=&#034;18&#034; height=&#034;30&#034; alt=&#034;\alpha&#034; title=&#034;\alpha&#034; /&gt; es mucho menor que uno y haces la aproximaci&#243;n &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/16d736361b3f217367a9a6cf33d709cb.png' style=&#034;vertical-align:middle;&#034; width=&#034;87&#034; height=&#034;63&#034; alt=&#034;\cancelto{1}{(1 - \alpha)}&#034; title=&#034;\cancelto{1}{(1 - \alpha)}&#034; /&gt;: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/61722e529ffa7a2a26c24c5be85dce58.png' style=&#034;vertical-align:middle;&#034; width=&#034;472&#034; height=&#034;55&#034; alt=&#034;K_b = c_0\alpha^2\ \to\ \alpha = \sqrt{\frac{K_b}{c_0}} = \sqrt{\frac{1.8\cdot 10^{-5}}{0.5}} = \color[RGB]{0,112,192}{\bf 6\cdot 10^{-3}}&#034; title=&#034;K_b = c_0\alpha^2\ \to\ \alpha = \sqrt{\frac{K_b}{c_0}} = \sqrt{\frac{1.8\cdot 10^{-5}}{0.5}} = \color[RGB]{0,112,192}{\bf 6\cdot 10^{-3}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; (La aproximaci&#243;n que has hecho, es buena por lo tanto). &lt;br/&gt; &lt;br/&gt; La concentraci&#243;n de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/d3ae778e752a042743bcfcf19b7a416d.png' style=&#034;vertical-align:middle;&#034; width=&#034;53&#034; height=&#034;23&#034; alt=&#034;[\ce{OH-}]&#034; title=&#034;[\ce{OH-}]&#034; /&gt; en el equilibrio ser&#225;: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/ce4e38990858fc0b9d402ab247c09579.png' style=&#034;vertical-align:middle;&#034; width=&#034;362&#034; height=&#034;25&#034; alt=&#034;[\ce{OH-}] = 0.5\ M\cdot 6\cdot 10^{-3} = \color[RGB]{0,112,192}{\bm{3\cdot 10^{-3}\ M}}&#034; title=&#034;[\ce{OH-}] = 0.5\ M\cdot 6\cdot 10^{-3} = \color[RGB]{0,112,192}{\bm{3\cdot 10^{-3}\ M}}&#034; /&gt;&lt;/p&gt;
&lt;p&gt;Los moles de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/d0c875b2dd9adbfe34c1e9a5bf242bff.png' style=&#034;vertical-align:middle;&#034; width=&#034;39&#034; height=&#034;15&#034; alt=&#034;\ce{OH-}&#034; title=&#034;\ce{OH-}&#034; /&gt; que aporta el amoniaco son: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/bdea65baa2f16ed2b76c903569fd5b4d.png' style=&#034;vertical-align:middle;&#034; width=&#034;416&#034; height=&#034;50&#034; alt=&#034;60\ \cancel{mL}\cdot \frac{3\cdot 10^{-3}\ mol}{10^3\ \cancel{mL}} = \color[RGB]{0,112,192}{\bm{1.8\cdot 10^{-4}}}\ \color[RGB]{0,112,192}{\textbf{mol \ce{OH-}}}&#034; title=&#034;60\ \cancel{mL}\cdot \frac{3\cdot 10^{-3}\ mol}{10^3\ \cancel{mL}} = \color[RGB]{0,112,192}{\bm{1.8\cdot 10^{-4}}}\ \color[RGB]{0,112,192}{\textbf{mol \ce{OH-}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Al haber m&#225;s protones que hidroxilos en disoluci&#243;n, el resultado ser&#225; una disoluci&#243;n &#225;cida de la que tienes que conocer su concentraci&#243;n molar para poder calcular el pH. Quedar&#225;n sin neutralizar: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/17c0d60b24924724e12b81200ac95298.png' style=&#034;vertical-align:middle;&#034; width=&#034;418&#034; height=&#034;25&#034; alt=&#034;n_{\ce{H+}} = (10^{-2} - 1.8\cdot 10^{-4}) = \color[RGB]{0,112,192}{\bm{9.82\cdot 10^{-3}\ mol}}&#034; title=&#034;n_{\ce{H+}} = (10^{-2} - 1.8\cdot 10^{-4}) = \color[RGB]{0,112,192}{\bm{9.82\cdot 10^{-3}\ mol}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La concentraci&#243;n molar la obtienes considerando que los vol&#250;menes mezclados son aditivos y el volumen final son 80 mL: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/778dd8e311a2ad810df454e5706c375f.png' style=&#034;vertical-align:middle;&#034; width=&#034;321&#034; height=&#034;48&#034; alt=&#034;[\ce{H+}] = \frac{9.82\cdot 10^{-3}\ mol}{8\cdot 10^{-2}\ L} = \color[RGB]{0,112,192}{\bf 0.123\ M}&#034; title=&#034;[\ce{H+}] = \frac{9.82\cdot 10^{-3}\ mol}{8\cdot 10^{-2}\ L} = \color[RGB]{0,112,192}{\bf 0.123\ M}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El pH de la disoluci&#243;n final ser&#225;: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/c010f7567c1887114f1a412727d0c684.png' style=&#034;vertical-align:middle;&#034; width=&#034;364&#034; height=&#034;27&#034; alt=&#034;pH = -log\ [\ce{H+}] = -log\ 0.123 = \fbox{\color[RGB]{192,0,0}{\bf 0.91}}&#034; title=&#034;pH = -log\ [\ce{H+}] = -log\ 0.123 = \fbox{\color[RGB]{192,0,0}{\bf 0.91}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>pH y diluci&#243;n Opci&#243;n A EBAU Andaluc&#237;a junio 2017</title>
		<link>https://ejercicios-fyq.com/pH-y-dilucion-Opcion-A-EBAU-Andalucia-junio-2017</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/pH-y-dilucion-Opcion-A-EBAU-Andalucia-junio-2017</guid>
		<dc:date>2017-06-17T07:46:48Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>pH</dc:subject>
		<dc:subject>Constante basicidad</dc:subject>
		<dc:subject>EBAU</dc:subject>
		<dc:subject>Diluci&#243;n</dc:subject>
		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>EvAU</dc:subject>

		<description>
&lt;p&gt;a) El grado de disociaci&#243;n de una disoluci&#243;n 0,03 M de hidr&#243;xido de amonio () es 0,024. Calcula la constante de disociaci&#243;n () del hidr&#243;xido de amonio y el pH de la disoluci&#243;n. b) Calcula el volumen de agua que hay que a&#241;adir a 100 mL de una disoluci&#243;n de NaOH 0,03 M para que el pH sea 11,5.&lt;/p&gt;


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		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;a) El grado de disociaci&#243;n de una disoluci&#243;n 0,03 M de hidr&#243;xido de amonio (&lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L80xH40/e1c6b3a2ece1335b94d9a625beb22b2e-1a9ca.png?1733219782' style='vertical-align:middle;' width='80' height='40' alt=&#034;NH_4OH&#034; title=&#034;NH_4OH&#034; /&gt;) es 0,024. Calcula la constante de disociaci&#243;n (&lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L28xH40/82babf9266d8cb6d7f42fffe47f40df2-4c891.png?1732967609' style='vertical-align:middle;' width='28' height='40' alt=&#034;K_b&#034; title=&#034;K_b&#034; /&gt;) del hidr&#243;xido de amonio y el pH de la disoluci&#243;n. &lt;br/&gt; b) Calcula el volumen de agua que hay que a&#241;adir a 100 mL de una disoluci&#243;n de NaOH 0,03 M para que el pH sea 11,5.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;En primer lugar debemos escribir el equilibrio y las concentraciones en el equilibrio: &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/23dfb48752b28a1e5aa067d8e2a91e1b.png' style=&#034;vertical-align:middle;&#034; width=&#034;238&#034; height=&#034;45&#034; alt=&#034;NH_4OH\ \rightleftharpoons\ NH_4^+ + OH^-&#034; title=&#034;NH_4OH\ \rightleftharpoons\ NH_4^+ + OH^-&#034; /&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/00008cc74a9f4add9f19e59c8a2b4b29.png' style=&#034;vertical-align:middle;&#034; width=&#034;227&#034; height=&#034;42&#034; alt=&#034;c_0(1-\alpha)\ \ \ \ \ \ \ c_0\alpha\ \ \ \ \ \ c_0\alpha&#034; title=&#034;c_0(1-\alpha)\ \ \ \ \ \ \ c_0\alpha\ \ \ \ \ \ c_0\alpha&#034; /&gt; &lt;br/&gt; Como conocemos el valor de la concentraci&#243;n inicial y el del grado de disociaci&#243;n, tenemos que las concentraciones en el equilibrio son: &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/ff79234df58f6ad20bec212f40809d34.png' style=&#034;vertical-align:middle;&#034; width=&#034;235&#034; height=&#034;47&#034; alt=&#034;[NH_4OH] = 2,93\cdot 10^{-2}\ M&#034; title=&#034;[NH_4OH] = 2,93\cdot 10^{-2}\ M&#034; /&gt; y &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a060439c5ccde2dc043f59f0a04cc104.png' style=&#034;vertical-align:middle;&#034; width=&#034;278&#034; height=&#034;47&#034; alt=&#034;[NH_4^+] = [OH^-] = 7,2\cdot 10^{-4}\ M&#034; title=&#034;[NH_4^+] = [OH^-] = 7,2\cdot 10^{-4}\ M&#034; /&gt; &lt;br/&gt; El valor de la constante &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/82babf9266d8cb6d7f42fffe47f40df2.png' style=&#034;vertical-align:middle;&#034; width=&#034;28&#034; height=&#034;40&#034; alt=&#034;K_b&#034; title=&#034;K_b&#034; /&gt; ser&#225;: &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/21ac045b945c895c46091b1ce703fdb1.png' style=&#034;vertical-align:middle;&#034; width=&#034;342&#034; height=&#034;72&#034; alt=&#034;K_b = \frac{(7,2\cdot 10^{-4}\ M)^2}{2,93\cdot 10^{-2}\ M} = \bf 1,77\cdot 10^{-5}\ M&#034; title=&#034;K_b = \frac{(7,2\cdot 10^{-4}\ M)^2}{2,93\cdot 10^{-2}\ M} = \bf 1,77\cdot 10^{-5}\ M&#034; /&gt;&lt;/p&gt; &lt;br/&gt;
Conociendo la concentraci&#243;n del iones hidroxilo podemos determinar el valor del pOH y, a partir de &#233;ste, el valor del pH: &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/bbf0456ddb7101207b8dfd0a8f43cfad.png' style=&#034;vertical-align:middle;&#034; width=&#034;397&#034; height=&#034;47&#034; alt=&#034;pOH = -log\ [OH^-] = -log\ 7,2\cdot 10^{-4} = 3,14&#034; title=&#034;pOH = -log\ [OH^-] = -log\ 7,2\cdot 10^{-4} = 3,14&#034; /&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/985dec86c350c3285b79fc944e7bdd8b.png' style=&#034;vertical-align:middle;&#034; width=&#034;403&#034; height=&#034;40&#034; alt=&#034;pH + pOH = 14\ \to\ pH = 14 - 3,14 = \bf 10,86&#034; title=&#034;pH + pOH = 14\ \to\ pH = 14 - 3,14 = \bf 10,86&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) Si el pH es 11,5 quiere decir que el pOH ha de ser: 14 - 11,5 = 2,5. A partir de este dato podemos determinar la concentraci&#243;n necesaria de NaOH: &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/d0fc83fff84da34b9ad458684c19f5e8.png' style=&#034;vertical-align:middle;&#034; alt=&#034;[OH^-] = 10^{-2,5} = 3,16\cdot 10^{-3}\ M&#034; title=&#034;[OH^-] = 10^{-2,5} = 3,16\cdot 10^{-3}\ M&#034; /&gt; &lt;br/&gt; Esto quiere decir que la disoluci&#243;n ha de contener &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/7326bd61c1cfc4d62952e044ad2ef560.png' style=&#034;vertical-align:middle;&#034; width=&#034;137&#034; height=&#034;47&#034; alt=&#034;3,16\cdot 10^{-3}\ mol&#034; title=&#034;3,16\cdot 10^{-3}\ mol&#034; /&gt; de NaOH por cada &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/72c4e54800fc49c7d60e35245249f389.png' style=&#034;vertical-align:middle;&#034; width=&#034;67&#034; height=&#034;20&#034; alt=&#034;10^3\ mL&#034; title=&#034;10^3\ mL&#034; /&gt; de disoluci&#243;n. &lt;br/&gt; La disoluci&#243;n de partida contiene: &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6f67c9ed056ed61915a1cbd729b3b7e1.png' style=&#034;vertical-align:middle;&#034; width=&#034;443&#034; height=&#034;72&#034; alt=&#034;100\ mL\cdot \frac{3\cdot 10^{-2}\ mol\ NaOH}{10^3\ mL} = 3\cdot 10^{3}\ mol\ NaOH&#034; title=&#034;100\ mL\cdot \frac{3\cdot 10^{-2}\ mol\ NaOH}{10^3\ mL} = 3\cdot 10^{3}\ mol\ NaOH&#034; /&gt; &lt;br/&gt; S&#243;lo nos queda hacer una proporci&#243;n:
&lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/db2aedddb26a4a00e8c207455f636103.png' style=&#034;vertical-align:middle;&#034; width=&#034;272&#034; height=&#034;72&#034; alt=&#034;\frac{3\cdot 10^{-3}\ mol}{3,16\cdot 10^{-3}\ mol\cdot L^{-1}} = 0,95\ L&#034; title=&#034;\frac{3\cdot 10^{-3}\ mol}{3,16\cdot 10^{-3}\ mol\cdot L^{-1}} = 0,95\ L&#034; /&gt; &lt;br/&gt;
Esto quiere decir que el volumen final ha de ser 950 mL, por lo tanto habr&#225; que a&#241;adir &lt;b&gt;850 mL de agua&lt;/b&gt; a los 100 mL iniciales.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>C&#225;lculo del pH de una disoluci&#243;n de una base d&#233;bil 0001</title>
		<link>https://ejercicios-fyq.com/Calculo-del-pH-de-una-disolucion-de-una-base-debil-0001</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Calculo-del-pH-de-una-disolucion-de-una-base-debil-0001</guid>
		<dc:date>2013-03-06T06:23:56Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>pH</dc:subject>
		<dc:subject>Constante basicidad</dc:subject>
		<dc:subject>&#193;cidos y bases</dc:subject>

		<description>
&lt;p&gt;Burbujeamos 7,33 L de sobre 10 L de a 25&#176;C y 1 atm. Sabiendo que es , calcular el pH de la disoluci&#243;n que se obtiene.&lt;/p&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Burbujeamos 7,33 L de &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L43xH40/12350202d6efb1554c9bcc571957851c-2907d.png?1732967609' style='vertical-align:middle;' width='43' height='40' alt=&#034;NH_3&#034; title=&#034;NH_3&#034; /&gt; sobre 10 L de &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L47xH40/45e46989e3704bc2ba0899724acdca5c-4d78d.png?1732957503' style='vertical-align:middle;' width='47' height='40' alt=&#034;H_2O&#034; title=&#034;H_2O&#034; /&gt; a 25&#176;C y 1 atm. Sabiendo que &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L28xH40/82babf9266d8cb6d7f42fffe47f40df2-4c891.png?1732967609' style='vertical-align:middle;' width='28' height='40' alt=&#034;K_b&#034; title=&#034;K_b&#034; /&gt; es &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L93xH47/22b6b6b22b6723005ff2b93ad902affc-f3e01.png?1732967609' style='vertical-align:middle;' width='93' height='47' alt=&#034;1,85\cdot 10^{-5}&#034; title=&#034;1,85\cdot 10^{-5}&#034; /&gt;, calcular el pH de la disoluci&#243;n que se obtiene.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;&lt;b&gt;pH = 10,86&lt;/b&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>EBAU Andaluc&#237;a: qu&#237;mica (junio 2011) - ejercicio A.5 (1463)</title>
		<link>https://ejercicios-fyq.com/EBAU-Andalucia-quimica-junio-2011-ejercicio-A-5-1463</link>
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		<dc:date>2011-08-16T12:37:26Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Constante basicidad</dc:subject>
		<dc:subject>&#193;cidos y bases</dc:subject>
		<dc:subject>Ionizaci&#243;n</dc:subject>
		<dc:subject>Molaridad</dc:subject>
		<dc:subject>EBAU</dc:subject>
		<dc:subject>Selectividad</dc:subject>
		<dc:subject>EvAU</dc:subject>

		<description>
&lt;p&gt;A una disoluci&#243;n acuosa de amoniaco contiene 0.17 g de este compuesto por litro y se encuentra disociado en un . Calcula: &lt;br class='autobr' /&gt;
a) La concentraci&#243;n de iones hidroxilo y amonio. &lt;br class='autobr' /&gt;
b) La constante de disociaci&#243;n. &lt;br class='autobr' /&gt;
Masas at&#243;micas: N = 14 ; H = 1.&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Molaridad" rel="tag"&gt;Molaridad&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/EBAU-329" rel="tag"&gt;EBAU&lt;/a&gt;, 
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		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;A &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L36xH13/be6d2d5f5497225d8529a0ceea40da82-38932.png?1732951859' style='vertical-align:middle;' width='36' height='13' alt=&#034;25 ^oC&#034; title=&#034;25 ^oC&#034; /&gt; una disoluci&#243;n acuosa de amoniaco contiene 0.17 g de este compuesto por litro y se encuentra disociado en un &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L36xH14/ecf8013feb5a005d1f2bbf750225240b-781b1.png?1733045717' style='vertical-align:middle;' width='36' height='14' alt=&#034;4.3\%&#034; title=&#034;4.3\%&#034; /&gt;. Calcula:&lt;/p&gt;
&lt;p&gt;a) La concentraci&#243;n de iones hidroxilo y amonio.&lt;/p&gt;
&lt;p&gt;b) La constante de disociaci&#243;n.&lt;/p&gt;
&lt;p&gt;Masas at&#243;micas: N = 14 ; H = 1.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;a) &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/b7d4d20527f7dcc75b52cf4d7a1bacd6.png' style=&#034;vertical-align:middle;&#034; width=&#034;263&#034; height=&#034;28&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bf{[\ce{NH4+}] = [\ce{OH-}]}}\color[RGB]{192,0,0}{\bm{= 4.3\cdot 10^{-4}\ M}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bf{[\ce{NH4+}] = [\ce{OH-}]}}\color[RGB]{192,0,0}{\bm{= 4.3\cdot 10^{-4}\ M}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; b) &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/4fc58682abdb263589e8d9ed52cebdd7.png' style=&#034;vertical-align:middle;&#034; width=&#034;166&#034; height=&#034;26&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{K_b = 1.93\cdot 10^{-5}\ M}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{K_b = 1.93\cdot 10^{-5}\ M}}}&#034; /&gt;&lt;/math&gt;&lt;/p&gt;
&lt;p&gt; &lt;br/&gt;&lt;/p&gt;
&lt;p&gt;&lt;u&gt;RESOLUCI&#211;N DEL PROBLEMA EN V&#205;DEO&lt;/u&gt;.&lt;/p&gt;
&lt;iframe width=&#034;560&#034; height=&#034;315&#034; src=&#034;https://www.youtube.com/embed/UUIP24RpDWw&#034; title=&#034;YouTube video player&#034; frameborder=&#034;0&#034; allow=&#034;accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture&#034; allowfullscreen&gt;&lt;/iframe&gt;&lt;/div&gt;
		
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	</item>
<item xml:lang="es">
		<title>Cuesti&#243;n &#225;cido-base 0023</title>
		<link>https://ejercicios-fyq.com/Cuestion-acido-base-0023</link>
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		<dc:date>2010-03-24T07:05:45Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Constante acidez</dc:subject>
		<dc:subject>Constante basicidad</dc:subject>
		<dc:subject>&#193;cidos y bases</dc:subject>
		<dc:subject>Par conjugado</dc:subject>

		<description>
&lt;p&gt;La constante de acidez del &#225;cido cianh&#237;drico, HCN, vale . Calcula la constante de basicidad del i&#243;n cianuro, .&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Par-conjugado" rel="tag"&gt;Par conjugado&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;La constante de acidez del &#225;cido cianh&#237;drico, HCN, vale &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L142xH47/8ccaddf452211f7c49b9685a9ea5ed7c-0e871.png?1733198358' style='vertical-align:middle;' width='142' height='47' alt=&#034;K_a = 4,9\cdot 10^{-10}&#034; title=&#034;K_a = 4,9\cdot 10^{-10}&#034; /&gt;. Calcula la constante de basicidad del i&#243;n cianuro, &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L48xH45/37d93f5911299ee363a5c6e6f6cff343-92c54.png?1733198358' style='vertical-align:middle;' width='48' height='45' alt=&#034;CN^-&#034; title=&#034;CN^-&#034; /&gt;.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/9c0673e45203e8680c25b57839f2de3d.png' style=&#034;vertical-align:middle;&#034; width=&#034;157&#034; height=&#034;47&#034; alt=&#034;\bf K_b = 2,05\cdot 10^{-5}&#034; title=&#034;\bf K_b = 2,05\cdot 10^{-5}&#034; /&gt;&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

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