pH and pOH of an acetic acid solution (8401)

, por F_y_Q

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Calculate the pH and pOH of a 0.001 mol/L acetic acid solution, given its ionization constant (\ce{K_a} = 1.8\cdot 10^{-5})

P.-S.

From the ionization constant value, you can calculate the concentration of ions at equilibrium:

\color[RGB]{2,112,20}{\textbf{\ce{CH3COOH + H2O <=> CH3COO^- + H3O^+}}}

The acidity constant follows the equation:

\color[RGB]{2,112,20}{\bm{K_a = \frac{[\ce{CH3COO^-}][\ce{H3O^+}]}{[\ce{CH3COOH}]}}}

The equilibrium concentrations, based on the initial concentration, are as follows:

[\ce{CH3COOH}] = \color[RGB]{0,112,192}{\bm{c_0(1-\alpha)}}
[\ce{CH3COO^-}] = [\ce{H3O^+}] = \color[RGB]{0,112,192}{\bm{c_0\alpha}}

Substitute these concentrations into the equilibrium constant equation to get:

K_a = \frac{10^{-3}\alpha^2}{(1-\alpha)} = 1.8\cdot 10^{-5}

You can do one of two things: solve the quadratic equation or assume that, given the small value of the constant, the denominator is very close to one. Here’s how to do each.

Solving the quadratic equation:

10^{-3}\alpha^2 + 1.8\cdot 10^{-5}\alpha - 1.8\cdot 10^{-5} = 0

You get the value \color[RGB]{0,112,192}{\bm{\alpha = 0.125}}. The other value obtained is negative and lacks chemical significance.

Approximating the denominator as one:

\alpha = \sqrt{\frac{1.8\cdot 10^{-5}}{10^{-3}}} = \color[RGB]{0,112,192}{\bf 0.134}

Since the acid concentration is low, it’s not a good idea to use the approximation, and it is preferable to solve the quadratic equation to avoid an error of over 7\ \%.

Taking the first calculated dissociation degree value, you can calculate the equilibrium hydronium concentration:

[\ce{H3O^+}]_{\ce{eq}} = 10^{-3}\ M\cdot 0.125 = \color[RGB]{0,112,192}{\bm{1.25\cdot 10^{-4}\ M}}

The pH calculation is straightforward:

pH = - log [\ce{H3O^+}] = -log\ 1.25\cdot 10^{-4} = \fbox{\color[RGB]{192,0,0}{\bf 3.9}}


You can calculate the pOH considering their relationship:

{\color[RGB]{2,112,20}{\bf pH + pOH = 14}}\ \to\ pOH = 14 - pH = (14 - 3.9) = \fbox{\color[RGB]{192,0,0}{\bf 10.1}}