## Examples

Calculate the pH of a 1.0 solution of sodium acetate (CH3CO2Na) considering that the Ka of acetic acid (CH3CO2H is 1.7 x 10-5.

CH3CO2Na is a salt of a strong base NaOH and a weak acid CH3CO2H therefore, the solution will be basic.

Dissociate the salt and write the reaction of its weak component with water.

CH3CO2Na → Na+ + CH3CO2

CH3CO2(aq) + H2O(l) → CH3CO2H(aq) + OH(aq)  (basic)

Next, determine the Kb from the Ka:

Ka · Kb = Kw = 10-14

Rearranging this, we get an expression for Ka:

${K_{{\rm{b}}}}\; = \,\frac{{{K_{\rm{w}}}}}{{{K_{\rm{a}}}}}\; = \;\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ – 14}}}}}}{{{\rm{1}}{\rm{.7}}\; \times \;{\rm{1}}{{\rm{0}}^{{\rm{ – 5}}}}}}\; = \;{\rm{5}}{\rm{.9 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 10}}}}$

Set up an ICE table for Kb assigning x mol/l for the ionization.

CH3CO2(aq) + H2O(l) → CH3CO2H(aq) + OH(aq)

 [CH3CO2–] [CH3CO2H] [OH–] Initial 1.0 0 0 Change -x +x +x Equil 1.0 – x x x

Write the expression for Kb using the equilibrium concentrations in the ICE table in order to determine [OH].

${K_{{\rm{b}}}}\; = \,\frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H][O}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{1}}{\rm{.0}}\;{\rm{ – }}\;{\rm{x}}}}\; \approx \;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{1}}{\rm{.0}}}}{\rm{ = }}\;{\rm{5}}{\rm{.9}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{ – 10}}}}$

We find that x = 2.4 x 10-5 mol/l.

Check if the approximation was valid:

$\% \, = \;\frac{{2.4\, \times \;{{10}^{ – 5}}}}{{1.0}}\; \times \;100\% \; = \;0.0024\%$

The approximation is valid, so the concentration of OH– ions at equilibrium is 3.0 x 10-6 M.

We can now calculate the pOH, and then pH using the pH + pH = 14 relationship:

pOH = -log 2.4 x 10-5 = 4.6, and therefore, the pH is:

pH = 14 – pOH = 14 – 4.6 = 9.4

And this answer is also reasonable because we predicted to have a basic solution.

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