Examples

Calculate the pH of a 0.35 solution of KNO2 (Ka = 4.0 x 10-4).

KNO2 is a salt of a strong base KOH and a weak acid HNO2 and therefore, the solution will be basic.

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

KNO2 → K+ + NO2

NO2(aq) + H2O(l) → HNO2(aq) + OH(aq)  (basic)

Next, determine the Kb from the Ka:

Ka · Kb = Kw = 10-14

Rearranging this, we get an expression for Ka:

${K_{{\rm{b}}}}\; = \,\frac{{{K_{\rm{w}}}}}{{{K_{\rm{a}}}}}\; = \;\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ – 14}}}}}}{{{\rm{4}}{\rm{.0}}\; \times \;{\rm{1}}{{\rm{0}}^{{\rm{ – 4}}}}}}\; = \;{\rm{2}}{\rm{.5 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 11}}}}$

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

 [NO2–] [HNO2] [OH–] Initial 0.35 0 0 Change -x +x +x Equil 0.35 – x x x

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

${K_{{\rm{b}}}}\; = \,\frac{{{\rm{[HN}}{{\rm{O}}_{\rm{2}}}{\rm{][O}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[N}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.35}}\;{\rm{ – }}\;{\rm{x}}}}\; \approx \;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.35}}}}{\rm{ = }}\;{\rm{2}}{\rm{.5}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{ – 11}}}}$

Solving for x, we find that x = 3.0 x 10-6 mol/L, and we check if the approximation was valid:

$\% \, = \;\frac{{3.0\, \times \;{{10}^{ – 6}}}}{{0.35}}\; \times \;100\% \; = \;0.00084\%$

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 3.0 x 10-6 = 5.5, and therefore, the pH is:

pH = 14 – pOH = 14 – 5.5 = 8.5

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

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