Examples

Calculate the pH of a 0.35 M 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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