KNO2 pH - Chemistry Steps

Acids and Bases

KNO2 pH

Calculate the pH of a 0.35 M solution of KNO₂ (K_a = 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 K_b from the K_a:

K_a · K_b = K_w = 10^-14

Rearranging this, we get an expression for K_a:

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

Set up an ICE table for K_b 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 K_b 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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