10.6: The pH of Weak Acid Solutions
COOH ] is 0.50 M and both [CH 3 COO – ] and [H 3 O + ] are zero. A small amount of CH 3 COOH will ionize; let’s call this x , making the change for [CH 3 COOH] “- x ”, increasing both [CH 3 COO – ] and [H 3 O + ] by the amount “+ x ”. Finally, the equilibrium concentration of [CH 3 COOH] will be (0.50 M – x) and both [CH 3 COO – ] and [H 3 O + ] will be x . The completed table is shown below.
For a solution of a strong acid, calculating the [H 3 O + ] concentration is simple; because the acid is 100% dissociated, the concentration of hydronium ions is equal to the molar concentration of the strong acid (this is, of course, only true for a monoprotic acid such as HCl or HNO 3 ; for H 2 SO 4 , [H 3 O + ] = 2 × [H 2 SO 4 ], etc.). For a weak acid, however, the hydronium ion concentration will be much, much less than the molar concentration of the acid and [H 3 O + ] must be calculated using the value of K a . We can approach this using an ICE table, like we did for previous equilibrium problems. If we prepared a solution of acetic acid that was exactly 0.50 M, then initially [CH 3| [CH 3 COOH] | [CH 3 COO – ] | [H 3 O] + | |
|---|---|---|---|
| === Initial === | 0.50 M | 0 | 0 |
|
Change
|
- x | + x | + x |
|
Equilibrium
|
0.50 M - x | x | x |
The expression for K a for acetic acid is given in equation in section 10.5 . Substituting for our equilibrium values:
\[K_{a}=1.8\times 10^{-5}=\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}=\frac{x^{2}}{0.50-x} \nonumber \]
\[x^{2}+9.0\times 10^{-6}x-1.8\times 10^{-5}=0 \nonumber \]
The above equation is a quadratic equation and we could solve it using the standard quadratic formula. This is not necessary, however, because acetic acid is a weak acid and by definition, very little of the dissociated form will exist in solution, making the quantity x very, very small. If x is much, much less than 0.50 M (our initial concentration of acetic acid), then (0.50 M – x) 0.50 M and the equation simplifies to:
\[K_{a}=1.8\times 10^{-5}=\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}=\frac{x^{2}}{0.50} \nonumber \]
\[x=[H_{3}O^{+}]=\sqrt{(1.8\times 10^{-5})\times 0.50}=3.0\times 10^{-3}M \nonumber \]
We can test our assumption by substituting for x ; (0.50 – 0.0030 ) = 0.497, which rounds to 0.50 to two significant figures. Because the concentration of hydronium ion is very small for a weak acid, for most typical solutions, the concentration of hydronium ion can be estimated simply as:
\[[H_{3}O^{+}]=\sqrt{(K_{a}\times C_{0}} \nonumber \]
where C 0 is the initial molar concentration of the weak acid.
- Nitrous acid (HNO 2 ) is a weak acid with a K a of 4.3 × 10 -4 . Estimate the hydronium ion concentration and the p H for a 0.50 M solution of nitrous acid in distilled water.
- Acetic acid is a weak acid with K a = 1.8 × 10 -5 . For a solution of acetic acid in water, the [H 3 O + ] is found to be 4.2 × 10 -3 M. What is the concentration of unionized acetic acid in this solution?
\[\ce{CH3COOH(aq) + H2O(l) <=> CH3COO^{–}(aq) + H3O^{+}(aq)} \nonumber\]
- A solution is prepared in which acetic acid is 0.700 M and its conjugate base, acetate anion is 0.600 M. As shown above, the K a of acetic acid is 1.8 x 10 -5 ; what will the p H of this solution be?
- What concentration of the weak acid, acetic acid ( K a = 1.8 × 10 -5 ) must you have in pure water in order for the final pH to be 2.38?