10.6: The pH of Weak Acid Solutions
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 79596
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}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.




=== Initial === 



Change




Equilibrium




The expression for K_{a} for acetic acid is given in equation in section 10.5. Substituting for our equilibrium values:
\[K_{a}=1.8\times 10^{5}=\frac{[H_{3}O^{+}][CH_{3}COO^{}]}{[CH_{3}COOH]}=\frac{x^{2}}{0.50x}\]
\[x^{2}+9.0\times 10^{6}x1.8\times 10^{5}=0\]
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:
\[K_{a}=1.8\times 10^{5}=\frac{[H_{3}O^{+}][CH_{3}COO^{}]}{[CH_{3}COOH]}=\frac{x^{2}}{0.50}\]
\[x=[H_{3}O^{+}]=\sqrt{(1.8\times 10^{5})\times 0.50}=3.0\times 10^{3}M\]
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:
\[[H_{3}O^{+}]=\sqrt{(K_{a}\times C_{0}}\]
where C_{0} is the initial molar concentration of the weak acid.
Exercise \(\PageIndex{1}\)
 Nitrous acid (HNO_{2}) is a weak acid with a K_{a} of 4.3 × 10^{4}. Estimate the hydronium ion concentration and the pH 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?
CH_{3}COOH(aq) + H_{2}O(l) ⇄ CH_{3}COO^{–}(aq) + H_{3}O^{+}(aq)
 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 pH 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?
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