10.6: The pH of Weak Acid Solutions
- Page ID
- 79596
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)For a solution of a strong acid, calculating the [H3O+] 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 HNO3; for H2SO4, [H3O+] = 2 × [H2SO4], etc.). For a weak acid, however, the hydronium ion concentration will be much, much less than the molar concentration of the acid and [H3O+] must be calculated using the value of Ka. 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 [CH3COOH] is 0.50 M and both [CH3COO–] and [H3O+] are zero. A small amount of CH3COOH will ionize; let’s call this x, making the change for [CH3COOH] “-x”, increasing both [CH3COO–] and [H3O+] by the amount “+x”. Finally, the equilibrium concentration of [CH3COOH] will be (0.50 M – x) and both [CH3COO–] and [H3O+] will be x. The completed table is shown below.
| [CH3COOH] | [CH3COO–] | [H3O]+ | |
|---|---|---|---|
| === Initial === | 0.50 M | 0 | 0 |
|
Change
|
- x | + x | + x |
|
Equilibrium
|
0.50 M - x | x | x |
The expression for Ka 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 C0 is the initial molar concentration of the weak acid.
- Nitrous acid (HNO2) is a weak acid with a Ka 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 Ka = 1.8 × 10-5. For a solution of acetic acid in water, the [H3O+] 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 Ka 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 (Ka = 1.8 × 10-5) must you have in pure water in order for the final pH to be 2.38?