16.7: Weak Bases
- Page ID
- 91286
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The pH of a solution of a weak base can be calculated in a way which is very similar to that used for a weak acid. Instead of an acid constant Ka, a base constant Kb must be used. If a weak base B accepts protons from water according to the equation
\[\text{B} + \text{ H}_{\text{2}}\text{O}\rightleftharpoons\text{BH}^{+} + \text{OH}^{-} \label{1} \]
then the base constant is defined by the expression
\[K_{b}=\dfrac{ \text{ BH}^{\text{+}} \text{ OH}^{-} }{ \text{ B } } \label{2} \]
A list of Kb values for selected bases arranged in order of strength is given in the table below. This table is part of our larger collection of acid-base resources.
Base | Formula and Ionization Equation | Kb | Molecular Shape |
---|---|---|---|
Ammonia | \(NH_3 + H_2O \rightleftharpoons NH^+_4 + OH^–\) | 1.77 × 10–5 | |
Aniline | \(C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH^+_3 + OH^–\) | 3.9 × 10–10 | |
Carbonate ion | \(CO_3^{2–} + H_2O \rightleftharpoons HCO^-_3 + OH^–\) | 2.1 × 10–4 | |
Hydrazine | \(N_2H_4 + H_2O \rightleftharpoons N_2H^+_5 + OH^–\) \(N_2H^+_5 + H_2O \rightleftharpoons N_2H_6^{2+} + OH^–\) |
K1 = 1.2 × 10–6 K2 = 1.3 × 10–15 |
|
Hydride ion | \(H^– + H_2O \rightarrow H_2 + OH^–\) | 1.0 | |
Phosphate ion | \(PO_4^{3–} + H_2O \rightleftharpoons HPO^{2-}_4 + OH^–\) | 5.9 × 10–3 | |
Pyridine | \(C_5H_5N + H_2O \rightleftharpoons C_5H_5NH^+ + OH^–\) | 1.6 × 10–9 |
To find the pH we follow the same general procedure as in the case of a weak acid. If the stoichiometric concentration of the base is indicated by cb, the result is entirely analogous to equation 4 in the section on the pH of weak acids; namely,
\[K_{b}=\dfrac{ [\text{OH}^{-}]^2}{c_{b}- [\text{ OH}^{-}] } \label{3} \]
Under most circumstances we can make the approximation
\[c_b – [OH^–] \approx c_b \nonumber \]
in which case Equation \ref{3} reduces to the approximation
\[[OH^–] ≈ \sqrt{K_{b}c_{b}} \label{4} \]
which is identical to the expression obtained in the acid case (approximation shown in equation 6 in the section on the pH of weak acids) except that OH– replaces H3O+ and b replaces a. Once we have found the hydroxide-ion concentration from this approximation, we can then easily find the pOH, and from it the pH.
Using the value for Kb listed in the table, find the pH of 0.100 M NH3.
Solution
It is not a bad idea to guess an approximate pH before embarking on the calculation. Since we have a dilute solution of a weak base, we expect the solution to be only mildly basic. A pH of 13 or 14 would be too basic, while a pH of 8 or 9 is too close to neutral. A pH of 10 or 11 seems reasonable. Using Equation \ref{4} we have
\[\begin{align*} [\text{ OH}^{-}] &=\sqrt{K_{b}c_{b}} \\[4pt] & =\sqrt{\text{1.8 }\times \text{ 10}^{-\text{5}}\text{ mol L}^{-\text{1}} \times \text{ 0.100 mol L}^{-\text{1}}} \\[4pt] &=\sqrt{\text{1.8 }\times \text{ 10}^{-\text{6}}\text{ mol}^{\text{2}}\text{ L}^{-2}} \\[4pt] &=\text{1.34 }\times \text{ 10}^{-\text{3}}\text{ mol L}^{-\text{1}} \end{align*} \nonumber \]
Checking the accuracy of the approximation, we find
\(\dfrac{ [\text{ OH}^{-} ]}{c_{\text{b}}}=\dfrac{\text{1.34 }\times \text{ 10}^{-\text{3}}}{\text{0.1}}\approx \text{1 percent}\)
The approximation is valid, and we thus proceed to find the pOH.
\(\text{pOH}=-\text{log}\dfrac{ [\text{ OH}^{-} ]}{\text{mol L}^{-\text{1}}}=-\text{log(1.34 }\times \text{ 10}^{-\text{3}}\text{)}=\text{2.87}\)
From which
\[pH = 14.00 – pOH = 14.00 – 2.87 = 11.13 \nonumber \]
This calculated value checks well with our initial guess.
Occasionally we will find that the approximation
\[c_b – [OH^{–}] ≈ c_b \nonumber \]
is not valid, in which case we must use a series of successive approximations similar to that outlined above for acids. The appropriate formula can be derived from Equation \ref{3} and reads
\[[OH^{-}] \approx \sqrt{K_{b} ( c_b - [OH^{-}] )} \nonumber \]