Question 7.6: The pH of Weak Bases Calculate the pH of a 1.0 M solution of...
The pH of Weak Bases
Calculate the pH of a 1.0 M solution of methylamine ( K_{b} = 4.38 ×10^{-4} ).
Since methylamine (CH _{3}NH _{2} ) is a weak base, the major species in solution are
CH _{3}NH _{2} and H _{2}O
Both are bases; however, since water can be neglected as a source of OH^{-} , the dominant equilibrium is
CH _{3}NH _{2} (aq) + H _{2}O (l) \xrightleftharpoons[]{} CH _{3}NH _{3} ^{+}(aq) + OH^{-} (aq)
and
K_{b} =4.38 ×10^{-4} =\frac{[CH _{3}NH _{3} ^{+}] [OH^{-}]}{[CH _{3}NH _{2}]}
The concentrations are as follows:
| Initial Concentration (mol/L) |
Equilibrium Concentration (mol/L) |
|
| [CH _{3}NH _{2}]_{0} =1.0 |
\xrightarrow[to reach equilibrium]{x mol / L CH _{3}NH _{2} reacts with H _{2}O } |
[CH _{3}NH _{2}] =1.0 -x |
| [CH _{3}NH _{3} ^{+}]_{0} =0 | [CH _{3}NH _{3} ^{+}]= x | |
| [OH ^{-}]_{0} ≈0 | [OH ^{-}]=x |
Or in shorthand form:
| CH _{3}NH _{2} (aq) | + | H _{2}O (l) | \xrightleftharpoons[]{} | CH _{3}NH _{3} ^{+}(aq) | + | OH ^{-} (aq) | |
| Initial: | 1.0 | — | 0 | 0 | |||
| Change: | -x | — | +x | +x | |||
| Equilibrium: | 1.0-x | — | x | x |
Substituting the equilibrium concentrations into the equilibrium expression and making the usual approximation gives
K_{b} = 4.38 ×10^{-4} =\frac{[CH _{3}NH _{3} ^{+}] [OH ^{-}]}{[CH _{3}NH _{2}]} = \frac{(x) (x)}{1.0-x} ≈ \frac{x²}{1.0}
x ≈ 2.1 × 10 ^{-2}
The approximation is valid by the 5% rule, so
[OH ^{-}] = x = 2.1 × 10 ^{-2} M and pOH= 1.68
Note that since [H ^{+}] [OH ^{-}] = 1.0 × 10 ^{-14} , pH + pOH =14 . Thus pH = 14.00 – 1.68 = 12.32 .