Question 7.14: Salts As Weak Acids II Calculate the pH of a 0.010 M AlCl3 s...
Salts As Weak Acids II
Calculate the pH of a 0.010 M AlCl _{3} solution. The K_{a} value for Al(H_{2}O)_{6} ^{3+} is 1.4 × 10^{-5} .
The major species in solution are
Al(H_{2}O)_{6} ^{3+} , Cl ^{-} , and H_{2}O
Since the Al(H_{2}O)_{6} ^{3+} ion is a stronger acid than water, the dominant equilibrium is
Al(H_{2}O)_{6} ^{3+} (aq) \xrightleftharpoons[]{} Al(OH)(H_{2}O) _{5} ^{2+} (aq) + H ^{+} (aq)
and
1.4 × 10^{-5} = K_{a} = \frac{[Al(OH)(H_{2}O) _{5} ^{2+}] [ H ^{+}] }{[Al(H_{2}O)_{6} ^{3+} ]}
This is a typical weak acid problem, which we can solve with the usual procedures.
| Initial Concentration (mol/L) |
Equilibrium Concentration (mol/L) |
|
| [Al(H_{2}O)_{6} ^{3+}]_{0} = 0.010 |
\xrightarrow[Al(H_{2}O)_{6} ^{3+} dissociates to reach equilibrium]{x mol/L } |
[Al(H_{2}O)_{6} ^{3+}] = 0.010 -x |
| [Al(OH)(H_{2}O) _{5} ^{2+}] _{0} = 0 | [Al(OH)(H_{2}O) _{5} ^{2+} ] = x | |
| [H^{+}]_{0} ≈ 0 | [H^{+}] = x |
Thus
1.4 × 10^{-5} = K_{a} = \frac{[Al(OH)(H_{2}O) _{5} ^{2+}] [ H ^{+}] }{[Al(H_{2}O)_{6} ^{3+} ]} =\frac{(x) (x)}{0.010 -x} ≈ \frac{x²}{0.010}
x ≈ 3.7 × 10^{-4}
Since the approximation is valid by the 5% rule,
[ H ^{-}] =x= 3.7 × 10^{-4} M and pH=3.43