Question 7.11: The pH of Sulfuric Acid II Calculate the pH of a 1.00 × 10^-...
The pH of Sulfuric Acid II
Calculate the pH of a 1.00 × 10 ^{-2} M H_{2} SO_{4} solution.
The major species in solution are
H^{+}, HSO_{4}^{-}, and H_{2}O
Proceeding as in Example 7.10, we consider the dissociation of HSO_{4}^{-}, which leads to the following concentrations:
| Initial
Concentration (mol/L) |
Equilibrium
Concentration (mol/L) |
|
| [ HSO _{4}^{-}]_{0} =0.0100 |
\xrightarrow[to reach equilibrium]{x mol / L HSO _{4} ^{-} dissociates} |
[ HSO _{4} ^{-}] =0.0100 -x |
| [SO _{4}^{2-}]_{0} =0 | [SO _{4}^{2-}]= x | |
| [H ^{+}]_{0} = 0.0100
\nearrow |
[H ^{+}]=0.0100 + x |
Substituting the equilibrium concentrations into the expression for K_{a_2} gives
1.2 × 10 ^{-2} = K_{a_2}= \frac{[H ^{+}] [SO _{4}^{2-} ]}{[HSO _{4} ^{-}]} = \frac{(0.0100 + x) (x)}{(0.0100 -x)}If we make the usual approximation, then 0.010 + x ≈0.010 and 0.010 – x ≈ 0.010, and we have
1.2 × 10 ^{-2} = \frac{(0.0100 + x) (x)}{(0.0100 -x)}≈\frac{(0.0100)x}{(0.0100)}The calculated value of x is
x= 1.2 × 10 ^{-2} = 0.012
This value is larger than 0.010, clearly a ridiculous result. Thus we cannot make the usual approximation and must instead solve the quadratic equation. The expression
1.2 × 10 ^{-2} = \frac{(0.0100 + x) (x)}{(0.0100 -x)}leads to
(1.2 × 10 ^{-2}) (0.0100 -x) = (0.0100 + x) (x)(1.2 × 10 ^{-4}) – (1.2 × 10 ^{-2}) x= (1.0 × 10 ^{-2})x + x²
x² +( 2.2 × 10 ^{-2}) x – (1.2 × 10 ^{-4}) = 0
This equation can be solved by using the quadratic formula,
x = \frac{- b \pm \sqrt{b² -4ac}}{2a}where a = 1, b = 2.2 × 10 ^{-2} and c = -1.2 × 10 ^{-4} . Use of the quadratic formula gives one negative root (which cannot be correct) and one positive root,
x= 4.5 × 10 ^{-3}
Thus [H^{+}] =0.0100 + x = 0.0100 + 0.0045= 0.0145
and pH = 1.84
Note that in this case the second dissociation step produces about half as many H^{+} ions as the initial step does.
This problem can also be solved by successive approximations, a method illustrated in Appendix A1.4.