Question 14.12: CALCULATING THE pH AND THE EQUILIBRIUM CONCENTRATIONS IN A S...
CALCULATING THE pH AND THE EQUILIBRIUM CONCENTRATIONS IN A SULFURIC ACID SOLUTION
Calculate the pH and the concentrations of all species present (HSO_{4}^{-}, SO_{4}^{2-}, H_{3}O^{+}, and OH^{-}) in 0.10 M H_{2}SO_{4}.
STRATEGY
Use the eight-step procedure summarized in Figure 14.6.
Steps 1–3. Because H_{2}SO_{4} is essentially 100% dissociated to give H_{3}O^{+} and HSO_{4}^{-} (K_{a1} is very large), the species present initially are H_{3}O^{+} (acid), HSO_{4}^{-} (acid), and H_{2}O (acid or base). Although H_{3}O^{+} is the strongest acid present, proton transfer from H_{3}O^{+} to H_{2}O to give H_{2}O and H_{3}O^{+} is not considered to be the principal reaction because it does not change any of the concentrations. The next strongest acid present is HSO_{4}^{-} (K_{a2} > K_{w}), so we consider the principal reaction to be dissociation of HSO_{4}^{-}.
Step 4.
Note that the second dissociation step takes place in the presence of 0.10 M H_{3}O^{+} from the first dissociation step.
Step 5. Substituting the equilibrium concentrations into the equilibrium equation for the principal reaction gives
K_{a2} = 1.2 × 10^{-2} = \frac{[H_{3}O^{+}][SO_{4}^{2-}]}{[HSO_{4}^{-}]} = \frac{(0.10 + x)(x)}{(0.10 – x)}Neglecting x compared with 0.10 and solving this equation would give x = K_{a2} = 0.012, which is not negligible compared with 0.10. Therefore, we use the quadratic equation to obtain the value of x:
0.0012 – 0.012x = 0.10x + x²
x² + 0.112x – 0.0012 = 0
x = \frac{-b ± \sqrt{b^{2} – 4ac}}{2a} = \frac{-0.112 ± \sqrt{(0.112)^{2} – 4(1)(-0.0012)}}{2(1)}= \frac{-0.112 ± 0.132}{2}
= +0.010 or -0.122
Because x is the SO_{4}^{2-} concentration, it must be positive. Therefore,
x = 0.010
Step 6. The big concentrations are
[SO_{4}^{2-}] = x = 0.010 M
[HSO_{4}^{-}] = 0.10 – x = 0.10 – 0.010 = 0.09 M
[H_{3}O^{+}] = 0.10 + x = 0.10 + 0.010 = 0.11 M
Step 7. The small concentration, [OH^{-}], is obtained from the subsidiary equilibrium, dissociation of water:
[OH^{-}] = \frac{K_{w}}{[H_{3}O^{+}]} = \frac{1.0 × 10^{-14}}{0.11} = 9.1 × 10^{-14} M
Step 8. pH = – log [H_{3}O^{+}] = – log 0.11 = 0.96
BALLPARK CHECK
The easiest check in this case is to substitute the big concentrations obtained in Step 6 into the equilibrium equation for the second dissociation step and show that the equilibrium constant expression equals K_{a2}:
Since [H_{3}O^{+}] is approximately 10^{-1} M, the pH should be about 1, in agreement with the detailed solution.
The brightness of the planet Venus is due in part to thick, highly reflective clouds in its upper atmosphere. These clouds consist of sulfur dioxide and droplets of sulfuric acid.
| Principal reaction HSO_{4}^{-}(aq) + H_{2}O(l) \rightleftharpoons H_{3}O^{+}(aq) + SO_{4}^{2-}(aq) |
| Initial concentration (M) 0.10 0.10 0
Change (M) -x +x +x Equilibrium concentration (M) 0.10 – x 0.10 + x x |
