Question 15.10: Calculate the pH and the concentrations of all species prese......
Calculate the pH and the concentrations of all species present (H_{3}O^{+}, F^{-}, HF, and OH^{-}) in 0.050 M HF.
STRATEGY
Follow the eight-step sequence outlined in Figure 15.7.
| IDENTIFY | |
| Known | Unknown |
| Concentration of HF (0.050 M) | pH |
| K_{a}=3.5\times 10^{-4} | Equilibrium concentrations of H_{3}O^{+}, F^{-}, HF, and OH^{-} |
Step 1. The species present initially are
\underset{Acid}{HF} \underset{Acid or base}{H_{2}O}Step 2. The possible proton-transfer reactions are
HF(aq)+H_{2}O(l)\xrightleftharpoons{}H_{3}O^{+}(aq)+F^{-}(aq) K_{a}=3.5\times 10^{-4}\\ H_{2}O(l)+H_{2}O(l)\xrightleftharpoons{}H_{3}O^{+}(aq)+OH^{-}(aq) K_{w}=1.0\times 10^{-14}Step 3. Since K_{a} >> K_{w}, the principal reaction is the dissociation of HF.
Step 4.
Table 1
Step 5. Substituting the equilibrium concentrations into the equilibrium equation for the principal reaction gives
K_{a}=3.5\times 10^{-4}=\frac{[H_{3}O^{+}][F^{-}]}{[HF]}=\frac{(x)(x)}{(0.050-x)}Making the usual approximation that x is negligible compared with the initial concentration of the acid, we assume that (0.050 – x) ≈ 0.050 and then solve for an approximate value of x:
x^{2}\approx(3.5\times 10^{-4})(0.050)x\approx 4.2\times 10^{-3}
Since the initial concentration of HF (0.050 M) is known to the third decimal place, x is negligible compared with the initial [HF] only if x is less than 0.001 M. Our approximate value of x (0.0042 M) is not negligible compared with 0.050 M because 0.050 M – 0.0042 M = 0.046 M. Therefore, our approximation, 0.050 – x ≈ 0.050, is invalid, and we must solve the quadratic equation without making approximations:
3.5\times 10^{-4}=\frac{x^{2}}{(0.050-x)}x^{2}+(3.5\times 10^{-4})x -(1.75\times 10^{-5})=0
We use the standard quadratic formula (Appendix A.4):
x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}=\frac{-(3.5\times 10^{-4})\pm\sqrt{(3.5\times 10^{-4})^{2}-4(1)(1.75\times 10^{-5})}}{2(1)}
=\frac{(-3.5\times 10^{-4})\pm (8.37\times 10^{-3})}{2}
= +4.0\times 10^{-3} or -4.4\times 10^{-3}
Of the two solutions for x, only the positive value has physical meaning, since x is the H_{3}O^{+} concentration. Therefore,
x= 4.0\times 10^{-3}Note that whether we must solve the quadratic equation depends on both the size of x and the number of significant figures in the initial concentration.
Step 6. The big concentrations are
[H_{3}O^{+}]=[F^{-}]= x =4.0\times 10^{-3} M[HF] = (0.050 – x) = (0.050 – 0.0040) = 0.046 M
Step 7. The small concentration, [OH^{-}], is obtained from the subsidiary equilibrium, the dissociation of water:
[OH^{-}]=\frac{K_{w}}{[H_{3}O^{+}]}=\frac{1.0\times 10^{-14}}{4.0 \times 10^{-3}}=2.5\times 10^{-12} MStep 8. pH = -log [H_{3}O^{+}]=-log (4.0\times 10^{-3})=2.40
Table 1
| Principal reaction | HF(aq) | + | H_{2}O(l) | \xrightleftharpoons{} | H_{3}O^{+}(aq) | + | F^{-}(aq) |
| Initial concentration (M) | 0.050 | ∼0 | 0 | ||||
| Change (M) | -x | +x | +x | ||||
| Equilibrium concentration (M) | 0.050-x | x | x |