Question 14.10: CALCULATING THE pH AND THE EQUILIBRIUM CONCENTRATIONS IN A S...

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)  \rightleftharpoons  H_{3}O^{+}(aq)  +  F^{-}(aq)              K_{a}  =  3.5  ×  10^{-4}

H_{2}O(l)  +  H_{2}O(l)  \rightleftharpoons  H_{3}O^{+}(aq)  +  OH^{-}(aq)            K_{w}  =  1.0  ×  10^{-14}

Step 3. Since K_{a}  >>  K_{w}, the principal reaction is the dissociation of HF.

Step 4.

Step 5. Substituting the equilibrium concentrations into the equilibrium equation for the principal reaction gives

K_{a}  =  3.5  ×  10^{-4}  =  \frac{[H_{3}O^{+}][F^{-}]}{[HF]}  =  \frac{(x)(x)}{(0.050  –  x)}

Making the usual approximation that 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:

x² ≈ (3.5 × 10^{-4})(0.050)

x ≈ 4.2 × 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 × 10^{-4}  =  \frac{x^{2}}{(0.050  –  x)}

x² + (3.5 × 10^{-4})x – (1.75 × 10^{-5}) = 0

We use the standard quadratic formula (Appendix A.4):

x = \frac{-b  ±  \sqrt{b^{2}  –  4ac}}{2a}

= \frac{-(3.5  ×  10^{-4})  ±  \sqrt{(3.5  ×  10^{-4})^{2}  –  4(1)(-1.75  ×  10^{-5})}}{2(1)}

= \frac{-(3.5  ×  10^{-4})  ±  (8.37  ×  10^{-3})}{2}

= +4.0 × 10^{-3}           or        -4.4 × 10^{-4}

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 × 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  ×  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  ×  10^{-14}}{4.0  ×  10^{-3}}  =  2.5  ×  10^{-12} M

Step 8. pH = -log [H_{3}O^{+}]  =  -log (4.0  ×  10^{-3}) = 2.40

BALLPARK CHECK
Arithmetic errors in solving quadratic equations are common, so it’s a good idea to check that the value of x obtained from the quadratic equation is reasonable. If the approximate value of x (4.2 × 10^{-3}) is fairly small compared to the initial concentration of the acid (0.050 M), as is the case in this problem, then the value of x obtained from the quadratic equation (4.2 × 10^{-3}) should be fairly close to the approximate value of x.
The approximate and more exact values of x agree.