Question 16.13: Calculate the pH and percentage of HF molecules ionized in a...
Calculate the pH and percentage of HF molecules ionized in a 0.10 M HF solution.
Analyze We are asked to calculate the percent ionization of a solution of HF. From Appendix D, we find K_{a} = 6.8 ×10^{-4}.
Plan We approach this problem as for previous equilibrium problems: We write the chemical equation for the equilibrium and tabulate the known and unknown concentrations of all species. We then substitute the equilibrium concentrations into the equilibrium expression and solve for the unknown concentration of H^{+}.
Solve
The equilibrium reaction and equilibrium concentrations are as follows:
The equilibrium expression is:
K_{a} =\frac{[H^{+}][F^{-}]}{[HF]}=\frac{(x)(x)}{0.10 – x}= 6.8 × 10^{-4}When we try solving this equation using the approximation 0.10 – x\simeq 0.10 (that is, by neglecting the concentration of acid that ionizes), we obtain:
x=8.2 \times 10^{-3} M
Because this approximation is greater than 5% of 0.10 M, however, we should work the problem in standard quadratic form. Rearranging, we have:
x² = (0.10 – x)(6.8 × 10^{-4})
= 6.8 × 10^{-5} – (6.8 × 10^{-4})x
x² + (6.8 × 10^{-4}) x – 6.8 × 10^{-5} = 0
Substituting these values in the standard quadratic formula gives:
x=\frac{- 6.8 × 10^{-4} \pm \sqrt{(6.8 × 10^{-4})^{2} – 4(-6.8 × 10^{-5})}}{2}
=\frac{- 6.8 × 10^{-4} \pm 1.6 × 10^{-2}}{2}
Of the two solutions, only the positive value for x is chemically reasonable. From that value, we can determine [H^{+}] and hence the pH:
x = [H^{+}] = [F^{-}] = 7.9 × 10^{-3} M, so pH = – log [H^{+}] = 2.10
From our result, we can calculate the percent of molecules ionized:
Percent ionization of HF =\frac{concentration ionized}{original concentration}× 100%
=\frac{7.9 × 10^{-3} M}{0.10 M }× 100% = 7.9%