Answer to Question #107419 in General Chemistry for kim

Question

Find the pH and percent ionization of each HF solution. (Ka for HF is 6.8×10^−4.)

1) Find the pH of a 0.250 M HF solution.
2) Find the percent dissociation of a 0.250 M HF solution.
3) Find the pH of a 0.130 M HF solution.
4) Find the percent dissociation of a 0.130 M HF solution.
5) Find the pH of a 4.00×10−2 M HF solution.
6) Find the percent dissociation of a 4.00×10−2 M HF solution.

Answer 1

Find the pH and percent ionization of each HF solution. (Ka for HF is 6.8×10^−4.)

1) Find the pH of a 0.250 M HF solution

$[H^+] = \sqrt (C * Ka)$

$[H^+] = \sqrt (0.250 * 6.8*10^{-4}) = 1.3*10^{-2}$

$pH = -log(H^+)$

$pH = -log(1.3*10^{-2}) = 1.9$

2) Find the percent dissociation of a 0.250 M HF solution.

$\Alpha = ([H^+]/[HF])*100\%$

$\Alpha = (1.3*10^{-2}/0.250)*100\% = 5.2\%$

3) Find the pH of a 0.130 M HF solution.

$[H^+] = \sqrt (0.130 * 6.8*10^{-4}) = 9.4*10^{-3}$

$pH = -log(9.4*10^{-3}) = 2.03$

4) Find the percent dissociation of a 0.130 M HF solution.

$\Alpha = (9.4*10^{-3}/0.130)*100\% = 7.2\%$

5) Find the pH of a 4.00×10−2 M HF solution.

$[H^+] = \sqrt (4.00*10^{-2} * 6.8*10^{-4}) = 5.2*10^{-3}$

$pH = -log(5.2*10^{-3}) = 2.28$

6) Find the percent dissociation of a 4.00×10−2 M HF solution.

$\Alpha = (5.2*10^{-3}/4.0*10^{-2})*100\% = 13\%$