Question

Find the pH and percent ionization of each HFHF solution. (KaKa for HFHF is 6.8×10−46.8×10−4.)

A. Find the percent dissociation of a 0.240 M HF solution. (Express your answer using two significant figures.)

B. Find the pHpH of a 0.100 M HF solution. (Express your answer to two decimal places.)

C. Find the pHpH of a 6.00×10⁻² M HF solution. (Express your answer to two decimal places.)

D. Find the percent dissociation of a 0.100 M HF solution. (Express your answer using two significant figures.)

E. Find the percent dissociation of a 6.00×10⁻² M HF solution. (Express your answer using two significant figures.)

Answer 1

(a) percent dissociation = 5.2 %

(b) pH = 2.10

(c) pH = 2.22

(d) percent dissociation = 7.9 %

(e) percent dissociation = 10. %

Explanation

(a) Given : concentration of HF = 0.240 M

K_a = 6.8 x 10^-4

ICE table	HF (aq)		H+ (aq)	F- (aq)
Initial conc.	0.240 M		0	0
Change	-x		+x	+x
Equilibrium conc.	0.240 M - x		+x	+x

K_a = [H⁺]_eq[F^-]_eq / [HF]_eq

6.8 x 10^-4 = [(x) * (x)] / (0.240 M - x)

solving for x, x = 0.0124 M

[H⁺] = x = 0.0124 M

percent ionization = ([H⁺] / C) * 100

where C = initial concentration of HF = 0.240 M

percent ionization = (0.0124 M / 0.240 M) * 100

percent ionization = 5.2 %

pH = -log[H⁺]

pH = -log(0.0124 M)

pH = 1.91