Question

What is the pH of solution if 40 mL 0.1 M HCl is mixed with 60 mL 0.2 M HAc?

Answer 1

Concentration HAc = 0.2 M

volume HAc = 60 mL

moles HAc = (concentration HAc) * (volume HAc)

moles HAc = (60 mL) * (0.2 M)

moles HAc = 12 mmol

Total volume = (volume HAc) + (volume HCl)

Total volume = (60 mL) + (40 mL)

Total volume = 100 mL

concentration of HAc = (moles HAc) / (Total volume)

concentration of HAc = (12 mmol) / (100 mL)

concentration of HAc = 0.12 M

Similarly, concentration of HCl = 0.04 M

Since HCl is a very strong acid, all HCl dissociates into H⁺ and Cl^-

initial concentration H⁺ = concentration of HCl = 0.04 M

HAc K_a = 1.8 x 10^-5

ICE table	HAc		H+	Ac-
Initial conc.	0.12 M		0.04 M	0
Change	-x		+x	+x
Equilibrium conc.	0.12 M - x		0.04 M + x	+x

K_a = [H⁺]_eq[Ac^-]_eq / [HAc]_eq

1.8 x 10^-5 = [(0.04 M + x) * (x)] / (0.12 M - x)

Solving for x, x = 5.4 x 10^-5 M

[H⁺]_eq = 0.04 M + x

[H⁺]_eq = 0.04 M + 5.4 x 10^-5 M

[H⁺]_eq = 0.040054 M

pH = -log[H⁺]

pH = -log(0.040054 M)

pH = 1.40