Question

Calculate the equilibrium concentration of Ni?that is 0.05 M in potassium hydrogen phthalate, KHP, and 0.15 M in potassium phthalate, KP. Phthalic acid: K 1 = 1.2x10-?; K 21 = 3.90x10

Answer 1

First we will calculate the pH of solution

KHP is a weak acid and K₂P is the conjugate base

K_a2 = 3.90 x 10^-6

pK_a2 = -log(K_a2)

pK_a2 = -log(3.90 x 10^-6)

pK_a2 = 4.51

According to Henderson - Hasselbalch equation,

pH = pK_a2 + log([conjugate base] / [weak acid])

pH = pK_a2 + log([K₂P] / [KHP])

Substituting the values,

pH = 4.51 + log(0.15 M / 0.05 M)

pH = 4.51 + log(3)

pH = 4.51 + 0.477

pH = 5.886

pOH = 14 - pH

pOH = 14 - 5.886

pOH = 8.114

[OH^-] = 10^-pOH

[OH^-] = 10^-8.114

[OH^-] = 7.7 x 10^-9 M

K_sp Ni(OH)₂ = [Ni²⁺]_eq[OH^-]_eq²

2.8 x 10^-16 = [Ni²⁺]_eq * (7.7 x 10^-9 M)²

[Ni²⁺]_eq = (2.8 x 10^-16) / ((7.7 x 10^-9 M)²

[Ni²⁺]_eq = 4.732 M

equilibrium concentration of Ni²⁺ is 4.732 M