Answer to Question #322817 in Chemistry for Anjali

Question

Calculate the pH of 0.01 M aqueous solution of sodium formate at 298 K. [Given:

Ka

(HCOOH) =1.7 × 10−4 at 298K]

Answer 1

Solution:

HCOONa(aq) + H₂O(l) ⇌ NaOH(aq) + HCOOH(aq)

HCOO⁻(aq) + H₂O(l) ⇌ OH⁻(aq) + HCOOH(aq)

ICE Table:

According to the ICE Table:

[OH⁻] = [HCOOH] = 0.01h

[HCOO⁻] = 0.01 × (1 − h)

The equlibrium-constant (K) expression for the reaction is:

Therefore,

K_a × [HCOOH] × [OH⁻] = K_w × [HCOO⁻]

K_a × (0.01h) × (0.01h) = K_w × [0.01 × (1 − h)]

Assume that h << 1, so 0.01 × (1 − h) ≈ 0.01

Thus,

K_a × (0.01²h²) = K_w × (0.01 )

(1.7×10⁻⁴) × (0.01²h²) = (1.0×10⁻¹⁴) × (0.01 )

h² = 5.88×10⁻⁹

h = (5.88×10⁻⁹)^1/2 = 7.67×10⁻⁵

h = 7.67×10⁻⁵

[OH⁻] = 0.01h = (0.01) × (7.67×10⁻⁵) = 7.67×10⁻⁷

pOH = −log[OH⁻] = −log(7.67×10⁻⁷) = 6.12

pH = 14 − pOH = 14 − 6.12 = 7.88

pH = 7.88

Answer: pH = 7.88