Answer to Question #268745 in Chemistry for eli

Question

What is the pH of a sodium formate solution prepared by adding 0.680 grams of sodium formate to 100.0 ml of water at 25.0°C? The Ka at 25.0°C for formic acid is 1.8 × 10-4.

Answer 1

Solution:

sodium formate = HCOONa

Calculate the moles of sodium formate:

The molar mass of HCOONa is 68 g/mol

Hence,

(0.680 g HCOONa) × (1 mol HCOONa / 68 g HCOONa) = 0.01 mol HCOONa

Calculate the molar concentration (C_M) of a sodium formate solution:

C_M(HCOONa) = moles of sodium formate / volume of solution

C_M(HCOONa) = (0.01 mol / 0.1 L) = 0.1 mol/L = 0.1 M

C_M(HCOO^–) = C_M(HCOONa) = 0.1 M

Сalculate the dissociation constant:

K = K_w / K_a

K = (1.0×10^-14) / (1.8×10^-4) = 5.56×10^-11

The equilibrium reaction for dissociation of is,

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

HCOO^–(aq) + H₂O(l) ⇌ HCOOH(aq) + OH^–(aq)

ICE Table:

[HCOO^–] = 0.1 - x

[HCOOH] = [OH^–] = x

K = [OH^–] × [HCOOH] / [HCOO^–]

K = (x]) × (x) / (0.1 - x) = (x²) / (0.1 - x) = 5.56×10^-11

By solving the terms, we get:

x = 2.358×10^-6

[OH^–] = x = 2.358×10^-6 M

From [OH^–], we can calculate pOH:

pOH = -log[OH^–] = -log(2.358×10^-6) = 5.627

 

From pOH, we can calculate pH:

pH + pOH = 14;

pH = 14 - pOH = 14 - 5.627 = 8.373 = 8.37

pH = 8.37

 

Answer: pH = 8.37