Question

7.
The amount of calcium formate which should be added in 500 ml of 0.1 M HCOOH to obtained a solution with
pH = 4 is -
[Given: K of HCOOH = 1.8 x 104]
(A) 4.36g
(B) 5.85 g
(C) 3.76g
(D) 5.12 g
​

Answer 1

5.85 g is the amount of calcium formate which should be added in 500 mL of 0.1 M HCOOH to obtain a solution with pH 4.

Given-

• Volume of solution = 500 mL
• Molarity of formic acid = 0.1 M
• pH of solution = 4
• Ka of HCOOH = 1.8 × 10⁻⁴

By applying Henderson–Hasselbalch equation for acidic buffer

pH = pKa + log [salt]/[acid]

Here salt is calcium formate. Formate ion comes from calcium formate acts as a salt.

By substituting the values we get

4 = (4 - log 1.8) + log [HCOO⁻]/[0.1]

By solving the above equation we get

[HCOO⁻] = 0.18 × 0.5 × 0.5 = 0.045 mol

Weight of salt = 0.045 × 130(molar mass) = 5.85 g