Question

If 200 ml of 0.50 M formic acid is added to 300 ml of water, what is the resulting pH of the solution? pKa= 3.75

Answer 1

For mono basic weak acids like formic acid,

pH = - log₁₀ (C*K_a)^1/2

Where C = concentration of weak acid in solution = 0.2M

Ka = dissociation constant

From the given data first we have to calculate concentration of solution

applying dilution law

C1 V1 = C2 V2

C1 = concentration of original solution = 0.5M

V1 = volume of original solution = 200ml

C2 = concentration of diluted solution = ?

V2 = volume of diluted solution = 200 + 300 = 500ml

C2 = C1V1 / V2 = 0.5 * 200 / 500 = 0.2M

pKa of formic acid = 3.75

pKa = - log Ka

Ka = 10^-3.75 = 1.778 x 10^-4

Now pH = - log₁₀ (C*K_a)^1/2

pH = - (1/2)log 0.2 * 1.778x10^-4 = 2.22448