Question

Calculate the pH of a 0.065 M formic acid (HCOOH) solution.
Formic acid is a weak acid with K_a = 1.8 × 10^–4 at 25°C.

Write your answer to two decimal places. Calculate the pH of a 0.065 M formic acid (HCOOH) solution. Formic acid is a weak acid with 1.8x 10at 25°C. pH 2.48

2.48 is WRONG answer!!!!!!

Answer 1

Dissociation of formic acid in solution is as shown

$HCOOH+H_2O\rightleftharpoons HCOO^-+H_3O^+$

Initial concentration (M) 0.65 M 0 0

Change -y + y + y

Equilibrium concentration (M) 0.065 M- y y y

Acid dissociation constant=Ka=1.8x10^-4

$\frac{ [HCOO^-][H_3O^+]}{[HCOOH]}$=1.8x10^-4

$\frac{ y\times y}{0.065 -y}$=1.8x10^-4

As the Ka value is very small, we can assume that y<<<<0.065 M and can ignore it in denominator

$\frac{ y\times y}{0.065}$=1.8x10^-4

y^{​​​​​2}=0.065x1.8x10^-4=0.117x10^-4

y=0.342x10^-2=$[H_3O^&plus;]=[HCOO^-]$

pH=$-log[H_3O^&plus;]$=$-log(0.342\times 10^{-2})$=1.53

So pH of the given formic acid is 1.53