Question

A 420 mL sample of a 0.100 M formate buffer, pH 3.75, is treated with 7 mL of 1.00 M KOH. What is the pH following this addition? (pKa for formic acid is 3.75)

Answer 1

Answer: The pH of the resulting solution will be 3.60

Explanation:

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}$ ......(1)

We are given:

Molarity of formic acid = 0.100 M

Molarity of potassium formate = 0.100 M

Volume of solution = 420 mL = 0.420 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$\text{Moles of formic acid}=(0.100mol/L\times 0.420L)=0.0420mol$

$\text{Moles of potassium formate}=(0.100mol/L\times 0.420L)=0.042mol$

Molarity of KOH = 1.00 M

Volume of solution = 7 mL = 0.007 L

Putting values in equation 1, we get:

$\text{Moles of KOH}=(1mol/L\times 0.007L)=0.007mol$

The chemical equation for the reaction of formic acid and KOH follows:

                 $HCOOH+KOH\rightleftharpoons HCOOK+H_2O$

I:                   0.042     0.007       0.042

C:                -0.007    -0.007     +0.007

E:                  0.035         -           0.049

Volume of solution = [420 + 7] = 427 mL = 0.427 L

To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:

$pH=pK_a+ \log \frac{\text{[conjugate base]}}{\text{[acid]}}$ .......(2)

Given values:

$[HCOOK]=\frac{0.049}{0.427}$

$[HCOOH]=\frac{0.035}{0.427}$

$pK_a=3.75$

Putting values in equation 2, we get:

$pH=3.75-\log \frac{(0.049/0.427)}{(0.035/0.427)}\\\\pH=3.75-0.146\\\\pH=3.60$

Hence, the pH of the resulting solution will be 3.60