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

Step-by-step 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* 0.420L)=0.0420mol`

`\text{Moles of potassium formate}=(0.100mol/L* 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* 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]=(0.049)/(0.427)`

`[HCOOH]=(0.035)/(0.427)`

`pK_a=3.75`

Putting values in equation 2, we get:

`pH=3.75-\log ((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