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)

The pH of the resulting solution will be 3.60

Explanation:

Molarity is calculated by using the equation:

......(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:

Molarity of KOH = 1.00 M

Volume of solution = 7 mL = 0.007 L

Putting values in equation 1, we get:

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

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:

.......(2)

Given values:

Putting values in equation 2, we get:

Hence, the pH of the resulting solution will be 3.60

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