<u>Answer:</u> The pH of the resulting solution will be 3.60
<u>Explanation:</u>
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:
![[HCOOK]=\frac{0.049}{0.427}](https://tex.z-dn.net/?f=%5BHCOOK%5D%3D%5Cfrac%7B0.049%7D%7B0.427%7D)
![[HCOOH]=\frac{0.035}{0.427}](https://tex.z-dn.net/?f=%5BHCOOH%5D%3D%5Cfrac%7B0.035%7D%7B0.427%7D)

Putting values in equation 2, we get:

Hence, the pH of the resulting solution will be 3.60