To determine the change in pH for each buffer upon the addition of 10 mL of 1.00 M HCl, we calculate the moles of HCl that react with the weak acid in each buffer and use the Henderson-Hasselbalch equation to find the new pH. Buffer A and Buffer B both have a pH of 3.74 initially, and the addition of HCl results in a negligible change in pH for both buffers.
To calculate the change in pH for each buffer upon the addition of 10 mL of 1.00 M HCl, we need to determine the moles of HCl that react with the buffer components. The reaction between HCl and the weak acid in the buffer will result in the formation of its conjugate base, and the pH will change accordingly. The reaction can be represented as follows:
HCl + HCOOH ⇌ HCOO- + H3O+
First, we need to determine the initial moles of the weak acid and its conjugate base in each buffer:
Buffer A:
Initial moles of HCOOH = 1.00 mol
Initial moles of HCOO- = 1.00 mol
Buffer B:
Initial moles of HCOOH = 0.10 mol
Initial moles of HCOO- = 0.10 mol
Next, we calculate the moles of HCl that react with the weak acid in each buffer:
Buffer A:
Buffer A:
Moles of HCOOH reacted = Initial moles of HCOOH - Moles of HCl reacting = 1.00 mol - (10 mL * 0.00100 mol/mL) = 0.99 mol
Buffer B:
Buffer B:
Moles of HCOOH reacted = Initial moles of HCOOH - Moles of HCl reacting = 0.10 mol - (10 mL * 0.00100 mol/mL) = 0.09 mol
Using the Henderson-Hasselbalch equation, we can calculate the new pH for each buffer:
Buffer A:
pH = pKa + log ([HCOO-] / [HCOOH])
pH = 3.74 + log (1.00 mol / 0.99 mol) = 3.74 + log (1.01) ≈ 3.75
Buffer B:
pH = pKa + log ([HCOO-] / [HCOOH])
pH = 3.74 + log (0.10 mol / 0.09 mol) = 3.74 + log (1.11) ≈ 3.75