Answer:
25,3mL of 0,450M HCl are the maximum volume you can add to the buffer.
Explanation:
The maximum buffer capacity that you can have with HCOOH-HCOO⁻ is when the acid and its conjugate base are in the same proportions, that means [HCOOH] = 0,090M and [HCOO⁻] = 0,090M.
In moles: 0,155L×0,090M = 0,01395 moles
Hendersson-Hasselbalch equation for this system is:
pH = 3,74 + log₁₀ [HCOO⁻] /[HCOOH]
The pH of this buffer is equal to pka = 3,74.
A buffer lost is buffering capacity when pH is pka-1 or pka+1. Under addition of HCl the buffer will lost is buffering capacity when pH = 2,74. Using Henderson-Hasselbalch equation:
2,74 = 3,74 + log₁₀ [HCOO⁻] /[HCOOH]
0,1 = [HCOO⁻] /[HCOOH] (1)
The reaction of HCl with HCOO⁻ is:
HCOO⁻ + HCl → HCOOH + Cl⁻
Using this reaction and moles of formic/formiate buffer:
0,1 = [0,01395-x] / [0,01395+x]
Being x the moles of HCl added in the reaction.
0,001395+0,1x = 0,01395 - x
1,1x = 0,012555
x = 0,0114 moles of HCl you need to add.
As the concentration of the HCl solution is 0,450M
0,0114 mol ×[tex]\frac{1L}{0,450mol}[/tex] = 0,0253L ≡ 25,3 mL that are the maximum volume you can add.
I hope it helps!