Answer:
The pH change in 0,206 units
Explanation:
When the acetic acid buffer is at pH 5,000; it is possible to obtain the acetate/acetic acid proportions using Henderson-Hasselbalch formula, thus:
pH = pka + log₁₀ [A⁻]/[HA] Where A⁻ is CH₃COO⁻ and HA is CH₃COOH.
Replacing:
5,000 = 4,740 + log₁₀ [A⁻]/[HA]
1,820 = [A⁻]/[HA] <em>(1)</em>
As buffer concentration is 0,100M:
[A⁻] + [HA] = 0,100 <em>(2)</em>
Replacing (2) in (1)
[HA] = 0,035M
And [A⁻] = 0,065M
As volume is 1,80x10²mL, moles of HA and A⁻ are:
0,180L × 0,035M = <em>6,3x10⁻³mol of HA</em>
0,180L × 0,065M = <em>1,17x10⁻²mol of A⁻</em>
The reaction of HCl with A⁻ is:
HCl + A⁻ → HA + Cl⁻
The add moles of HCl are:
0,0065L×0,330M = 2,145x10⁻³ moles of HCl that are equivalent to moles of A⁻ consumed and moles of HA produced.
Thus, moles of HA after addition of HCl are:
6,3x10⁻³mol + 2,145x10⁻³ mol = <em>8,445x10⁻³ moles of HA</em>
And moles of A⁻ are:
1,17x10⁻²mol - 2,145x10⁻³ mol = <em>9,555x10⁻³ moles of A⁻</em>
Replacing these values in Henderson-Hasselbalch formula:
pH = 4,740 + log₁₀ [9,555x10⁻³ ]/[8,445x10⁻³ ]
pH = 4,794
<em>The pH change in </em>5,000-4,794 <em>= 0,206 units</em>
I hope it helps!
