Answer:
ΔpH = 0.296
Explanation:
The equilibrium of acetic acid (CH₃COOH) in water is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Henderson-Hasselbalch formula to find pH in a buffer is:
pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]
Replacing with known values:
5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]
0.260 = log₁₀ [CH₃COO⁻] / [CH₃COOH]
1.820 = [CH₃COO⁻] / [CH₃COOH] (1)
As total molarity of buffer is 0.100M:
[CH₃COO⁻] + [CH₃COOH] = 0.100M (2)
Replacing (2) in (1):
1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]
1.820[CH₃COOH] = 0.100M - [CH₃COOH]
2.820[CH₃COOH] = 0.100M
[CH₃COOH] = 0.100M / 2.820
[CH₃COOH] = 0.035M
Thus: [CH₃COO⁻] = 0.100M - 0.035M = 0.065M
5.40 mL of a 0.490 M HCl are:
0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.
Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles
HCl reacts with CH₃COO⁻ thus:
HCl + CH₃COO⁻ → CH₃COOH
After reaction, moles of CH₃COO⁻ are:
0.0101 moles - 2.646x10⁻³ moles = 7.429x10⁻³ moles of CH₃COO⁻
Moles of CH₃COOH before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:
5.425x10⁻³ moles + 2.646x10⁻³ moles = 8.071x10⁻³ moles of CH₃COOH. Replacing these values in Henderson-Hasselbalch formula:
pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]
pH = 4.704
As initial pH was 5.000, change in pH is:
ΔpH = 5.000 - 4.740 = 0.296
