Use the Henderson-Hasselbalch equation to perform the following calculations?
2018-03-12 10:14 pm
The Ka of acetic acid is 1.8 x 10–5. Buffer A: Calculate the mass of solid sodium acetate required to mix with 100.0 mL of 0.1 M acetic acid to prepare a pH 4 buffer. Buffer B: Calculate the mass of solid sodium acetate required to mix with 100.0 mL of 1.0 M acetic acid to prepare a pH 4 buffer.
回答 (1)
2018-03-12 10:50 pm
Buffer A:
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) …. Ka = 1.8 × 10⁻⁵
Henderson-Hasselbalch equation: pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
4 = -log(1.8 × 10⁻⁵) + log([CH₃COO⁻]/0.1)
log([CH₃COO⁻]/0.1) = 4 + log(1.8 × 10⁻⁵)
[CH₃COO⁻]/0.1 = 10^[4 + log(1.8 × 10⁻⁵)]
[CH₃COO⁻] = 0.1 × 10^[4 + log(1.8 × 10⁻⁵)] M
[CH₃COO⁻] = 0.018 M
[CH₃COONa] = 0.018 M
Molar mass of CH₃COONa = (12.0×2 + 1.0×3 + 16.0×2 + 23.0) g/mol = 82.0 g/mol
Mass of CH₃COONa required = (0.018 mol/L) × (82.0 g/mol) × (100.0/1000 L) = 0.148 g
====
Buffer B:
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) …. Ka = 1.8 × 10⁻⁵
Henderson-Hasselbalch equation: pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
4 = -log(1.8 × 10⁻⁵) + log([CH₃COO⁻]/1.0)
log([CH₃COO⁻]/1.0) = 4 + log(1.8 × 10⁻⁵)
[CH₃COO⁻]/1.0 = 10^[4 + log(1.8 × 10⁻⁵)]
[CH₃COO⁻] = 1.0 × 10^[4 + log(1.8 × 10⁻⁵)] M
[CH₃COO⁻] = 0.18 M
[CH₃COONa] = 0.18 M
Molar mass of CH₃COONa = (12.0×2 + 1.0×3 + 16.0×2 + 23.0) g/mol = 82.0 g/mol
Mass of CH₃COONa required = (0.18 mol/L) × (82.0 g/mol) × (100.0/1000 L) = 1.48 g
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) …. Ka = 1.8 × 10⁻⁵
Henderson-Hasselbalch equation: pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
4 = -log(1.8 × 10⁻⁵) + log([CH₃COO⁻]/0.1)
log([CH₃COO⁻]/0.1) = 4 + log(1.8 × 10⁻⁵)
[CH₃COO⁻]/0.1 = 10^[4 + log(1.8 × 10⁻⁵)]
[CH₃COO⁻] = 0.1 × 10^[4 + log(1.8 × 10⁻⁵)] M
[CH₃COO⁻] = 0.018 M
[CH₃COONa] = 0.018 M
Molar mass of CH₃COONa = (12.0×2 + 1.0×3 + 16.0×2 + 23.0) g/mol = 82.0 g/mol
Mass of CH₃COONa required = (0.018 mol/L) × (82.0 g/mol) × (100.0/1000 L) = 0.148 g
====
Buffer B:
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) …. Ka = 1.8 × 10⁻⁵
Henderson-Hasselbalch equation: pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
4 = -log(1.8 × 10⁻⁵) + log([CH₃COO⁻]/1.0)
log([CH₃COO⁻]/1.0) = 4 + log(1.8 × 10⁻⁵)
[CH₃COO⁻]/1.0 = 10^[4 + log(1.8 × 10⁻⁵)]
[CH₃COO⁻] = 1.0 × 10^[4 + log(1.8 × 10⁻⁵)] M
[CH₃COO⁻] = 0.18 M
[CH₃COONa] = 0.18 M
Molar mass of CH₃COONa = (12.0×2 + 1.0×3 + 16.0×2 + 23.0) g/mol = 82.0 g/mol
Mass of CH₃COONa required = (0.18 mol/L) × (82.0 g/mol) × (100.0/1000 L) = 1.48 g
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