This question:
Calculate the mass of sodium benzoate, NaC6H5COO, that must be added to 500.0 mL of 0.100 M Benzoic acid, [C6H5COOH, Ka=6.3 x 10^-5] to form a pH 4.10 buffer solution. Assume no bolume change upon addition of the salt.
I'm not sure how to solve this.
Help with buffering?
2019-03-19 11:53 pm
回答 (1)
2019-03-20 12:13 am
Moles of C₆H₅COOH in the solution = (0.100 mol/L) × (500.0/1000 L) = 0.05 mol
Let y g be the mass of NaC₆H₅COO added.
Molar mass of NaC₆H₅COO = (23.0 + 12.0×7 + 1.0×5 + 16.0×2) g/mol = 144.0 g/mol
Moles of NaC₆H₅COO added = (y g) / (144.0 g/mol) = y/144.0 mol
Moles of C₆H₅COO⁻ in the solution = y/144.0 mol
After addition of NaC₆H₅COO, in the solution:
[C₆H₅COO⁻]/[C₆H₅COOH] = (Moles of C₆H₅COO⁻)/(Moles of C₆H₅COOH) = (y/144.0)/0.05 = y/7.2
Consider the dissociation of C₆H₅COOH:
C₆H₅COOH(aq) + H₂O(l) ⇌ C₆H₅COO⁻(aq) + H₃O⁺(aq) …… Ka = 6.3 × 10⁻⁵
Henderson-Hasselbalch equation: pH = pKa + log([C₆H₅COO⁻]/[C₆H₅COOH])
4.10 = -log(6.3 × 10⁻⁵) + log(y/7.2)
log(y/7.2) = -0.1
y/7.2 = 10⁻⁰˙¹
y = 7.2 × 10⁻⁰˙¹ = 5.72
Mass of NaC₆H₅COO added = 5.72 g
Let y g be the mass of NaC₆H₅COO added.
Molar mass of NaC₆H₅COO = (23.0 + 12.0×7 + 1.0×5 + 16.0×2) g/mol = 144.0 g/mol
Moles of NaC₆H₅COO added = (y g) / (144.0 g/mol) = y/144.0 mol
Moles of C₆H₅COO⁻ in the solution = y/144.0 mol
After addition of NaC₆H₅COO, in the solution:
[C₆H₅COO⁻]/[C₆H₅COOH] = (Moles of C₆H₅COO⁻)/(Moles of C₆H₅COOH) = (y/144.0)/0.05 = y/7.2
Consider the dissociation of C₆H₅COOH:
C₆H₅COOH(aq) + H₂O(l) ⇌ C₆H₅COO⁻(aq) + H₃O⁺(aq) …… Ka = 6.3 × 10⁻⁵
Henderson-Hasselbalch equation: pH = pKa + log([C₆H₅COO⁻]/[C₆H₅COOH])
4.10 = -log(6.3 × 10⁻⁵) + log(y/7.2)
log(y/7.2) = -0.1
y/7.2 = 10⁻⁰˙¹
y = 7.2 × 10⁻⁰˙¹ = 5.72
Mass of NaC₆H₅COO added = 5.72 g
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