Acetic Acid Sodium acetate buffer? PLEASE HELP!?

2017-11-05 4:45 am

You are to assume that the buffer is made up by mixing volumes of 0.100 M acetic acid and 0.100 M sodium acetate solutions.

Calculate the volume of 0.100 M acetic acid required to prepare 60.0 mL of a buffer of pH 4.6
Calculate the volume of 0.100 M sodium acetate required to prepare 60.0 mL of a buffer of pH 4.6.
Calculate the concentration [HA] in this buffer.
Calculate the concentration [A-] in this buffer

回答 (1)

micatkie

2017-11-05 5:15 am

✔ 最佳答案

Let V mL of 0.100 M CH₃COOH solution required.
Then, volume of 0.100 M CH₃COONa solution required = (60 - V) mL

In the buffer :
No. of moles of CH₃COOH = (0.100 mol/L) × (V/1000 L)
No. of moles of CH₃COO⁻ = (0.100 mol/L) × [(60.0 - V)/1000 L]
[CH₃COO⁻]/[CH₃COOH] = (60.0 - V)/V

CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) …. Ka = 1.8 × 10⁻⁵
pH = pKa + log{[CH₃COO⁻]/[CH₃COOH]}
4.6 = -log(1.8 × 10⁻⁵) + log{(60.0 - V)/V}
4.6 = 4.745 + log{(60.0 - V)/V}
log{(60.0 - V)/V} = -0.145
(60.0 - V)/V = 10⁻⁰·¹⁴⁵
60.0 - V = 10⁻⁰·¹⁴⁵V
(10⁻⁰·¹⁴⁵ + 1)V = 60.0
V = 60.0/(10⁻⁰·¹⁴⁵ + 1)
V = 35.0

Volume of 0.100 M acetic acid required = 35.0 mL
Volume of 0.100 M sodium acetate required = (60.0 - 35.0) mL = 25.0 mL
[HA] in the buffer = (0.100 M) × (35.0/60.0) = 0.0583 M
[A⁻] in the buffer = (0.100 M) × (25.0/60.0) = 0.0417 M

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