Question

What weight of dry sodium acetate (FW = 82 g/mol) and volume of acetic acid (MW = 60 g/mol; density = 1.049 ml/g are required for 1 liter of 0.1 M sodium acetate, pH 4.50. (pKa= 4.73)?

Answer: 3.6 mL acetic acid, 3.0 g sodium acetate

Please provide steps to get the above answer.

Answer 1

Henderson - Hasselbalch whic give pH of a buffer solution is

pH = pKa + log([A^-]/[HA])

where,

A^- = conjucate base , CH3COO^-

HA = weak acid , CH3COOH

pKa of CH3COOH = 4.73

required pH = 4.50

4.50 = 4.73 + log([CH3COO^-]/[CH3COOH])

log([CH3COO^-]/[CH3COOH]) =- 0.23

[CH3COO^-] /[CH3COOH] = 0.5888

moles of CH3COO^- / moles of CH3COOH = 0.5888

moles of CH3COO^- = 0.5888 × moles of CH3COOH

Total moles of phospate species = 0.1mol

moles of CH3COO^- + moles of CH3COOH = 0.10mol

(0.5888 × moles of CH3COOH) + ( moles of CH3COOH) = 0.10mol

1.5888 × moles of CH3COOH = 0.10mol

moles of CH3COOH = 0.06294mol

moles of CH3COO^- = 0.10mol - 0.06294mol = 0.03706mol

moles of acetic acid required = 0.06294mol

moles of sodium acetate required = 0.03706mol

mass of acetic acid required = 0.06294mol × 60g/mol = 3.7764g

mass of sodium acetate required = 0.03706mol × 82g/mol = 3.0g

Volume = mass/density

Volume of acetic acid requred = 3.7764g /1.049g/ml = 3.6ml

Therefore,

3.6 ml acetic acid and 3.0g sodium acetate