Question

Buffers Concentration Volume Weak Acid Conjugate Base sodium acetate pH Ka acetic acid Buffer 1 500 mL 4.75

Calculate the amounts in grams of weak acid and conjugate base needed to make the buffer.

Answer 1

Answer

Weak acid needed = 1.7850g

Conjucate base needed = 0.4346g

Explanation

i) Henderson-Hasselbalch equation is

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

A^- = conjucta base , CH3COO^-

HA = weak acid , CH3COOH

target pH= 4.00

pKa of acetic acid = 4.75

substituting the values

4.00 = 4.75 + log([CH3COO^-]/[CH3COOH])

log([CH3COO^-] /[CH3COOH]) = -0.75

[CH3COO^-] /[CH3COOH] = 0.1778

[CH3COO^-] =0.1778× [CH3COOH]

ii) moles of CH3COO^- = 0.1778×moles of CH3COOH

Total moles of buffer = (0.07mol/1000ml)×500ml = 0.035mol

moles of CH3COOH + moles of CH3COO^- = 0.035mol

moles of CH3COOH + 0.1778× moles of CH3COOH = 0.035mol

1.1778×moles of CH3COOH = 0.035mol

moles of CH3COOH = 0.0297mol

moles of CH3COO^- = 0.0350mol - 0.0297mol = 0.0053mol

iii) Mass = Molar mass × no of moles

mass of CH3COOH reaquired = 60.1g/mol × 0.0297mol = 1.7850g

mass of CH3COONa required = 82.0g/mol × 0.0053mol = 0.4346g