Question

A beaker with 105 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.30 mL of a 0.470 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Express your answer numerically to two decimal places. Use a minus ( − ) sign if the pH has decreased.

Answer 1

∆pH = -0.68

Explanation

Henderson - Hasslbalch equation is

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

where,

pKa = pKa of weak acid

A^- = conjucate base, CH3COO^-

HA = weak acid , CH3COOH

substituting the values

5.000 = 4.740 + log([CH3COO^-]/[CH3COOH])

log([CH3COO^-]/[CH3COOH]) = 0.260

[CH3COO^-]/[CH3COOH] =1.82

[CH3COO^-] = [CH3COOH] × 1.82

Molarity of buffer = 0.100M

[CH3COO^-] + [CH3COOH] = 1.82

([CH3COOH] × 1.82 ) + [CH3COOH] = 0.100M

2.82 × [CH3COOH]= 0.100M

[CH3COOH] = 0.03546M

[CH3COO^-] = 0.100M - 0.03546M = 0.06454M

moles of CH3COOH = (0.03546mol/1000ml) × 105ml = 0.003723mol

moles of CH3COO^- = ( 0.06454mol/1000ml)! × 105ml = 0.0067767 mol

moles of HCl added = ( 0.470mol/1000ml) × 8.30ml = 0.003901mol

As an acid HCl react with conjucate base CH3COO^-

CH3COO^- (aq) + H⁺(aq) ---------> CH3COOH(aq)

0.003901moles of HCl react with 0.003901moles of CH3COO^- to give 0.003901moles of CH3COOH

After adding HCl

moles of CH3COOH = 0.003723mol + 0.003901mol = 0.007624mol

moles of CH3COO^- = 0.0067767mol - 0.003901mol = 0.0028757mol

Total volume = 105ml + 8.30ml = 113.3ml

[CH3COOH] = ( 0.007624mol/113.3ml) ×1000ml = 0.06729M

[CH3COO^-] = (0.0028757mol/113.3ml) ×1000ml = 0.02538M

Applying Henderson-Hasselbalch equation

pH = 4.740 + log(0.02538M/0.06729M)

pH = 4.740 - 0.420

pH = 4.320

∆pH = 4.320 - 5.000 = -0.68