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 5.70 mL of a 0.480 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

Answer

∆pH = - 0.46

Explanation

Henderson-Hasselblach equation is

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

5.000 = 4.740 + log( [CH3COO^-] /[CH3OOH])

log([CH3COO^-]/[CH3COOH]) = 0.26

[CH3COO^-]/[CH3COOH] = 1.820

[CH3COO^-] = 1.820 × [CH3COOH]

moles of CH3COO^- = 1.820 × moles of CH3COOH

moles of buffer = (0.100mol/1000ml) × 105ml = 0.0105mol

moles of CH3COO^- + moles of CH3COOH = 0.0105mol

1.820 × moles of CH3COOH + moles of CH3COOH = 0.0105mol

2.820× moles of CH3COOH = 0.0105mol

moles of CH3COOH = 0.003723mol

moles of CH3COO^- = 0.0105mol - 0.003723mol = 0.006777mol

moles of HCl added = ( 0.480mol/1000ml) × 5.70ml = 0.002736mol

HCl react with conjucate base , CH3COO^-

CH3COO^- + H⁺ ------> CH3COOH

1:1molar reaction

0.002736moles of HCl react with 0.002736moles of CH3COO^- to give 0.002736 moles of CH3COOH

After addition of HCl

moles of CH3COO^- = 0.006777mol - 0.002736mol = 0.004041mol

moles of CH3COOH = 0.003723mol + 0.002736mol = 0.006459mol

Total volume = 105ml + 5.70ml = 110.7ml

[CH3COOH] = ( 0.006459mol/110.7ml)×1000ml = 0.05835M

[CH3COO^-] = ( 0.004041mol/110.7ml) × 1000ml = 0.03650M

Applying Henderson - Hasselbalch equation

pH = pKa + log(0.03650M/0.05835M)

pH = 4.74 - 20

pH = 4.54

∆pH = 4.54 - 5.00 = - 0.46