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 8.70 mL of a 0.490 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

Given the pH of the acetic acid buffer = 5.000

Applying Hendersen equation

pH = pKa + log[CH3COO-] / [CH3COOH]

=> 5.000 = 4.74 + log[CH3COO-] / [CH3COOH]

=>  log[CH3COO-] / [CH3COOH] = 0.26

=> [CH3COO-] / [CH3COOH] = 10^0.26 = 1.82 ------- (1)

Given [CH3COO-] + [CH3COOH] = 0.100 M ------ (2)

By solving eqn(1) and (2) we get

[CH3COOH] = 0.03546 M and

[CH3COO-] = 0.06454 M

Hence initial mol of CH3COOH = MxV = 0.03546 M x 0.105 L = 0.0037233 mol

initial mol of CH3COO- = MxV = 0.06454 M x 0.105 L = 0.0067767 mol

Volume of 0.490 M HCl added = 8.70 mL = 0.00870 L

Hence moles of HCl added = MxV = 0.490 mol/L x 0.00870 L = 0.004263 mol

Now 0.004263 mol HCl will react with 0.004263 mol of CH3COO- to form 0.004263 mol of CH3COOH.

moles of CH3COO- left after reaction = 0.0067767 mol - 0.004263 mol = 0.0025137 mol

moles of CH3COOH left after the reaction = 0.0037233 mol + 0.004263 mol = 0.0079863 mol

Now applying Hendersen equation

pH = pKa + log[CH3COO-] / [CH3COOH]

=> pH = 4.74 + log(moles of CH3COO- / moles of CH3COOH) [ Since Volume is same for both]

=> pH = 4.74 + log(0.0025137 mol / 0.0079863 mol) = 4.238

Hence change in pH = 4.238 - 5.000 = - 0.762