Question

A beaker with 195 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 5.30 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.

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

Answer 1

Given the pH of the acetic acid buffer = 5.000

We can apply Hendersen equation to calculate the ratio of the moles of conjugate base and the acid.

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)

Also 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 0.195 L = 0.0069147 mol

initial mol of CH3COO- = MxV = 0.06454 M x 0.195 L = 0.0125853 mol

Volume of 0.480 M HCl added = 5.30 mL = 0.00530 L

Hence moles of HCl added = MxV = 0.480 mol/L x 0.00530 L = 0.002544 mol

Now 0.002544 mol HCl will react with 0.002544 mol of CH3COO- to form 0.002544 mol of CH3COOH.

moles of CH3COO- left after reaction = 0.0125853 mol - 0.002544 mol = 0.0100413 mol

moles of CH3COOH left after the reaction = 0.0069147 mol + 0.002544 mol = 0.0094587 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(0.0100413 mol / 0.0094587 mol) = 4.766

Hence change in pH = 4.766 - 5.000 = - 0.234 (answer)