Question

Part A

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 4.50 mL of a 0.430 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

Moles of HCl added =0.43 mol/L *4.5mL/1000mL = 0.00194 moles

Calculate the moles of acetate and acetic acid initially present. pka of acetate buffer = 4.74 (refer to your textbook)

pH = pka + log [salt/acid]

5 = 4.74 + log [salt/acid]

[salt/acid] = 1.82

decimal fraction of salt = 1.82/ 1+ 1.82 = 0.65

decimal fraction of acid = 1/1+ 1.82 = 0.35

Molarity of acetate =0.1*0.65 = 0.065 M

Number of moles of acetate = 0.065 M * 195mL/1000mL = 0.013 moles

Molarity of acetic acid = 0.1 * 0.35 M

Number of moles of acetic acid = 0.1 * 0.35 M * 195mL/1000mL = 0.195 moles

HCl added will react with acetate to convert it to acetic acid. So, you have to substract the number of moles of acetate and add it to number of moles of acetic acid.

Moles of acetate after adding HCl = 0.013 moles - 0.00194 moles = 0.011 moles

Molarity of acetate in after adding HCl = 0.011 moles *1000ml/(5 + 195mL) = 0.055 M

Moles of acetic acid afetr adding HCl = 0.195+ 0.00194 =0.197 moles

Molarity of acetic acid after adding HCl = 0.197 moles *1000mL/200mL = 0.98 M

pH = pKa + log [salt/acid]

= 4.74 + log [0.055/0.98] = 3.49