Question

A beaker with 125 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 7.00 mL of a 0.360 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

Answer – There are given, volume of buffer = 125 mL , pH = 5.00

[CH₃COOH] + [CH₃COO^-] = 0.100 M

Volume of HCl = 7.0 mL [HCl] = 0.360 M

First we need to calculate the [CH₃COOH] and [CH₃COO^-]

We know Henderson Hasselbalch equation –

pH = pKa + log [CH₃COO^-] / [CH₃COOH]

so, 5.00 = 4.74 + log [CH₃COO^-] / [CH₃COOH]

so, log [CH₃COO^-] / [CH₃COOH] = 5.00-4.74 = 0.26

taking antilog from both side

[CH₃COO^-] / [CH₃COOH] = 0.549

So, [CH₃COO^-] = 0.549 * [CH₃COOH]

We know

[CH₃COOH] + [CH₃COO^-] = 0.100 M

So, [CH₃COOH] = 0.100 - [CH₃COO^-]

So, [CH₃COO^-] = 0.549 * (0.100 - [CH₃COO^-] )

[CH₃COO^-] = 0.0549 – 0.549 [CH₃COO^-]

So, [CH₃COO^-] + 0.549 [CH₃COO^-] = 0.0549

1.549 [CH₃COO^-] = 0.0549

So, [CH₃COO^-] = 0.0549 / 1.549

                         = 0.0355 M

So [CH₃COOH] = 0.100 - [CH₃COO^-]

                             = 0.100 – 0.0355

                             = 0.0645 M

Now we need to calculate moles of each

Moles of CH₃COOH = 0.0645 M * 0.125 L = 0.00807 moles

CH₃COO^- = 0.0355 M * 0.125 L = 0.0443 moles

Moles of acid HCl added = 0.007 L *0.360 M = 0.00252 moles

We know when we added acid there is moles of acid increase and base gets decreased

So moles of CH₃COOH = 0.00807 + 0.00252 = 0.0106 moles

Moles of CH₃COO^- = 0.00443 - 0.00252 = 0.00191 moles

Total volume = 125 +7 = 132 mL = 0.132 L

So, new [CH₃COOH] = 0.0106 moles / 0.132 L = 0.0802 M

[CH₃COO^-] = 0.00191 moles / 0.132 L = 0.0145 M

So using the Henderson Hasselbalch equation –

pH = pKa + log [CH₃COO^-] / [CH₃COOH]

pH = 4.74 + log 0.0145 / 0.0802

      = 4.74 + (-0.743)

      = 3.99

So change in pH = 5.00 – 3.99

                            = 1.00