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 5.00 mL of a 0.420 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

Answer – In this problem, we are given, pH of buffer solution is 5.00 and volume = 195 mL

There are also given total molarity of acid and conjugate base is 0.100 M

Step 1) Calculate the molarity of acid and its conjugate base

From the given pH and pKa for acetic acid buffer and using the Henderson Hasselbalch equation

pH = pKa + log [CH₃COO^-] / [CH₃COOH]

total molarity 0.100 M,

So we assume moles of acid = x,

moles of conjugate base = 0.100-x

5.00 = 4.740 + log ((0.100-x)/0.195 L ) / (x)/0.195L))

5.00-4.740 = log 0.100-x/x

0.26 = log 0.100-x / x

Taking anilog from both side

1.82 = 0.100-x/0x

1.82x = 0.100-x

1.82x +x = 0.100

2.82x = 0.100

x = 0.0355

x = 0.0355 moles of acid

Moles of conjugate base = 0.100- 0.0355

                                            = 0.0645 moles

When we added 5.00 mL of 0.420M HCl, then moles of acid increase and moles of conjugate base decrease

Moles of HCl = 0.420 L * 0.005 L = 0.0021 moles

New moles of CH₃COOH = 0.0355 + 0.0021 = 0.0376 moles

Moles of CH₃COO^- = 0.0645 moles - 0.0021 = 0.0624 moles

Total volume = 195+5.0 = 200 mL

[CH₃COOH] = 0.0376 moles / 0.200 L = 0.188 M

[CH₃COO^-] = 0.0624 moles/0.200 L = 0.312 M

Plugging these value in to the Henderson Hasselbalch equation

pH = pKa + log [CH₃COO^-] / [CH₃COOH]

pH = 4.70 + log (0.312 / 0.188)

      = 4.740 + 0.221

       = 4.96

So, change in pH = 4.96 -5.00

                         = -0.0393