Question

A beaker with 155 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 5.20 mL of a 0.310 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

initially

pH = pKa + log(acetate/acid)

5 = 4.74 + log(ratio)

ratio = 10^(5-4.74) = 1.8197

[Acetate] = 1.819[Acid]

[Acetate] + [acid] = 0.1

then

1.819[Acid] + [Acid] = 0.1

2.819[Acid] = 0.1

[Acid] = 0.1/2.819 = 0.03547

[ACetate] = 0.03547*1.819 = 0.06451993

then

mmol of acid added = MV = 0.31*5.2 = 1.612 mmol of HCl

after addition:

mmol of acetate = MV - 1.612 = 155*0.06451993 - 1.612 = 8.388

mmol of acid = MV+ 1.612 = 155*0.03547+1.612 = 7.10985

then

pH = 4.74+ log(8.388/7.10985) =4.811

dpH = 5-4.811 = 0.189

decrease of 0.189 units of pH