Question

A beaker with 1.00×102 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 8.20 mL of a 0.470 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

First we shall determine the initial concentration of acid and salt.

Let, concentration of acid in initial solution = x

So, concentration of salt in initial solution = 0.1-x

According to Henderson's equation:

pH = pKa + log [salt]/[acid]

or, 5 = 4.74 + log (0.1-x / x)

or, x = 0.035 M

So, [acid]_initial = 0.035 M

[salt]_initial = 0.1-0.035 = 0.065 M

From this we can determine the initial no of moles of acid and base in the solution.

Moles of acid in solution = 0.1 L * 0.035 M = 0.0035 moles

Moles of salt in solution = 0.1 L * 0.065 M = 0.0065 moles

Now come to the second part. HCl will react with salt to form acid, as a result moles of salt in solution will decrease and moles of acid in solution will increase.

CH₃COO^- + HCl ---> CH₃COOH + Cl^-

No of moles of HCl added = 0.0082 L * 0.47 M = 0.0039 moles

Final volume of solution = 100 mL + 8.2 mL = 108.2 mL

Final no of moles of acid = 0.0035 + 0.0039 = 0.0074 moles;

and concentration [acid]_final = 0.0074 moles * 1000 / 108.2 = 0.068 M

Final no of moles of salt = 0.0065 - 0.0039 = 0.0026 moles;

and concentration [salt]_final = 0.0026 moles * 1000 / 108.2 = 0.024 M

pH = pKa + log[salt]/[acid]

= 4.74 + log (0.024/0.068)

= 4.28

ΔpH = 4.28 - 5 = - 0.72