Question

What is the change in pH after addition of 10.0 mL of 1.0 M sodium hydroxide to 90.0 mL of a 1.0 M NH3/1.0 M NH₄⁺ buffer? [K_b for ammonia is 1.8 x 10^-5]

Answer 1

pH of 90 mL of 1 M NH₃/1 M NH₄⁺ buffer = pKa + Log (n_NH3/n_NH4+)

Here, K_b = 1.8 x 10^-5.

We know that: pK_b = -Log(K_b)

= -Log(1.8 x 10^-5)

= 4.74

We also know that: pK_a + pK_b = 14

i.e. pK_a = 14 - 4.74

pK_a = 9.26

And n_NH3 = 1*90, i.e 90 mmol., n_NH4+ = 90 mmol, i.e. n_NH3/n_NH4+ = 1

Therefore, pH = 9.26

No. of mmol. of NaOH added = 10*1, i.e. 10 mmol.

i.e. NH₃ (90 mmol) + NH₄⁺ (90 mmol) + NaOH (10 mmol) = NH₃ (100 mmol) + NH₄⁺ (80 mmol) + NaX (10 mmol)

pH = pKa + Log (n_NH3/n_NH4+)

= 9.26 + Log(100/80)

= 9.26 + 0.097

Therefore, the required change in pH = (9.26 + 0.097) - 9.26, i.e. 0.097 ~ 0.1.