Question

K_b of base imidazole at 25°C is 8.8 × 10^–8. 

(a) In what amounts should 0.02 M HCl and 0.02 M imidazole be mixed 100 mL of a buffer at pH = 7? 

(b) When the resulting buffer is diluted to one litre, calculate pH of the diluted buffer.

Answer 1

(a) As pH = 7, pOH = 14 – 7 = 7 (at 25°C), pK_b = –log K_b = – log (8.8 × 10^–8) = 7.0555

Applying

Suppose V₁ ml of HCl is mixed with V₂ ml of imidazole (base) to make the buffer. 

millimole of HCl = 0.02 V₁ 

millimole of imidazole = 0.02 V₂ 

As the buffer is of the base and its salt, 0.02 millimole of HCl will combine with 0.02 millimole of base to give 0.02 millimole of salt. 

∴ millimole of salt = millimole of HCl

= 0.02 V₁ 

and m.m. of base left = 0.02 V₂ – 0.02 V₁ 

Given that V₁ + V₂ = 100 mL .....(3) 

From (2) and (3) we get, V₁ = 31.84 mL and V₂ = 68.15 mL 

(b) pH shall remain the same on dilution as both Kb and [salt]/[base] will not change.