Question

Calculate the pH of a solution obtained by adding 12.0 mL of 0.25 M H2SO4 to 6.0 mL of 1.0 M NH3.

Answer 1

pKb of NH3 = 9.25

moles of NH3 present = M*V = 1.0 M * 6.0 mL =6.0 mmol
moles of H+ added= 2*mol of H2SO4
= 2*0.25 M *12 mL
= 6.0 mmol

both will react to form 6 mmol of NH4+
[NH4+] = mol / total volume
= 6 mmol/ (12 mL+6 mL)
= 1/3 M

Kb of NH3 = 1.8*10^-5
Ka of NH4+ = 10^-14 / Kb
= 10^-14 /(1.8*10^-5)
=5.56*10^-10

NH4+ <----------> NH3 + H+
1/3 0 0 (initial)
1/3 - x x x (at equilibrium)

Ka = x*x/ (1/3-x)
since Ka is small, x will be small and it can be ignored as compared to 1/3

5.56*10^-10 = x^2 / (1/3)
x = 1.361*10^-5 M

So,[H+] = x = 1.361*10^-5

pH = -log [H+]
= -log (1.361*10^-5 )
= 4.87
Answer: 4.87