Grade 10Physical Chemistry
The Ph of 0.1M solution of NH4Cl will be (take pKb value of NH3 as 5.0)
The Ph of 0.1M solution of NH4Cl will be (take pKb value of NH3 as 5.0)
1 Answer
Arun
Since NH4Cl is made by reacting a weak base (NH3) with a strong acid (HCl), the salt will produce a slightly acidic solution when dissolved in water. That is, the NH4+ ion will hydrolyze to produce equally small amounts of H3O+ and NH3.
Molarity . . . . .NH4+ + H2O H3O+ + NH3
Initial . . . . . . . .0.1 . . . . . . . . . . . . . .0 . . . . . .0
Change . . . . . .-x . . . . . . . . . . . . . . .x . . . . . .x
At Equil. . . . .0.1-x . . . . . . . . . . . . . .x . . . . . .x
The Kb for NH3 = 1.8 x 10^-5. Ka for NH4+ ion = (Kw / Kb) = (1 x 10^-14 / 1.8 x 10^-5) = 5.6 x 10^-10
Ka = [H3O+][NH3] / [NH4+] = x^2 / (0.1 - x) = 5.6 x 10^-10 . . .becuse Ka is so small, x will be small compared to 0.1 so we can delete it from the 0.1 - x term.
x^2 / 0.1 = 5.6 x 10^-10
x^2 = 5.6 x 10^-11
x = 7.5 x 10^-6 = [H3O+]
pH = -log[H3O+] = -log (7.5 x 10^-6) = 5.1
Molarity . . . . .NH4+ + H2O H3O+ + NH3
Initial . . . . . . . .0.1 . . . . . . . . . . . . . .0 . . . . . .0
Change . . . . . .-x . . . . . . . . . . . . . . .x . . . . . .x
At Equil. . . . .0.1-x . . . . . . . . . . . . . .x . . . . . .x
The Kb for NH3 = 1.8 x 10^-5. Ka for NH4+ ion = (Kw / Kb) = (1 x 10^-14 / 1.8 x 10^-5) = 5.6 x 10^-10
Ka = [H3O+][NH3] / [NH4+] = x^2 / (0.1 - x) = 5.6 x 10^-10 . . .becuse Ka is so small, x will be small compared to 0.1 so we can delete it from the 0.1 - x term.
x^2 / 0.1 = 5.6 x 10^-10
x^2 = 5.6 x 10^-11
x = 7.5 x 10^-6 = [H3O+]
pH = -log[H3O+] = -log (7.5 x 10^-6) = 5.1
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