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The Ph of 0.1M solution of NH4Cl will be (take pKb value of NH3 as 5.0)
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
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