CHemistryyyy pH

Question

NH₃ is a weak base (𝐾b=1.8×10−5) and so the salt NH₄Cl acts as a weak acid. What is the pH of a solution that is 0.076 M in NH₄Cl at 25 °C?

Answer 1

NH₄Cl + H₂O ==> NH₃ + HCl .. hydrolysis of NH₄Cl

Since NH₄Cl is a weak acid, we need to write the Ka expression

Ka = [NH₃][HCl] / [NH₄Cl]

Recall that KaKb = 1x10^-14

Ka = 1x10^-14 / 1.8x10^-5 = 5.56x10^-10

5.56x10^-10 = [NH₃][HCl] / [NH₄Cl]

5.56x10^-10 = (x)(x) / 0.076-x (assume x is small relative to 0.078 and ignore it in the denominator)

5.56x10^-10= x2 / 0.076

x² = 4.22x10^-11

x = 6.50x10^-6 (this is very small relative to 0.078 so above assumption was valid)

[NH₃] = [HCl] = 6.50x10^-6

pH = -log [H⁺] = -log 6.50x10^-6

pH = 5.19