Question

A. 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         0.030M in NH4Cl at 25 °C?

B. HClO is a weak acid (Ka = 4.0 × 10–8) and so the salt NaClO acts as a weak base. What is the pH of a solution that is 0.045 M in NaClO at 25 °C?

Answer 1

Answer – A) We are given, [NH­_4­Cl] = 0.030 M ,

[NH­_4­Cl] =[NH₄⁺] = 0.030 M

Kb = 1.8*10^-5

We need to put ICE chart

NH₄⁺ + H₂O -----> NH₃ + H₃O⁺

I 0.030                      0           0

C -x                          +x           +x

E 0.030-x                  +x           +x

We know, Kb for NH₃ = 1.8*10^-5

So, Ka = 1*10^-14 / 1.8*10^-5

            = 5.56*10^-10

Ka = [H₃O⁺] [NH₃] /[NH₄⁺]

5.56*10^-10 = x *x / (0.030-x)

We can neglect x in the 0.030-x, since Ka value is too small

So, x² = 5.56*10^-10 *0.030

x = 4.08*10^-6 M

so, x = [H₃O⁺] = 4.08*10^-6 M

pH = -log [H₃O⁺]

      = - log 4.08*10^-6 M

      = 5.39

B) We are given, [NaClO] = 0.045 M ,

[NaClO] =[ClO^-] = 0.03 M

Ka = 4.8*10^-8

We need to put ICE chart

ClO^- + H₂O -----> HClO + OH^-

I 0.045                      0           0

C -x                          +x           +x

E 0.045-x                  +x           +x

We know, Ka for HClO = 4.8*10^-6

So, Kb = 1*10^-14 / 4.8*10^-6

            = 2.08*10^-9

Kb = [OH^-] [HClO] /[ClO^-]

2.08*10^-9 = x *x / (0.045-x)

We can neglect x in the 0.045-x, since Kb value is too small

So, x² = 2.08*10^-9 *0.045

x = 9.68*10^-6 M

so, x = [OH^-] = 4.08*10^-6 M

pOH = -log [OH^-]

      = - log 9.68*10^-6 M

      = 5.01

So, pH = 14-pOH

            = 14 -5.01

            = 8.98