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 0.038 M in NH4Cl at 25 °C?

Answer 1

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 0.038 M in NH4Cl at 25 °C?

First calculate the value of Ka as follows:

Ka =Kw/Kb = 1.00*10^- 14/1.8* 10^-5

= 5.56*10^-10

in the solution of NH4Cl

NH4Cl = NH4+   + Cl-

0.038       0.038

0.038 NH4+ reacts with water as follows:

                            NH4+ +H2O-----> NH3 + H3O+

Now make ICE table

                            NH4+ +H2O-----> NH3 + H3O+

I                            0.038                    0        0

C                          -x                          +x    +x

E                 0.038 –x                        x        x

Now equilibrium constant ;

Ka = [NH3][H3O+]/[NH4+]

5.56*10^-10 = x*x / 0.0338 –x

Due to small value of 5.56*10^-10 we can write 0.0338 –x 0.0338

Therefore;

5.56*10^-10 = x^2 / 0.0338

x^2= 2.11*10^-11

x= 4.595 *10^-6 M

so pH=-log[H+]

= - log (4.595 *10^-6)

= -(-5.34)
pH=5.34