In: Chemistry
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?
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