Question

A 10.00 mL solution of 0.1900 M NH₄Cl is titrated with 0.1300 M NaOH to the equivalence point.

Volume of NaOH = 14.62mL

Calculate the pH of the solution at the equivalence point.

Calculate the pH of a simple 0.0772 M aqueous solution of ammonia.

Answer 1

pH in equivalence point

total V = 10+14.62 = 24.62 mL

so..

[NH4+] = M1V1 /V2 = 10*0.19 / 14.62 = 0.12995 M

so..

NH4+ in equilbirium with water, hydrolysis

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

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

Ka = Kw/Kb = (10^-14)/(1.77*10^-5) = 5.55*10^-10

in equilibrium:

[NH3]= [H3O+]= x

[NH4+] = M-x = 0.12995 -x

substitut ein Kb

5.55*10^-10 = x*x/(.12995 -x)

x = 8.49*10^-6

pH = -log(H3O+) = -log( 8.49*10^-6) = 5.0710923

B)

of

NH3 = 0.0772 M

so

NH3 + H3O+ <-> NH4+ +OH-

Kb = [NH4+][OH-]/[NH3]

in euqilibrium

[NH4+] = x= [OH-]

[NH3] = 0.0772-x

substitute

1.77*10^-5 = x*x/( 0.0772-x)

x = OH = 0.00116

pOH = -log(0.00116) = 2.935

pH = 14-2.935 = 11.065