Question

Calculate the pH of a 0.050 M solution of NH4NO3.

Calculate the pH of a 0.29 M solution of KC7H5O2.

Answer 1

NH4NO3 --> NH4+(aq) + NO3-(aq)

NH4+(aq) + H2O(l) --< NH3 + H3O+

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

Ka = Kw/Kb = ((10^-14)/(1.8*10^-5) = 5.55*10^-10

[NH3 ] = [H3O+] = x

[NH4+] = M-x = 0.05-x

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

5.55*10^-10 = x*x/(0.05-x)

x = 5.267*10^-6

[H+] = x =  5.267*10^-6

pH = -log(h) = -log( 5.267*10^-6 = 5.2784

b=

KC7H5O2 --> K + C7H5O2-

H2O +  C7H5O2- <--> OH- +  C7H5O2H

Kb = [C7H5O2H][OH-]/[C7H5O2-]

Ka = 6.46*10^-5

Kb = (10^-14)/(6.46*10^-5) = 1.5479*10^-10

[C7H5O2H] = [OH-] = x

[C7H5O2-] = M-x = 0.29-x

1.5479*10^-10 = x*x/(0.29-x)

x = 6.7*10^-6

pOH = -log(6.7*10^-6) = 5.17392

pH = 14-5.17392 = 8.82608