Question

calculate the ph of .50 M NH4CL solution for NH3 , Kb= 1.8x10^-5 what is the net ionic equation for the reaction

Answer 1

The net ionic equation is:

NH4+ (aq) —> NH3 (aq) + H+ (aq)

we have below equation to be used:

Ka = Kw/Kb

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/1.8*10^-5

Ka = 5.556*10^-10

NH4+ + H2O -----> NH3 + H+

0.5 0 0

0.5-x x x

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

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.556*10^-10)*0.5) = 1.667*10^-5

since c is much greater than x, our assumption is correct

so, x = 1.667*10^-5 M

So, [H+] = x = 1.667*10^-5 M

we have below equation to be used:

pH = -log [H+]

= -log (1.667*10^-5)

= 4.78

Answer: 4.78