What is the #pH# of a #0.10*mol*L^-1# solution of ammonium ion....for which #pK_a=9.26#?

1 Answer
Mar 31, 2017

Answer:

And the #pK_a# of ammonium ion is..............?

Explanation:

We interrogate the equilibrium,

#NH_4^+ + H_2O(l) rightleftharpoons NH_3 + H_3O^(+)#

For which, #K_a=([NH_3][H_3O^+])/([NH_4^+])=10^(-9.25)#

So #10^(-9.25)=x^2/(0.10-x)#, a quadratic in #x#, that we can solve by successive approximation, i.e. assuming that #0.1>x#:

Thus #x~=sqrt(10^(-9.25)xx0.10)#

#x_1=7.5xx10^-6#

#x_2=7.5xx10^-6#

Since #x=[H_3O^+]#, #pH=-log_10(7.5xx10^-6)=5.12#