What is #pH# of an aqueous solution for which #[NH_4^+]=1.0*mol*L^-1#?

1 Answer
Mar 23, 2017

Answer:

It should be under #7#, should it not?

Explanation:

#pH=-log_10[H_3O^+]#.

And we know that ammonium salts are weakly acidic, and undergo the acid-base reaction:

#NH_4^(+) + H_2O(l) rightleftharpoonsNH_3(aq) + H_3O^+#

For ammonium ion, #K_a=1.80xx10^-5#

And if there were a #1.0*mol*L^-1# solution of ammonium ion, we would solve the equilibrium expression in the usual way.

i.e. #K_a=1.80xx10^-5=([H_3O^+][NH_3(aq)])/([NH_4^(+)])#

And if #x*mol*L^-1# ammonium ion ionizes, then.........

#K_a=1.80xx10^-5=x^2/((1.0-x)*mol*L^-1)#, a quadratic in #x#, which may be simplified given the reasonable assumption that #1.0">>"x#, and thus #1.0-x~=1.0#.

And thus #x~=sqrt(1.80^-5)#,

so #x_1=4.2xx10^-3#, and recycling this approximation of #x# back into the expression:

#x_2=4.2xx10^-3#, and since #x# has converged, we may accept this value.

Thus #[H_3O^+]=4.2xx10^-3*mol*L^-1#, and #pH=-log_10[H_3O^+]=-log_10(4.2xx10^-3)=2.37#.