What is the concentration of ammonium ion in a solution of #NH_3(aq)# that is nominally #0.157*mol*L^-1#?

1 Answer
Apr 10, 2017

Answer:

#[NH_4^+]=1.67xx10^-3*mol*L^-1#

Explanation:

We interrogate the equilibrium:

#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + ""^(-)OH#,

for which #K_b=1.8xx10^-5=([NH_4^+][HO^-])/([NH_3(aq)])#

Initially, #[NH_3]=0.157*mol*L^-1#, and we assume some ammonia accepts a proton from water to give stoichiometric #NH_4^+# and #HO^-#. If we let the amount of protonation be #x#, then we write:

#K_b=1.8xx10^-5=(x^2)/(0.157-x)#

This is a quadratic in #x#, which we could solve exactly, but we make that approximation that #0.157-x~=0.157#, and thus........

#x_1~=sqrt(1.8xx10^-5xx0.157)=1.68xx10^-3#

This is indeed small compared to #0.157#, but we could plug it in again, now that we have an approx. for #x#.

#x_2=1.67xx10^-3#, and since the values have converged, I am willing to accept this value for our answer.

Now #x=[NH_4^+]=[HO^-]# by our definition. What is the #pH# of this solution?

I hope this agrees with your model answer............If you want something re-explained or rephrased, ax, and someone will help you.