What is the #pH# of a solution whose concentration is #0.450*mol*L^-1# in ammonia; #K_b=1.80xx10^-5#?

1 Answer
Mar 6, 2017

Answer:

#pH=11.5#

Explanation:

We interrogate the equilibrium:

#NH_3(aq) + H_2O(l) rightleftharpoons NH_4^(+) + HO^-#.

We set up the equilibrium in the usual way:

#K_b=1.80xx10^-5=([NH_4^+][HO^-])/([NH_3(aq)])#

And we call the concentration of ammonia that hydrolyzes #x#.

And thus: #K_b=1.80xx10^-5=x^2/(0.450-x)#

If, we assume that #0.450>x#, then #x_1=sqrt(1.80xx10^-5xx0.450)#

#=2.85xx10^-3#, which is indeed small compared to #0.450#, but since we have an estimate for #x_1#, we can put it thru the washing again, and come up with:

#x_2=sqrt(1.80xx10^-5xx(0.450-2.85xx10^-3))=2.84xx10^-3#

#x_3=sqrt(1.80xx10^-5xx(0.450-2.84xx10^-3))=2.84xx10^-3#

Clearly I am making a meal of the successive approximations, but I think you can get the gist of the operation. Approximate, then substitute, then resubstitute, till you reach a consistent answer. Alternatively, you could have used the quadratic equation, but here there are more terms to include, and more mistakes (potentially) to make.

And thus at equilibrium,

#[NH_4^+]=[HO^-]=2.84xx10^-3*mol*L^-1#

#pOH=-log_10[HO^-]=-log_10(2.84xx10^-3)=2.55#

And, since in aqueous solution, #pH+pOH=14#, #pH=11.45#.

If I were you, I would apply this general method to other #pH# problems. You can get very efficient at their solutions, and you must practise how to manipulate your calculator.