What is the pH of a solution whose concentration is 0.450⋅mol⋅L−1 in ammonia; Kb=1.80×10−5?

1 Answer
anor277
Mar 6, 2017

pH=11.5

Explanation:

We interrogate the equilibrium:

NH3(aq)+H2O(l)⇌NH+4+HO−.

We set up the equilibrium in the usual way:

Kb=1.80×10−5=[NH+4][HO−][NH3(aq)]

And we call the concentration of ammonia that hydrolyzes x.

And thus: Kb=1.80×10−5=x20.450−x

If, we assume that 0.450>x, then x1=√1.80×10−5×0.450

=2.85×10−3, which is indeed small compared to 0.450, but since we have an estimate for x1, we can put it thru the washing again, and come up with:

x2=√1.80×10−5×(0.450−2.85×10−3)=2.84×10−3

x3=√1.80×10−5×(0.450−2.84×10−3)=2.84×10−3

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−]=2.84×10−3⋅mol⋅L−1

pOH=−log10[HO−]=−log10(2.84×10−3)=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.

Related questions
See all questions in pH calculations
Impact of this question
5561 views around the world
You can reuse this answer
Creative Commons License