A chemistry problem by Suneetha Poluri
Chemistry
Level
2
A 25ml of solution of conjugate base of nitric acid was treated with excess of Al. The ammonia gas os passed into 50ml of 0.15N HCl. Excess unreacted acid HCl is back titrated with 32.1ml 0.1N NaOH Solution calculate molarity of the nitric acid's conjugate base in the original solution.
The answer is 0.172.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Firstly the ammonia gas released in the first reaction will have the same no. of equivalents as that of the conjugate base of HNO3.
Secondly, for the simple acid base titration we can write up
Milliequivalents of acid (HCl) = Milliequivalent of NaOH (base) + Milliequivalent of NH3 (base)
which equals
50 x 0.15 = 32.1 x 0.1 + 25 x N
giving N = 0.1716
The molarity is the same as n-factor of base is 1.