Question

100 mL of 1N $N{H_4}OH$ ( ${K_b} = 5 \times {10^{ - 5}}$) is neutralized to the equivalence point by 1N HCl. The pH of solution at equivalence point is
A. 2.0
B. 2.5
C. 3.0
D. 5.0

Answer 1

Hint: In this question, the ammonium hydroxide reacts with hydrochloric acid to form ammonium chloride and water. At equivalence point the number of moles of acid and base become equals as the salt is formed.

Complete step by step answer:
It is given that the volume of ammonium hydroxide solution is 100 mL and the normality of the ammonium hydroxide solution is 1N.
The normality of HCl solution is 1N.
Here, a neutralization reaction is taking place. In neutralization reaction, acid reacts with base to form salt by eliminating water.
Ammonium hydroxide is the base which reacts with hydrochloric acid to form ammonium chloride and water.
The equivalence point is defined as the point in titration where the number of moles of acid is equal to the number of moles. At an equivalence point no more titrant is added as the product is formed.
$M.eq\;of\;N{H_4}OH = M.eq\;of\;HCl = M.eq\;of\;N{H_4}Cl$
$ \Rightarrow 1 \times 100 = 1 \times V = M.eq\;of\;N{H_4}OH$
Also, the volume is 100 mL.
$ \Rightarrow \;N{H_4}Cl = \dfrac{{1 \times 100}}{{200}}$
$ \Rightarrow \;N{H_4}Cl = \dfrac{1}{2}$
As ammonium chloride is a salt it rapidly dissolves in water and dissociates into ions.
Due to hydrolysis of ammonium chloride.
Where,
${K_w}$ is the constant used for water.
${K_b}$ is dissociation constant.
C is the concentration.
Substitute the value in the above expression.
$ \Rightarrow [{H^ + }] = \sqrt {\dfrac{{{{10}^{ - 14}} \times 1}}{{2 \times 5 \times {{10}^{ - 5}}}}} $
$ \Rightarrow [{H^ + }] = 5$
The pH of solution at equivalence point when the salt is formed is 5.

Note:
Normally the pH of the salt is 7, but here the ammonium hydroxide is the weak base and hydrochloric acid is a strong acid, therefore the salt ammonium chloride is acidic salt and the pH is below 5.