Question

0.5g of impure ammonium chloride was heated with caustic soda solution to evolve ammonia gas; the gas is absorbed in 150mL of $\dfrac{N}{5}\text{ }{{\text{H}}_{2}}S{{O}_{4}}$ solution. Excess sulphuric acid required 20mL of 1N NaOH for complete neutralisation. The percentage of $N{{H}_{3}}$ in ammonium chloride is:
[A] 68%
[B] 34%
[C] 48%
[D] 17%

Answer 1

Hint: To solve this question using the concept of law of equivalence. Calculate the milliequivalents of the base and the acid and use the law. The law provides us with the ratio of reactant and the products which we can use here to solve this.

Complete step by step solution:
In the question, it is given that ammonium chloride was heated with caustic soda and ammonia gas was evolved. We know that caustic soda is nothing but sodium hydroxide. So we can write the reaction as-
\[N{{H}_{4}}Cl+NaOH\to NaCl+N{{H}_{3}}\uparrow +{{H}_{2}}O\]
The evolved ammonia gas was then absorbed by sulphuric acid. Then the excess sulphuric acid was completely neutralised by sodium hydroxide.
Here we can use the concept of law of equivalence which is the fundamental basis of all titrations, to find out the required percentage of ammonia in ammonium chloride.
According to the law of equivalence, at the endpoint of the titration, the volume of the two titrants reacted has the same number of equivalents or milliequivalents. We can write it as-
Equivalents of solute = normality $\times $ volume (in litre).
Therefore, following this law, for the given reaction we can write that-
Milliequivalents of ammonium salt = milliequivalent of ${{H}_{2}}S{{O}_{4}}$ - milliequivalent of NaOH.
Here, sulphuric acid is in excess therefore we subtract the milliequivalents of sodium hydroxide from it which will give us the milliequivalents of ammonia.
Therefore, we can write that-
Milliequivalent of ammonia = milliequivalent of ${{H}_{2}}S{{O}_{4}}$ - milliequivalent of NaOH.
Now, milliequivalent = normality $\times $ volume (in mL)
So, milliequivalent of ammonia = $\left( \dfrac{1}{5}\times 150 \right)-\left( 1\times 20 \right)=30-20=10$.
Now we have to find the weight of ammonia.
We know that ${{M}_{eq}}=\dfrac{Weight}{Equivalent\text{ weight}}$
Equivalent weight of ammonia = 17 and milliequivalent of ammonia = 10.
Therefore, weight of ammonia = $\dfrac{10}{1000}\times 17=0.17$
So, percentage of ammonia in ammonium chloride is 0.17 $\times $ 100 = 17%

Therefore, the correct answer is option [D] 17%.

Note: In the above question, the concept of back titration in the volumetric analysis is used. We can even use it to find out the percentage purity of a sample. We also use it to find out the excess of reagent added by titrating it with a suitable reagent. In an acid-base titration, if we have added excess base in the acid mixture, we can titrate the base with another acid of known strength to find out the excess base added.