Question

A quantity of ammonium chloride was heated with 100 mL of 0.8 N NaOH solution till the reaction was complete. The excess of NaOH was neutralized with 12.5 mL of 0.75 N $H_{2}S{O}_4$. Calculate the quantity of ammonium chloride.

Answer 1

Hint:The reaction where an acid compound reacts with the base compound, the resulting salt is formed with the elimination of water. This reaction is known as a neutralization reaction. In a neutralization reaction, the same amount of base will neutralize the same amount of acid.

Complete step by step answer:
Given
The volume of NaOH is 100 mL.
The normality of NaOH is 0.8 N.
The volume of $H_{2}S{O}_4$ is 12.5 mL.
The normality of $H_{2}S{O}_4$ is 0.75 N.
12.5 mL of 0.8 N NaOH neutralizes 12.5 mL of 0.75 N $H_{2}S{O}_4$.
Let us assume that the volume of excess 0.8 N NaOH after reacting with ammonium chloride be V mL.
Thus,
V mL of 0.8 N NaOH will neutralize 12.5 mL of 0.75 N $H_{2}S{O}_4$.
$VmLof0.8NaOH\cong 12.5mLof0.75N{H}_{2}S{O}_4$
To calculate the volume of NaOH, the formula used is shown below.
$N_1{V}_1=N_2{V}_2$
Substitute, the values in above equation
$V\times 0.8=12.5\times 0.75$
$\Rightarrow V\times 0.8=9.375$
$\Rightarrow V={\displaystyle \frac{9.375}{0.8}}$
$\Rightarrow V=11.71$
Volume of 0.8 N NaOH consumed by $N{H}_{4}Cl$$=100-11.71$
Volume of 0.8 N NaOH consumed by $N{H}_{4}Cl$=88.29
The mass equivalent of NaOH consumed by $N{H}_{4}Cl$= mass equivalent of $N{H}_{4}Cl$ used
The mass of ammonium chloride is calculated by the formula as shown below.
$m={\displaystyle \frac{E\times N\times V}{1000}}$
Where
*m is the mass
*E is the equivalent weight
*N is the normality
To calculate the mass of ammonium chloride, substitute the value in the above equation.
$m={\displaystyle \frac{53.5\times 0.8\times 88.28}{1000}}$
$\Rightarrow m=3.779g$
Therefore, the mass of ammonium chloride is 3.779 g.


Note:
The normality of the solution is defined as the amount of gram equivalent of solute dissolved in one liter solution. The number of gram equivalents is calculated as shown below.
$No.ofg.eq=massofsolute\times E.Wofsolute$