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

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 ${H_2}S{O_4}$.
$V\;mL\;of\;0.8\;NaOH \cong 12.5mL\;of\;0.75\;N\;{H_2}S{O_4}$
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
$V \times 0.8 = 12.5 \times 0.75$
$\Rightarrow V \times 0.8 = 9.375$
$\Rightarrow V = \dfrac{{9.375}}{{0.8}}$
$\Rightarrow V = 11.71$
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.
$m = \dfrac{{E \times N \times V}}{{1000}}$
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.
$m = \dfrac{{53.5 \times 0.8 \times 88.28}}{{1000}}$
$\Rightarrow m = 3.779\;g$
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.
$No.\;of\;g.eq = mass\;of\;solute \times E.W\;of\;solute$