Question

$0.2\,liters$ of ammonia at ${27^{\circ}}C$ and $2\;atm$ pressure is neutralized by $160\;ml$ of sulphuric acid. Find the normality of sulphuric acid.
A. 0.01 N
B. 0.2 N
C. 2 N
D. 0.1 N

Answer 1

When we gradually add a solution of unknown concentration and volume with another solution of unknown concentration until the reaction reaches its neutralization, we call this process titration. This can be used here to find the normality of sulphuric acid.

Complete step by step solution:
In question, we are given that
Volume of ammonia $= 0.2\,liters$
Temperature of ammonia = ${27^{\circ}}C\; = 273 + 27 = 300\,K$
Pressure of ammonia $= 2\;atm$
And now we can find the number of moles present in it using the Ideal gas equation.
i.e., $PV = nRT$
Where
$P$ is the pressure
$V$ is the volume
$n$ is the number of moles present
$R$ is the universal gas constant
$T$ is the temperature
Now to find number of moles
$n = \dfrac{{PV}}{{RT}}$
Substituting the values, $R\; = \;0.0821\;Lat{m^{ - 1}}{K^{ - 1}}mo{l^{ - 1}}$
$n = \dfrac{{2 \times 0.2}}{{0.0821 \times 300}}$
$\Rightarrow n = 0.0162$
Now, we have the number of moles of ammonia present.
By the equation in titration
${N_1}{V_1} = {N_2}{V_2}$
Where ${{\text{N}}_1}$ and ${{\text{N}}_2}$ are the normalities and ${{\text{V}}_1}$ and ${V_2}$ are its volumes.
Here ${{\text{N}}_1}$ is ammonia and ${{\text{N}}_2}$ is sulphuric acid.
So we know that ammonia has only capacity to take one ${H^ + }$ ion, therefore equivalents of ammonia equals to its moles.
i.e.,$\dfrac{{0.0162}}{{{V_1}}} \times {V_1} = {N_2}{V_2}$
The volume of sulphuric acid is given as $160\,ml\; = \;0.160\,L$
$\Rightarrow {N_2} = \dfrac{{0.0162}}{{0.160}} = 0.1\,N$

**Therefore, the normality of sulphuric acid is equal to 0.1 N i.e., option (d) is correct

Note:**
While applying the value of the universal gas constant, R we should check the units of the remaining parameters. And most appropriate has to be used. Here we used the value 0.0821, but the temperature was given in degree Celsius therefore, we had to change that in Kelvin.