Question

To $1.0L$ of $0.01M$ $N{H}_3$ solution, the following quantities of $1.0M$ $HCl$ is added. ($p{K}_b\left(N{H}_3\right)=4.74$)
List-I contains questions and List-II contains their answers.
$\begin{array}{cc}\text{List-I}& \text{List-II}\\ \text{Volume of HCl added}& \text{pOH}\\ \text{(I) 2.0 mL}& \text{(P) 11.0}\\ \text{(II) 5.0 mL}& \text{(Q) 4.14}\\ \text{(III) 10.0 mL}& \text{(R) 4.74}\\ \text{(IV) 11.0 mL}& \text{(S) 8.37}\\ & \text{(T) 8.20}\\ & \text{(U) 11.35}\end{array}$

• A. $\left(I\right),\left(S\right)and\left(III\right),\left(P\right)$
• B. $\left(I\right),\left(T\right)and\left(III\right),\left(U\right)$
• C. $\left(I\right),\left(R\right)and\left(III\right),\left(Q\right)$
• D. $\left(I\right),\left(Q\right)and\left(III\right),\left(S\right)$

Answer 1

The correct option is D $\left(I\right),\left(Q\right)and\left(III\right),\left(S\right)$

(I) $1.0Lof0.01MN{H}_3=0.01molofN{H}_3=10mmolofN{H}_3$
$2.0mLof1MHCl=2mmolofHCl$
$\therefore \text{mmol of}N{H}_4^{+}formed=2$
mmol of $N{H}_3$ left =$10-2=8$
Since it is a basic solution
$pOH=p{K}_b+log\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_3\right]}$
$=4.74+log\frac{2}{8}=4.14$

(III) Total volume of the solution, $1L+5mL\approx 1L$
$10mLof1MHCl=10mmolofHCl$
So the concentration of $\left[N{H}_{4}Cl\right]=0.01mol/L$
We know, $pH=\frac{1}{2}(p{K}_w-p{K}_b-logC)$
$=\frac{1}{2}(14-4.74-log{10}^{-2})=5.63$
$pOH=14-5.63=8.37$