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. $\left(II\right),\left(P\right)and\left(IV\right),\left(S\right)$
• B. $\left(II\right),\left(R\right)and\left(IV\right),\left(P\right)$
• C. $\left(II\right),\left(Q\right)and\left(IV\right),\left(T\right)$
• D. $\left(II\right),\left(U\right)and\left(IV\right),\left(P\right)$

Answer 1

The correct option is B $\left(II\right),\left(R\right)and\left(IV\right),\left(P\right)$

II) $1.0Lof0.01MN{H}_3=0.01molofN{H}_3=10mmolofN{H}_3$
$5.0mLof1MHCl=5mmolofHCl$
$\therefore \text{mmol of}N{H}_4^{+}formed=5$
$\therefore pOH=p{K}_b+log\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_3\right]}$
$=4.74+log\frac{5}{5}=4.74$

(IV) $11.0mLof1MHCl=11mmolofHCl$
So when 11.0 mL of HCl is added, all the $N{H}_3$ is neutralised and 1.0 mmol of $HCl$ will be excess.
$\therefore \left[H^{+}\right]=\frac{1\times {10}^{-3}mol}{1L}={10}^{-3}mol/L$
$\therefore pH=-log{10}^{-3}=3$
$\Rightarrow pOH=14-3=11$