Question

Calculate the concentration of $N{H}_{3}andN{H}_4^{+}C{l}^{-}$ present in the buffer solution of pH = 9, when total concentration of buffering reagents is 0.6 $mol{L}^{-1}.$ Take $p{K}_b$ for $N{H}_3=4.7,log2=0.3$

• A. $\left[N{H}_4^{+}C{l}^{-}\right]=0.1M\text{and}\left[N{H}_3\right]=0.3M$
• B. $\left[N{H}_4^{+}C{l}^{-}\right]=0.6M\text{and}\left[N{H}_3\right]=0.6M$
• C. $\left[N{H}_4^{+}C{l}^{-}\right]=0.8M\text{and}\left[N{H}_3\right]=0.2M$
• D. $\left[N{H}_4^{+}C{l}^{-}\right]=0.4M\text{and}\left[N{H}_3\right]=0.2M$

Answer 1

The correct option is D $\left[N{H}_4^{+}C{l}^{-}\right]=0.4M\text{and}\left[N{H}_3\right]=0.2M$

Given, $p{K}_{b}ofN{H}_3=4.7,pH=9$
total concentration of buffer = 0.6 $mol{L}^{-1}.$
Let $\left[N{H}_3\right]=b,\left[N{H}_4^{+}C{l}^{-}\right]=a$
$\therefore \left[N{H}_3\right]+\left[N{H}_4^{+}C{l}^{-}\right]=a+b$
Using the Henderson-Hasselbalch equation:
$pOH=p{K}_b+log\left[\frac{\left[\text{conjugate acid}\right]}{\left[\text{base}\right]}\right]$
$pOH=p{K}_b+log\frac{\left[N{H}_4^{+}C{l}^{-}\right]}{\left[N{H}_3\right]}$$\Rightarrow 14-pH=4.7+log\frac{a}{b}$
$\Rightarrow 14-pH=4.7+log\frac{a}{b}$
$\Rightarrow 14-9=4.7+log\frac{a}{b}$
$\Rightarrow 14-9=4.7+log\frac{a}{b}$
$\Rightarrow 5=4.7+log\frac{a}{b}$
$\Rightarrow log\frac{a}{b}=0.3$
$\Rightarrow \frac{a}{b}=2\therefore a=2b$
Given $a+b=0.6;2b+b=0.6$
$\therefore 3b=0.6orb=0.2mol$ and $a+b=0.6mol$
$a+0.2=0.6\Rightarrow a=0.4mol$
Thus, $\left[N{H}_4^{+}C{l}^{-}\right]=0.4M\text{and}\left[N{H}_3\right]=0.2M$