Question

Calculate the change in pH of one litre of buffer solution containing 0.10 mole each of $N{H}_3$ & $N{H}_{4}Cl$ upon addition of, (i) 0.02 mole of dissolved gaseous $HCl$ (ii) 0.02 mole of dissolved $NaOH$. Assume no change in solution volume. $(K_{b}forN{H}_3=1.8\times {10}^{-5})$

• A. $4.57,4.92$
• B. $4.25,4.57$
• C. $4.83,4.64$
• D. $3.25,3.9$

Answer 1

Answer: (C)

$4.83,4.64$

(i) $0.02$ mole of $HCl$ neutralizes $0.02$ moles of $N{H}_3$. So remaining concentration of $N{H}_3$ is $0.08$ moles per litre.

Now, mixture of $N{H}_3$ and $N{H}_{4}Cl$ behaves as buffer.

$pOH=p{K}_b+\log \frac{\left[N{H}_{4}Cl\right]}{\left[N{H}_3\right]}=4.74+\log \frac{0.1}{0.08}=4.83$

$\therefore pH=14-4.83=9.16$ (final $pH$)

(ii) $0.02$ mole of $NaOH$ neutralizes $0.02$ mole of $N{H}_{4}Cl$. So remaining concentration of $N{H}_{4}Cl$ will be $0.08$ moles per litre.

$\therefore pOH=p{K}_b+\log \frac{\left[N{H}_{4}Cl\right]}{\left[N{H}_3\right]}=4.74+\log \frac{0.08}{0.1}=4.64$

$\therefore$ Final $pH=14-4.64=9.35$

Now initial $pH=14-(p{K}_b+\log \frac{\left[N{H}_{4}Cl\right]}{\left[N{H}_3\right]})$

$=14-(4.74+0)=9.26$

So, case (i) change in $pH=9.26-9.16=0.1$

case (ii) change in $pH=9.35-9.28=0.09$