Question

$pH$ of the solution containing $50.0mL$ of $0.3M$ $HCl$ and $50.0mL$ of $0.4M{NH}_3$ is:

$\left[p{K}_a\right({NH}_4^{+})=9.26]$

• A. $4.74$
• B. $9.26$
• C. $8.78$
• D. $4.63$

Answer 1

The correct option is C $8.78$

$50ml$ of $0.3M$ $HCl$ $+$ $50ml$ of $0.4M$ $N{H}_3$

Moles of $HCl=0.3\times 50=15milli$ $moles$

Moles of $N{H}_3=50\times 0.4=20milli$ $moles$

$N{H}_3+HCl⟶N{H}_{4}Cl$

$1mol$ $1mol$ $1mol$

$15mol$ $15mol$ $15mol$

$N{H}_3$ left unreacted $⟶5milli$ $moles$

Concentration of $N{H}_3=\frac{5}{100}=0.05M$

$p^{OH}=p^{k_b}+\log \frac{\left[salt\right]}{Base]}$ $[\because p^{k_a}+p^{k_b}=14\Rightarrow 9.26+p^{k_b}=14\Rightarrow p^{k_b}=4.74]$

$\Rightarrow p^{OH}=4.74+\log \frac{\left(0.15\right)}{\left(0.05\right)}$

$\Rightarrow p^{OH}=4.74+\log 3=5.22$

$\Rightarrow p^{OH}=5.22$

$\Rightarrow p^H+p^{OH}=14$

$\Rightarrow p^H=14-p^{OH}=14-5.22=8.78$