Question

$100 \,ml$ of $0.5 \,M$ hydrazoic acid $\left( HN _{3}, K _{ a }=3.6 \times 10^{-4}\right)$ and $400 \,ml$ of $0.1 M$ cyanic acid $\left( HOCN , K _{ a }=\right.$ $8 \times 10^{-4}$ ) are mixed. which of the following is (are) true for the final solution?

• A. $\left[ H ^{+}\right]=10^{-2}$
• B. $\left[ N _{3}{ }^{-}\right]=3.6 \times 10^{-3}$
• C. $\left[ OCN ^{-}\right]=4.571 \times 10^{-3}$
• D. $\left[ H ^{+}\right]=1.4 \times 10^{-2}$

Answer 1

Answer: (A), (B)

$\left[ H ^{+}\right]=\sqrt{ C _{1} K _{ a _{1}}+ C _{2} K _{ a _{2}}}$
$=\sqrt{10^{-1} \times 3.6 \times 10^{-4}+8 \times 10^{-2} \times 8 \times 10^{-4}}$
$=\sqrt{3.6 \times 10^{-5}+6.4 \times 10^{-5}}$
$=\sqrt{10^{-4}}=10^{-2} M$
$\underset{0.1}{HN _{3}} \rightleftharpoons \underset{10^{-2}}{H ^{+}} + NH _{3}^{-}$
so, $\left[ N _{3}^{-}\right]=3.6 \times 10^{-2} \times \frac{10^{-1}}{10^{-2}}$
$=3.6 \times 10^{-3} M$
$\left[ OCN ^{-}\right]=8 \times 10^{-4} \times \frac{8 \times 10^{-2}}{10^{-2}}$
$=6.4 \times 10^{-3} M$