Question

$HCl$ gas is passed through an aqueous solution of $0.1\text{M}$ $1-aminorpopane$ $\left(PrN{H}_2\right)$ till the $pH$ reaches $9.71$. Calculate the ratio of $\left[PrN{H}_2\right]/[PrN{H}_3{]}^{+}$ in this solution:
$\text{Given:}{K}_a\left(PrN{H}_3^{+}\right)=1.96\times {10}^{-11}\log (-9.71)=1.95\times {10}^{-10}$

• A. $0.01$
• B. $10$
• C. $0.1$
• D. ${10}^{-3}$

Answer 1

The correct option is C $0.1$

Given
$pH=9.71\therefore \left[H^{+}\right]=1.95\times {10}^{-10}PrN{H}_2+HCl\rightleftharpoons PrN{H}_3^{+}+C{l}^{-}{K}_a=\frac{\left[PrN{H}_2\right]\left[H^{+}\right]}{\left[PrN{H}_3^{+}\right]}\frac{K_a}{\left[H^{+}\right]}=\frac{\left[PrN{H}_2\right]}{PrN{H}_3^{+}}\Rightarrow \frac{K_a}{\left[H^{+}\right]}=\frac{\left[PrN{H}_2\right]}{\left[PrN{H}_3^{+}\right]}=\frac{1.96\times {10}^{-11}}{1.95\times {10}^{-10}}=0.1$