Question

The equilibrium constant for the ionization of $RN{H}_2\left(g\right)$ in water :

$RN{H}_2\left(g\right)+H_{2}O\left(l\right)\rightleftharpoons RN{H}_3^{+}\left(aq\right)+OH\left(aq\right)$

is ${10}^{-6}$ at ${25}^{o}C$. Select the correct statement(s) :

• A. pH of solution is 11 at $P_{RN{H}_2\left(g\right)}=1bar$
• B. Forward reaction is favoured by addition of HCl(aq)
• C. Forward reaction is favoured by addition of $H_{2}O\left(l\right)$
• D. none

Answer 1

Answer: (A), (B), (C)

pH of solution is 11 at $P_{RN{H}_2\left(g\right)}=1bar$
Forward reaction is favoured by addition of HCl(aq)
Forward reaction is favoured by addition of $H_{2}O\left(l\right)$

$K=\frac{\left[RN{H}_3^{+}\right]\left[O{H}^{-}\right]}{P_{RNH2}}$

$[RNH3+]=\left[O{H}^{-}\right]$

${10}^{-6}=\frac{\left[O{H}^{-}\right]\left[O{H}^{-}\right]}{1}$

$\left[O{H}^{-}\right]={10}^{-3}$ M

$\left[H^{+}\right]=\frac{Kw}{\left[O{H}^{-}\right]}$

$\left[H^{+}\right]=\frac{{10}^{-14}}{{10}^{-3}}={10}^{-11}$

$pH=-log\left[H^{+}\right]=-log{10}^{-11}=11$

When either HCl or water is added, the equilibrium will shift in the forward reaction. Added HCl will provide protons which will react with amine to form alkyl ammonium ion.

$RN{H}_2+H^{+}\to RN{H}_3^{+}$

Similarly, added water will react with more amount of alkyl amine to form products.

Hence, option A, B and C are correct.