Question

The value of $\left[O{H}^{-}\right]$in a solution made by dissolving $0.005$ mol each of ammonia and pyridine $\left(C_5{H}_{5}N\right)$ in enough water to make $200c{m}^3$ of a solution is
$(K_b\left(N{H}_{4}OH\right)=1.8\times 10^{-5}$ and $K_b\left(C_5{H}_{5}N\right)=1.52\times 10^{-9})$

• A. $5.7\times 10^{-8}M$
• B. $7\times 10^{-4}M$
• C. $3.32\times 10^{-9}M$
• D. $5.7\times 10^{-9}M$

Answer 1

Answer: (B)

The two equilibria existing in the same solution are
$N{H}_3+H_{2}O\rightleftharpoons N{H}_4^{+}+O{H}^{-}$
$K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_3\right]}=1.8\times 10^{-9}$
And, $C_5{H}_{5}N+H_{2}O\rightleftharpoons C_5{H}_{5}N{H}^{+}+O{H}^{-}$
$K_b=\frac{\left[C_5{H}_{5}N{H}^{+}\right]\left[O{H}^{-}\right]}{\left[C_5{H}_{5}N\right]}=1.52\times 10^{-9}$
Considering the first reaction, the concentration of ammonia is
$=\frac{0.005}{0.200L}=0.025M$
Therefore, $K_b=1.8\times 10^{-5}=\frac{x^2}{0.025}$
or $\times =6.7\times 10^{-4}M=\left[N{H}_4^{+}\right]$
$=[O{H}^{-}]$
Now, since the dissociation constant of pyridine is much less than that of ammonia, the $O{H}^{-}$obtained from pyridine will be negligible as compared to those obtained from $N{H}_{4}OH$. Therefore, we need not consider the second equilibrium and report the value as $6.7\times 10^{-4}M$.