1. Tardigrade
  2. Question
  3. Chemistry
  4. The value of [ OH -]in a solution made by dissolving 0.005 mol each of ammonia and pyridine ( C 5 H 5 N ) in enough water to make 200 cm 3 of a solution is (K b ( NH 4 OH )=1.8 × 10-5. and .K b ( C 5 H 5 N )=1.52 × 10-9)

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Q. The value of $\left[ OH ^{-}\right]$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 $200 \,cm ^{3}$ of a solution is
$\left(K_{ b }\left( NH _{4} OH \right)=1.8 \times 10^{-5}\right.$ and $\left.K_{ b }\left( C _{5} H _{5} N \right)=1.52 \times 10^{-9}\right)$

Equilibrium

Solution:

The two equilibria existing in the same solution are
$NH _{3}+ H _{2} O \rightleftharpoons NH _{4}^{+}+ OH ^{-}$
$K_{ b }=\frac{\left[ NH _{4}^{+}\right]\left[ OH ^{-}\right]}{\left[ NH _{3}\right]}=1.8 \times 10^{-9}$
And, $C _{5} H _{5} N + H _{2} O \rightleftharpoons C _{5} H _{5} NH ^{+}+ OH ^{-}$
$K_{ b }=\frac{\left[ C _{5} H _{5} NH ^{+}\right]\left[ OH ^{-}\right]}{\left[ C _{5} H _{5} N \right]}=1.52 \times 10^{-9}$
Considering the first reaction, the concentration of ammonia is
$=\frac{0.005}{0.200 L }=0.025\, M$
Therefore, $K_{ b }=1.8 \times 10^{-5}=\frac{x^{2}}{0.025}$
or $\times=6.7 \times 10^{-4} M =\left[ NH _{4}^{+}\right]$
$=[ OH ^-]$
Now, since the dissociation constant of pyridine is much less than that of ammonia, the $OH ^{-}$obtained from pyridine will be negligible as compared to those obtained from $NH _{4} OH$. Therefore, we need not consider the second equilibrium and report the value as $6.7 \times 10^{-4}\, M$.