Question

When $0.100mole$ of ammonia ($N{H_3}$), is dissolved in sufficient water to make $1.0L$ of solution, the solution is found to have a hydroxide ion concentration of $1.34 \times {10^{ - 3}}M$. Calculate ${K_b}$ for ammonia.

Answer 1

Hint: The ${K_b}$ is known as the dissociation constant of base. The dissociation constant for a reaction can be given as the ratio of the multiplication of concentration of products to the multiplication of concentration of reactants.

Complete step by step answer:
As we know that the strong acids or bases dissociate or break apart completely but the weak acids or weak bases like ammonia do not dissociate or break down into their corresponding ions. The dissociation constants are used to measure how well an acid or base dissociates.
The ${K_b}$ is the dissociation constant of base which is used to measure how well a base dissociates in the solution. The ${K_b}$ for the dissociation reaction of a base is given as the ratio of product of concentration of substance formed to that of the concentration of reacting product.
For the given question, when ammonia reacts with water then this reaction can be represented as:
$N{H_3} + {H_2}O \rightleftharpoons NH_4^ + + O{H^ - }$
For the given reaction, the dissociation constant (${K_b}$) can be written as:
${K_b} = \dfrac{{[NH_4^ + ][O{H^ - }]}}{{[N{H_3}]}}$ $ - (1)$
Remember that here the concentration of water is not taken in the formula because the concentration of water is absorbed into the value of ${K_b}$.
As we know that initially the concentration of ammonia ($N{H_3}$) is $0.100mole$ and concentration of the reactants will be zero. After some time the hydroxide ion concentration of hydroxide ion is $1.34 \times {10^{ - 3}}M$. Therefore, the concentration of ammonia ($N{H_3}$) is $0.100 - 1.34 \times {10^{ - 3}}$ and the concentration of $NH_4^ + $ will be same as that of hydroxide ion because when one mole hydroxide ion produce then the one mole of $NH_4^ + $ is also produced.
Hence, it can be written as:
\[
  {\text{ }}N{H_3}{\text{ }} + {\text{ }}{H_2}O \rightleftharpoons {\text{ }}NH_4^ + {\text{ }} + {\text{ }}O{H^ - } \\
  t = 0,{\text{ }} 0.1{\text{ }} 0{\text{ }} 0 \\
  t = t{\text{ }} 0.1 - 1.34 \times {10^{ - 3}}{\text{ }} 1.34 \times {10^{ - 3}}{\text{ }}1.34 \times {10^{ - 3}} \\
 \]
Now, we can see that :
$[N{H_3}] = 0.1 - 1.34 \times {10^{ - 3}}M$
$[NH_4^ + ] = 1.34 \times {10^{ - 3}}M$
$[O{H^ - }] = 1.34 \times {10^{ - 3}}M$
Now, the equation $ - (1)$ can be written as:
$
  {K_b} = \dfrac{{[NH_4^ + ][O{H^ - }]}}{{[N{H_3}]}} \\
  {K_b} = \dfrac{{[1.34 \times {{10}^{ - 3}}][1.34 \times {{10}^{ - 3}}]}}{{[0.1 - 1.34 \times {{10}^{ - 3}}]}} \\
  {K_b} = \dfrac{{1.7956 \times {{10}^{ - 6}}}}{{0.098}} \\
  {K_b} = 1.82 \times {10^{ - 5}} \\
 $

Hence, the value of required dissociation constant ${K_b}$ is $1.82 \times {10^{ - 5}}$.

Note:
Always remember that while calculating dissociation constant for acid (${K_a}$) or dissociation constant for the base (${K_b}$), we will take the concentration of water in the reactant as constant and the concentration term of water $[{H_2}O]$ is not included in their formulas.