Question

The dissociation constant of $N{H_4}OH$ is $1.8 \times {10^{ - 5}}$, The concentration of $^ - OH$ ions, in $mol{L^{ - 1}}$ of $0 \cdot 1{\text{ M}}$ $N{H_4}OH$ is:
A. $1.8 \times {10^{ - 6}}$
B. $1.34 \times {10^{ - 3}}$
C.$4.20 \times {10^{ - 2}}$
D. $5.0 \times {10^{ - 2}}$

Answer 1

Hint: The given compound is basic in nature which shows it will dissociate the $^ - OH$ ions in the solution. The values of dissociation constant and the values of molarity are given. One can relate this information to find out the concentration of ions by using the relevant formula.

Complete step by step answer:
- First of all we will learn about the concept of dissociation constant which is a type of equilibrium constant which measures the capacity of a larger object to dissociate into smaller components.
- Now let's analyze the given values in the question which gives the values of dissociation constant and molarity of ammonium hydroxide which can be put into the formula of concentration of $^ - OH$ ions. The formula is as follows,
${\text{Concentration of}}{{\text{ }}^ - }{\text{OH ions = }}{{\text{C}}_a} = \sqrt {{K_b} \times C} = \sqrt {1.8 \times {{10}^{ - 5}} \times 0.1} = 1.34 \times {10^{ - 3}}mol{L^{ - 1}}$
Where ${K_b}$ value is for dissociation constant and $C$ stands for concentration in a molar.
Hence, The concentration of $^ - OH$ ions $0 \cdot 1{\text{ M}}$ $N{H_4}OH$ is $1.34 \times {10^{ - 3}}mol{L^{ - 1}}$ which shows option B as the correct choice of answer.

Note: In the dissociation process a complex breaks apart into its component molecules and salt breaks apart into its component ions. The dissociation reaction is a reversible type of reaction as the salt breaks into its ion components they again associate to form the complex again. For some types of reactions, the value of the dissociation constant can be called as an ionization constant. The ion concentration in a reaction is totally different from the molar concentration of that reaction. The ionic concentration depends on the value of the dissociation constant that’s why we multiply the molar value with a dissociation constant.