Question

When 0.05 M dimethyl amine is dissolve in 0.1M NaOH solution then the percentage dissociation of dimethylamine is$({{K}_{b}}){{(C{{H}_{2}})}_{2}}NH=5 \times {{10}^{-4}}$
(A) $5 \times {{10}^{-5}}$
(B) $5 \times {{10}^{-3}}$
(C) $5 \times{{10}^{-1}}$
(D) $5 \times{{10}^{-2}}$

Answer 1

Hint: To calculate the percentage dissociation of a base we can divide the mass of the dissociated ions by the total mass of the dissociated and the undissociated species and then multiply it with 100 to convert it into percentage.

Complete step by step solution:
The base dissociation constant or the ionization constant is the measurement of the basic strength of the solution. Weak base is a base which dissociates partially into ions in water or in the aqueous solution. The ionization reaction of either a weak acid or a weak base is irreversible.
Given in the question:
Molarity of dimethyl amine = 0.05M
Molarity of the NaOH solution = 0.1M
The value of $({{K}_{b}}){{(C{{H}_{2}})}_{2}}NH=5 \times {{10}^{-4}}$
Dimethyl amine is a weak base so it dissociates partially so let’s assume that dimethyl amine dissociates $\alpha $after dissolving.
The concentration of the hydroxyl ion after dissociation will be the sum of the hydroxyl ions from NaOH and the sum of the hydroxyl ions by dimethyl amine
\[[O{{H}^{-}}]=0.1+0.05\alpha \]
0.05 is so small that it can be neglected in front of 0.1, so the concentration of the hydroxyl ion will be = 0.1
Now the dissociation constant will be ${{K}_{b}}=\dfrac{[C\alpha ][O{{H}^{-}}]}{[C-C\alpha ]}$
\[\begin{align}
& (5)({{10}^{-4}})=\dfrac{(\alpha )(0.1)}{1-\alpha } \\
& \alpha =(5)({{10}^{-1}}) \\
\end{align}\]

Hence the correct answer is option ©.

Note: Dissociation is basically the separation of ions that occurs when a solid compound dissolves. Strong acids and the strong bases completely dissociates and the value of degree of dissociation for strong acids and strong bases is approximately equal to 1. The ionization reaction of the acids and bases are reversible which means that the products can combine again and form the reactants.