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MHTCET 22 Apr 2025 Shift 2 Paper

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Question : 70 of 150

Marks: +1, -0

A weak monoacidic base dissociates to 1.5\% in 0.001M solution at 298K . Calculate the dissociation constant of weak base.

$2.25 \times 10^{-7}$
$3.05 \times 10^{-7}$
$2.5 \times 10^{-5}$
$3.725 \times 10^{-6}$
We are tasked to calculate the dissociation constant $K_{b}$ of a weak monoacidic base.
Step 1: Information given
\item[-] Base concentration: $C=0.001 \mathrm{~N}$
\item[-] Degree of dissociation: $\alpha=1.5 \%=0.015$
Step 2: Expression for base dissociation constant
For a weak base:
$B+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{BH}^{+}+\mathrm{OH}^{-}$
The dissociation constant $\left(K_{b}\right)$ is:
$K_{b}=\frac{\left[B H^{+}\right]\left[O H^{-}\right]}{[B]}$
With degree of dissociation:
\item[-] Initial concentration of base: $C$
\item[-] At equilibrium:
\item[-] $\left[O H^{-}\right]=C \alpha$
\item[-] $\left[B H^{+}\right]=C \alpha$
\item[-] $[B]=C(1-\alpha) \approx C$ (since $\alpha \ll 1$ )
Thus:
$K_{b} \approx \frac{C \alpha^{2}}{1}=C \alpha^{2}$
Step 3: Substitution
$\begin{aligned} K_{b} & =0.001 \times(0.015)^{2} \\ K_{b} & =0.001 \times 0.000225 \\ K_{b} & =2.25 \times 10^{-7} \end{aligned}$
✓ Final Answer:
$2.25 \times 10^{-7} \quad \text { (Option A) }$

Solution:

Concept:

For a weak monoacidic base,

K_b = C\alpha^2

when

\alpha \ll 1

.

Explanation:

Given concentration,

C = 0.001\ \mathrm{M}

and degree of dissociation,

\alpha = 1.5\% = 0.015

.

For the equilibrium of a weak base,

K_b = \frac{[BH^+][OH^-]}{[B]} \approx C\alpha^2

Substituting the values:

K_b = 0.001 \times (0.015)^2

K_b = 0.001 \times 0.000225

K_b = 2.25 \times 10^{-7}

Answer:

2.25 \times 10^{-7}

Option A

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