Question

If concentration of two weak bases are same and degree of dissociation $\left(\alpha \right)$ are very less, then their relative strength can be compared by

• A. $\frac{[O{H}^{-}{]}_1}{[O{H}^{-}{]}_2}$
• B. $\frac{K_{b_1}}{K_{b_2}}$
• C. $\frac{{\alpha }_1}{{\alpha }_2}$
• D. $\frac{\sqrt{K_{b_1}}}{\sqrt{K_{b_2}}}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\frac{[O{H}^{-}{]}_1}{[O{H}^{-}{]}_2}$
C $\frac{{\alpha }_1}{{\alpha }_2}$
D $\frac{\sqrt{K_{b_1}}}{\sqrt{K_{b_2}}}$

Relative strength between two bases can be compared by the concentration ratio of $[OH{]}^{-}$ of both the bases. Since, it depends on the value of $K_b$. Larger the $K_b$, stronger will be the base.

So, relative strength= $\frac{[O{H}^{-}{]}_1}{[O{H}^1{]}_2}$

Let one base be $MOH$ and the other be $SOH$ where ${}^{\prime }{M}^{\prime }$ and

${}^{\prime }{S}^{\prime }$ both are positive monovalent ions.

Now , $MOH\to M^{+}+O{H}^{-}$
$C_1(1-{\alpha }_1)$ $\left(C_1{\alpha }_1\right)$ $\left(C_1{\alpha }_1\right)$

$K_{b1}=\frac{\left[M^{+}\right]\left[O{H}^{-}\right]}{\left[MOH\right]}=\frac{\left(C_1{\alpha }_1\right)\left(C_1{\alpha }_1\right)}{C_1(1-{\alpha }_1)}$

Since, degree of dissociation is very small, $C_1(1-{\alpha }_1)=C_1$....

(${\alpha }_1$ can be neglected)

$K_{b1}=\frac{C_1^2{\alpha }_1^2}{C_1}=C_1{\alpha }_1^2$

${\alpha }_1=\sqrt{\frac{K_{b1}}{C_1}}$.......(1)

Second case $SOH\to S^{+}+O{H}^{-}$
$C_2(1-{\alpha }_2)$ $\left(C_2{\alpha }_2\right)$ $\left(C_2{\alpha }_2\right)$

$K_{b2}=\frac{\left[S^{+}\right]\left[O{H}^{-}\right]}{\left[SOH\right]}=\frac{\left(C_2{\alpha }_2\right)\left(C_2{\alpha }_2\right)}{C_2(1-{\alpha }_2)}$

Since, degree of dissociation is very small, $C_2(1-{\alpha }_2)=C_2$....(${\alpha }_2$ can be neglected)

$K_{b2}=\frac{C_2^2{\alpha }_2^2}{C_2}=C_2{\alpha }_2^2$

${\alpha }_2=\sqrt{\frac{K_{b2}}{C_2}}$.......(2)

Relative strength =$\frac{[O{H}^{-}{]}_1}{[O{H}^1{]}_2}$

So, from 1 and 2 reaction , we can say that

Relative strength =$\frac{C_1{\alpha }_1}{C_2{\alpha }_2}$......(3)

Put (1) and (2) in (3), we get

Relative strength =$\frac{C_1\sqrt{\frac{K_{b1}}{C_1}}}{C_2\sqrt{\frac{K_{b2}}{C_2}}}$

Since $C_1=C_2$,

Relative strength =$\frac{\sqrt{K_{b1}}}{\sqrt{K_{b2}}}=\frac{{\alpha }_1}{{\alpha }_2}$

Hence, option $A$ and $D$ both are correct answers.