Question

The relative strength of two weak bases at same concentration may be given as
$\begin{align}
& \text{A}\text{. }\dfrac{{{K}_{{{b}_{1}}}}}{{{K}_{{{b}_{2}}}}} \\
& \text{B}\text{. }\dfrac{{{\left[ O{{H}^{-}} \right]}_{1}}}{{{\left[ O{{H}^{-}} \right]}_{2}}} \\
& \text{C}\text{. }\dfrac{{{\alpha }_{1}}}{{{\alpha }_{2}}} \\
& \text{D}\text{. }\dfrac{\sqrt{{{K}_{{{b}_{1}}}}}}{\sqrt{{{K}_{{{b}_{2}}}}}} \\
\end{align}$

Answer 1

Hint: The relative strengths of acids and bases is assessed as per their dissociation or ionization constants. The ionization constant is the tendency of a substance to separate (or dissociate) in solution into smaller components or parts.

Complete step by step solution:
The Relative Strength of a Base is the extent up to which it can ionise when it is dissolved in water. If the dissociation is complete, the base is termed as ‘strong’; if relatively less dissociation occurs, the base is ‘weak’.

* Let us derive the formula for the relative strength of the bases:
      $\text{BOH}$$\rightleftharpoons$ ${{B}^{+}}+O{{H}^{-}}$
 So, the dissociation constant of the base BOH will be:
  \[{{K}_{b}}=\dfrac{\left[ {{B}^{+}} \right]\left[ O{{H}^{-}} \right]}{\left[ BOH \right]}\]
 Let C moles per litre is the concentration of the base and $\alpha $ its degree of dissociation. Then,
\[\begin{align}
  & \left[ O{{H}^{-}} \right]=C\alpha \\
 & \left[ {{B}^{+}} \right]=C\alpha \\
 & \left[ BOH \right]=C(1-\alpha ) \\
\end{align}\]
Substitute the values in the equilibrium expression, we have
${{K}_{b}}=\dfrac{C\alpha \times C\alpha }{C(1-\alpha )}$
${{K}_{b}}$ can be resolved into
${{K}_{b}}=\dfrac{C{{\alpha }^{2}}}{(1-\alpha )}$
For weak bases, the degree of dissociation is very low so it will not dissociate completely, thus,$1-\alpha \simeq 1$. Therefore, ${{K}_{b}}=C{{\alpha }^{2}}$
For two different bases, 1 and 2, let the degree of dissociation be ${{\alpha }_{1}}$and${{\alpha }_{2}}$; and the dissociation constants ${{K}_{1}}$ and${{K}_{2}}$. Then,
Base 1, ${{K}_{1}}=C{{\alpha }_{1}}^{2}$ …… (i)
Base 2, ${{K}_{2}}=C{{\alpha }_{2}}^{2}$ …… (ii)
Divide equation (i) by (ii), we get,
  $\dfrac{{{\alpha }_{1}}}{{{\alpha }_{2}}}=\sqrt{\dfrac{{{K}_{1}}}{{{K}_{2}}}}$
Since$\left[ O{{H}^{-}} \right]$is the measure of the basic strength and it depends on the degree of dissociation$\alpha $, we can write,
 $\dfrac{S\text{trength of base 1}}{\text{Strength of base 2}}=\sqrt{\dfrac{{{K}_{1}}}{{{K}_{2}}}}$
The relative strength of bases can be written in other different forms:
(1) $\dfrac{S\text{trength of base 1}}{\text{Strength of base 2}}=\sqrt{\dfrac{{{K}_{1}}}{{{K}_{2}}}}$
The relative strength of a base is the magnitude of its base-dissociation constant (${{K}_{b}}$) in aqueous solutions. In the aqueous solutions of same concentration, stronger bases ionise to a greater extent, which gives higher hydroxide ion concentrations than weak bases.
(2) We have above derived, $\dfrac{{{\alpha }_{1}}}{{{\alpha }_{2}}}=\sqrt{\dfrac{{{K}_{1}}}{{{K}_{2}}}}$ …..(i)
  and the $\dfrac{S\text{trength of base 1}}{\text{Strength of base 2}}=\sqrt{\dfrac{{{K}_{1}}}{{{K}_{2}}}}$ …..(ii) by the equation (i) and (ii),
  we can write,
 \[\dfrac{\text{Strength of base 1}}{\text{Strength of base 2}}=\] $\dfrac{{{\alpha }_{1}}}{{{\alpha }_{2}}}$
(3) The concentration of $\left[ O{{H}^{-}} \right]$ is$C\alpha $. So,${{\left[ O{{H}^{-}} \right]}_{1}}=C{{\alpha }_{1}}$and ${{\left[ O{{H}^{-}} \right]}_{2}}=C{{\alpha }_{2}}$, we obtain the equation, $\dfrac{{{\left[ O{{H}^{-}} \right]}_{1}}}{{{\left[ O{{H}^{-}} \right]}_{2}}}=\dfrac{C{{\alpha }_{1}}}{C{{\alpha }_{2}}}$. The relation in strength of the bases will be now,
 \[\dfrac{\text{Strength of base 1}}{\text{Strength of base 2}}=\]$\dfrac{{{\left[ O{{H}^{-}} \right]}_{1}}}{{{\left[ O{{H}^{-}} \right]}_{2}}}$
The final answer of the question is,
* \[\dfrac{\text{Strength of base 1}}{\text{Strength of base 2}}=\]$\dfrac{{{\left[ O{{H}^{-}} \right]}_{1}}}{{{\left[ O{{H}^{-}} \right]}_{2}}}$, which is option B.
* \[\dfrac{\text{Strength of base 1}}{\text{Strength of base 2}}=\]$\dfrac{{{\alpha }_{1}}}{{{\alpha }_{2}}}$, which is option C.
* $\dfrac{S\text{trength of base 1}}{\text{Strength of base 2}}=\sqrt{\dfrac{{{K}_{1}}}{{{K}_{2}}}}$, which is option D.
Additional Information: The derivation and formula for the acids of the relative strength is the same as that of the bases.
* $\dfrac{\text{Strength of acid 1}}{\text{Strength of acid 2}}=$$\dfrac{{{\alpha }_{1}}}{{{\alpha }_{2}}}$
* $\dfrac{\text{Strength of acid 1}}{\text{Strength of acid 2}}=$$\dfrac{{{\left[ {{H}^{+}} \right]}_{1}}}{{{\left[ {{H}^{+}} \right]}_{2}}}$
* $\dfrac{\text{Strength of acid 1}}{\text{Strength of acid 2}}=$$\sqrt{\dfrac{{{K}_{1}}}{{{K}_{2}}}}$


Note: There is a difference between strength and relative strength of bases. Strength is defined for a base. But relative strength is the comparison between two bases on the basis of their individual basic strength.