Question

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

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,

& \left[ O{{H}^{-}} \right]=C\alpha \\\ & \left[ {{B}^{+}} \right]=C\alpha \\\ & \left[ BOH \right]=C(1-\alpha ) \\\ \end{aligned}$$ Substitute the values in the equilibrium expression, we have ${{K}_{b}}=\dfrac{C\alpha \times C\alpha }{C(1-\alpha )}$ ${{K}_{b}}$ can be resolved into ${{K}_{b}}=\dfrac{C{{\alpha }^{2}}}{(1-\alpha )}$ For weak bases, the degree of dissociation is very low so it will not dissociate completely, thus,$1-\alpha \simeq 1$. Therefore, ${{K}_{b}}=C{{\alpha }^{2}}$ For two different bases, 1 and 2, let the degree of dissociation be ${{\alpha }_{1}}$and${{\alpha }_{2}}$; and the dissociation constants ${{K}_{1}}$ and${{K}_{2}}$. Then, Base 1, ${{K}_{1}}=C{{\alpha }_{1}}^{2}$ …… (i) Base 2, ${{K}_{2}}=C{{\alpha }_{2}}^{2}$ …… (ii) Divide equation (i) by (ii), we get, $\dfrac{{{\alpha }_{1}}}{{{\alpha }_{2}}}=\sqrt{\dfrac{{{K}_{1}}}{{{K}_{2}}}}$ Since$\left[ O{{H}^{-}} \right]$is the measure of the basic strength and it depends on the degree of dissociation$\alpha $, we can write, $\dfrac{S\text{trength of base 1}}{\text{Strength of base 2}}=\sqrt{\dfrac{{{K}_{1}}}{{{K}_{2}}}}$ The relative strength of bases can be written in other different forms: (1) $\dfrac{S\text{trength of base 1}}{\text{Strength of base 2}}=\sqrt{\dfrac{{{K}_{1}}}{{{K}_{2}}}}$ The relative strength of a base is the magnitude of its base-dissociation constant (${{K}_{b}}$) in aqueous solutions. In the aqueous solutions of same concentration, stronger bases ionise to a greater extent, which gives higher hydroxide ion concentrations than weak bases. (2) We have above derived, $\dfrac{{{\alpha }_{1}}}{{{\alpha }_{2}}}=\sqrt{\dfrac{{{K}_{1}}}{{{K}_{2}}}}$ …..(i) and the $\dfrac{S\text{trength of base 1}}{\text{Strength of base 2}}=\sqrt{\dfrac{{{K}_{1}}}{{{K}_{2}}}}$ …..(ii) by the equation (i) and (ii), we can write, $$\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.