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# 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}

Last updated date: 20th Jun 2024
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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] isC\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.