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Question: Specific conductance of pure water at 25\(^{o}C\) is \(0.58\times {{10}^{-7}}mhoc{{m}^{-1}}\) . The ...

Specific conductance of pure water at 25oC^{o}C is 0.58×10−7mhocm−10.58\times {{10}^{-7}}mhoc{{m}^{-1}} . The ionic product of water (KW{{K}_{W}} ) if ionic conductances of H+{{H}^{+}} and OH−O{{H}^{-}} ions at infinite dilution are 350 and 198 mho cm2mho\text{ }c{{m}^{2}} respectively at 25oC^{o}C is:
A. 1×10−14(mole/litre)21\times {{10}^{-14}}{{(mole/litre)}^{2}}
B. 1×10−13(mole/litre)21\times {{10}^{-13}}{{(mole/litre)}^{2}}
C. 1×10−12(mole/litre)21\times {{10}^{-12}}{{(mole/litre)}^{2}}
D. None of these

Explanation

Solution

There is a formula to calculate the molar conductivity of a solution at particular dilution and molar conductivity at infinite dilution and they are as follows.
Molar conductivity of a solution at particular dilution (Λm{{\Lambda }_{m}}) = Kv×{{K}_{v}}\times molecular weight of the solvent.
Here Kv{{K}_{v}} = specific conductance.
Molar conductivity at infinite dilution Λmo=Λ[H+]+Λ[OH−]{{\Lambda }_{mo}}=\Lambda [{{H}^{+}}]+\Lambda [O{{H}^{-}}]
Here Λ[H+],Λ[OH−]\Lambda [{{H}^{+}}],\Lambda [O{{H}^{-}}] = ionic conductances of H+{{H}^{+}} and OH−O{{H}^{-}}

Complete step by step answer:
- In the question it is given to calculate the ionic product of water from the given data.
- Molar conductivity of a solution at particular dilution (Λm{{\Lambda }_{m}} ) = Kv×{{K}_{v}}\times molecular weight of the solvent.
Here Kv{{K}_{v}} = specific conductance = 0.58×10−7mhocm−10.58\times {{10}^{-7}}mhoc{{m}^{-1}}
Molecular weight of water = 18.
- Substitute the known values in the above formula to get the specific conductance of solution and it is as follows.
Molar conductivity of a solution at particular dilution (Λm{{\Lambda }_{m}} ) = 0.58×10−7×18=10.44×10−70.58\times {{10}^{-7}}\times 18=10.44\times {{10}^{-7}} .
- Then Molar conductivity at infinite dilution Λmo=Λ[H+]+Λ[OH−]{{\Lambda }_{mo}}=\Lambda [{{H}^{+}}]+\Lambda [O{{H}^{-}}]
Here Λ[H+],Λ[OH−]\Lambda [{{H}^{+}}],\Lambda [O{{H}^{-}}] = ionic conductances of H+{{H}^{+}} and OH−O{{H}^{-}} are 350and 198.
- Substitute the known values in the above formula to get the specific conductance of solution and it is as follows.

& {{\Lambda }_{mo}}=\Lambda [{{H}^{+}}]+\Lambda [O{{H}^{-}}] \\\ & {{\Lambda }_{mo}}=350+198 \\\ & {{\Lambda }_{mo}}=548 \\\ \end{aligned}$$ \- We know that the degree of dissociation of water $\alpha ={{\Lambda }_{m}}\times {{\Lambda }_{mo}}$ . \- We know the vales of ${{\Lambda }_{m}},{{\Lambda }_{mo}}$ . Then substitute the known values in the above formula to get the values of degree of dissociation and it is as follows. $$\begin{aligned} & \alpha ={{\Lambda }_{m}}\times {{\Lambda }_{mo}} \\\ & \alpha =548\times 10.44\times {{10}^{-7}} \\\ & \alpha =0.019\times {{10}^{-7}} \\\ \end{aligned}$$ \- The concentration of water = $$\dfrac{1000}{18}\text{ }=\text{ }55.55$$ . \- We know that $$\begin{aligned} & [O{{H}^{-}}]=[{{H}^{+}}]=\alpha \times 55.55=0.019\times {{10}^{-7}}\times 55.55 \\\ & [O{{H}^{-}}]=[{{H}^{+}}]=1\times {{10}^{-7}} \\\ \end{aligned}$$ \- Then Ionic product of water = $[O{{H}^{-}}][{{H}^{+}}]=[1\times {{10}^{-7}}][1\times {{10}^{-7}}]=1\times {{10}^{-14}}{{(mole/litre)}^{2}}$ **The correct answer is option “A” .** **Note:** First we should know the relationship between specific conductance and particular dilution and at infinite dilution with degree of dissociation and later by using the degree of dissociation we can calculate the ionic product of water.