Question

Specific conductance of pure water at ${25}^{\circ }$ is $0.58\times {10}^{-7}mhoc{m}^{-1}$. The ionic product of water ($K_W$) if ionic conductances of $H^{+}$ and $O{H}^{-}$ ions at infinite dilution are 350 and $198mhoc{m}^2$ respectively at ${25}^{\circ }C$ is:

• A. $1\times {10}^{-14}(mole/litre{)}^2$
• B. $1\times {10}^{-13}(mole/litre{)}^2$
• C. $1\times {10}^{-12}(mole/litre{)}^2$
• D. $Noneofthese$

Answer 1

The correct option is A $1\times {10}^{-14}(mole/litre{)}^2$

$Molarconductivityofsolution\left({\Lambda }_m\right)=K_v\times 18$

${\Lambda }_m=0.58\times {10}^{-7}\times 18{\Lambda }_m=10.44\times {10}^{-7}$

$Molarconductivityatinfinitedilution\left({\Lambda }_m^0\right)=\Lambda \left[H^{+}\right]+\Lambda [O{H}^{-}]{\Lambda }_m^0=350+198{\Lambda }_m^0=548$

$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^0}$

$\alpha =\frac{10.44\times {10}^{-7}}{548}\alpha =0.019\times {10}^{-7}$

$Conc.ofwater=1000/18=55.55\left[O{H}^{-}\right]=\left[H^{+}\right]=\alpha \times 55.55=0.019\times {10}^{-7}\times 55.55=1\times {10}^{-7}$

$Ionicproductofwater=\left[H^{+}\right]\left[O{H}^{-}\right]=1\times {10}^{-7}\times 1\times {10}^{-7}=1\times {10}^{-14}$

So answer is A.