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(B) The molar conductivity of water at infinite dilution is given by $\lambda_{H_2O}^\infty = \lambda_{H^{+}}^\infty + \lambda_{OH^{-}}^\infty = 350 +

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The molar conductivity of water at infinite dilution is given by $\lambda_{H_2O}^\infty = \lambda_{H^{+}}^\infty + \lambda_{OH^{-}}^\infty = 350 + 200 = 550 \ S \ cm^2 \ eq^{-1}$.
The relationship between conductivity $\kappa$,molar conductivity $\lambda$,and concentration $C$ is $\lambda = \frac{\kappa 	imes 1000}{C}$.
Substituting the values: $550 = \frac{5.5 	imes 10^{-7} 	imes 1000}{C_{H^{+}}}$.
Solving for $C_{H^{+}}$: $C_{H^{+}} = \frac{5.5 	imes 10^{-4}}{550} = 10^{-6} \ M$.
Since $pH = -\log[H^{+}]$,we get $pH = -\log(10^{-6}) = 6$.

At a certain temperature $T$,if the conductivity of pure water is $5.5 \times 10^{-7} \ S \ cm^{-1}$,calculate the $pH$ of water at the given temperature. Given: $\lambda_{H^{+}}^\infty = 350 \ S \ cm^2 \ eq^{-1}$ and $\lambda_{OH^{-}}^\infty = 200 \ S \ cm^2 \ eq^{-1}$.

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