Choose the correct representation of conductometric titration of benzoic acid vs sodium hydroxide.

Which one of the following statements is correct for electrolysis of brine solution?

The logarithm of equilibrium constant for the reaction ${\mathrm{Pd}}^{2+}+4{\mathrm{Cl}}^{-}\rightleftharpoons {\mathrm{PdCl}}_4^{2-}$ is
(Nearest integer)

Given: $\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{V}$

${\mathrm{Pd}}_{\left(\mathrm{aq}\right)}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Pd}\left(\mathrm{s}\right) {\mathrm{E}}^o=0.83 \mathrm{V}$

${\mathrm{PdCl}}_4^{2-}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Pd}\left(\mathrm{s}\right)+4{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

${\mathrm{E}}^{\mathrm{o}}=0.65 \mathrm{V}$

$1\times {10}^{-5} \mathrm{M} {\mathrm{AgNO}}_3$ is added to $1 \mathrm{L}$ of saturated solution of $\mathrm{AgBr}$. The conductivity of this solution at $298 \mathrm{K}$ is $\_\_\_\_\_\_\_\times {10}^{-8} \mathrm{S} {\mathrm{m}}^{-1}$.

[Given : ${\mathrm{K}}_{\mathrm{sp}}\left(\mathrm{AgBr}\right)=4.9\times {10}^{-13}$ at $298 \mathrm{K}$

${\lambda }_{{\mathrm{Ag}}^{+}}^0=6\times {10}^{-3}{\mathrm{Sm}}^2 {\mathrm{mol}}^{-1}$

${\lambda }_{{\mathrm{Br}}^{-}}^0=8\times {10}^{-3}{\mathrm{Sm}}^2 {\mathrm{mol}}^{-1}$

$\left.{\lambda }_{{\mathrm{NO}}_3^{-}}^0=7\times {10}^{-3}{\mathrm{Sm}}^2 {\mathrm{mol}}^{-1}\right]$