A volume of x mL of 5 M NaHCO_3 solution was mixed with 10 mL of 2 M H_2 CO_3 solution to make an electrolytic buffer. If the same buffer was used in…

A volume of $\displaystyle x \mathrm{~mL}$ of $\displaystyle 5 \mathrm{M} \mathrm{NaHCO}_3$ solution was mixed with $\displaystyle 10$ mL of $\displaystyle 2 \mathrm{M} \mathrm{H}_2 \mathrm{CO}_3$ solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of $\displaystyle 235.3$ mV , then the value of $\displaystyle x=$ $\displaystyle \_\_\_\_$ mL (nearest integer). $\displaystyle \mathrm{Sn}(\mathrm{s})\left|\mathrm{Sn}(\mathrm{OH})_6{ }^{2-}(0.5 \mathrm{M})\right| \mathrm{HSnO}_2{ }^{-}(0.05 \mathrm{M})\left|\mathrm{OH}^{-}\right| \mathrm{Bi}_2 \mathrm{O}_3(\mathrm{~s}) \mid \mathrm{Bi}(\mathrm{s})$ Consider upto one place of decimal for intermediate calculations $$\left[\begin{array}{ll} \text { Given : } & \mathrm{E}^{\mathrm{o}}{ }_{\mathrm{HSnO}_2^{-} \mid \mathrm{Sn}(\mathrm{OH})_6^{2-}}=-0.9 \mathrm{~V} \\ & \mathrm{E}^{\mathrm{o}}{ }_{\mathrm{Bi}_2 \mathrm{O}_3 \mid \mathrm{Bi}}=-0.44 \mathrm{~V} \\ & \mathrm{pKa}_{\left(\mathrm{H}_2 \mathrm{CO}_3\right)}=6.11 \\ & \frac{2.303 \mathrm{RT}}{\mathrm{~F}}=0.059 \mathrm{~V} \\ & \text { Antilog }(1.29)=19.5 \end{array}\right] $$