A volume of x\,mL of 5\,M\,NaHCO_3 solution was mixed with 10\,mL...

A volume of $x\,\mathrm{mL}$ of $5\,\mathrm{M}\,\mathrm{NaHCO_3}$ solution was mixed with $10\,\mathrm{mL}$ of $2\,\mathrm{M}\,\mathrm{H_2CO_3}$ solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of $235.3\,\mathrm{mV}$, then the value of $x = \_\_\_\_$ mL (nearest integer).

$\mathrm{Sn(s)} \; \big| \; \mathrm{Sn(OH)_6^{2-}}(0.5\,\mathrm{M}) \; \big| \; \mathrm{HSnO_2^-}(0.05\,\mathrm{M}) \; \big| \; \mathrm{OH^-} \; \big| \; \mathrm{Bi_2O_3(s)} \; \big| \; \mathrm{Bi(s)}$

Consider up to one place of decimal for intermediate calculations.

$$\begin{array}{ll}\text{Given:} & E^\circ_{\mathrm{[Sn(OH)_6]^{2-}/HSnO_2^-}} = -0.90\,\mathrm{V} \\[4pt]& E^\circ_{\mathrm{Bi_2O_3/Bi}} = -0.44\,\mathrm{V} \\[4pt]& \mathrm{p}K_a(\mathrm{H_2CO_3}) = 6.11 \\[4pt]& \dfrac{2.303\,RT}{F} = 0.059\,\mathrm{V} \\[6pt]& \text{Antilog}(1.29) = 19.5\end{array}$$