$A$ volume of $x \ mL$ of $5 \ M \ NaHCO_3$ solution was mixed with $10 \ mL$ of $2 \ M \ H_2CO_3$ solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of $235.3 \ mV$, then the value of $x = . . . . . . \ mL$ (nearest integer).
$Sn_{(s)} \mid Sn(OH)_6^{2-}(0.5 \ M) \mid HSnO_2^{-}(0.05 \ M) \mid OH^{-} \mid Bi_2O_{3(s)} \mid Bi_{(s)}$
Consider upto one place of decimal for intermediate calculations.

Given :	$E^{\circ}_{HSnO_2^{-} \mid Sn(OH)_6^{2-}} = -0.9 \ V$
	$E^{\circ}_{Bi_2O_3 \mid Bi} = -0.44 \ V$
	$pKa_{(H_2CO_3)} = 6.11$
	$\frac{2.303 \ RT}{F} = 0.059 \ V$
	$Antilog(1.29) = 19.5$

• A
  $70$
• B
  $75$
• C
  $78$
• D
  $80$