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 the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x=_______ mL (nearest integer).
$$Sn(s)|Sn(OH)_{6}^{2-}(0.5 M)|HSnO_{2}^{-}(0.05 M)|OH^{-}|Bi_{2}O_{3}(s)|Bi(s)$$
Consider upto one place of decimal for intermediate calculations

We need to determine the volume $$x\text{ mL}$$ of $$5\text{ M NaHCO}_3$$ required to form an electrolytic buffer used in the given electrochemical cell.

1. Determine Standard Cell Potential ($$E^\circ_{\text{cell}}$$):

• The overall cell reaction consists of:
  • Anode (Oxidation): $$3\text{HSnO}_2^- + 3\text{OH}^- + 6\text{H}_2\text{O} \rightarrow 3[\text{Sn(OH)}_6]^{2-} + 6e^- \quad (E^\circ_{\text{ox}} = +0.9\text{ V})$$
  • Cathode (Reduction): $$\text{Bi}_2\text{O}_3 + 3\text{H}_2\text{O} + 6e^- \rightarrow 2\text{Bi} + 6\text{OH}^- \quad (E^\circ_{\text{red}} = -0.44\text{ V})$$
• Net Cell Reaction:
  $$3\text{HSnO}_2^-(\text{aq}) + \text{Bi}_2\text{O}_3(\text{s}) + 9\text{H}_2\text{O}(\text{l}) \rightarrow 3[\text{Sn(OH)}_6]^{2-}(\text{aq}) + 2\text{Bi}(\text{s}) + 3\text{OH}^-(\text{aq})$$
• Standard Cell Potential:
  $$E^\circ_{\text{cell}} = E^\circ_{\text{ox}} + E^\circ_{\text{red}} = 0.9\text{ V} - 0.44\text{ V} = 0.46\text{ V}$$

2. Apply Nernst Equation to Calculate pOH and pH:

Using the Nernst equation at $$298\text{ K}$$ for $$n = 6$$ electrons transferred:

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{6} \log\left(\frac{[\text{Sn(OH)}_6^{2-}]^3 \cdot [\text{OH}^-]^3}{[\text{HSnO}_2^-]^3}\right)$$

Substitute the given concentrations ($$[\text{Sn(OH)}_6^{2-}] = 0.5\text{ M}$$, $$[\text{HSnO}_2^-] = 0.05\text{ M}$$) and the observed cell potential ($$E_{\text{cell}} = 0.2353\text{ V}$$):

$$0.2353 = 0.46 - \frac{0.059}{6} \log\left(\left(\frac{0.5}{0.05}\right)^3 \cdot [\text{OH}^-]^3\right)$$

$$0.2353 = 0.46 - \frac{0.059}{6} \cdot 3 \log\left(\frac{10}{[\text{OH}^-]}\right)$$

$$0.2353 = 0.46 - \frac{0.059}{2} \log\left(\frac{10}{[\text{OH}^-]}\right)$$

$$\frac{0.059}{2} \log\left(\frac{10}{[\text{OH}^-]}\right) = 0.46 - 0.2353 = 0.2247$$

$$\log\left(\frac{10}{[\text{OH}^-]}\right) = \frac{2 \times 0.2247}{0.059} = 7.6$$

$$\log(10) - \log[\text{OH}^-] = 7.6 \implies 1 + \text{pOH} = 7.6 \implies \text{pOH} = 6.6$$

$$\text{pH} = 14 - \text{pOH} = 14 - 6.6 = 7.4$$

3. Apply Henderson-Hasselbalch Equation for the Buffer:

The electrolytic buffer is formed by weak acid $$\text{H}_2\text{CO}_3$$ and its conjugate base $$\text{NaHCO}_3$$:

$$\text{pH} = pK_a + \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right)$$

$$7.4 = 6.11 + \log\left(\frac{\text{Millimoles of NaHCO}_3}{\text{Millimoles of H}_2\text{CO}_3}\right)$$

$$1.29 = \log\left(\frac{5x}{10 \times 2}\right)$$

$$\frac{5x}{20} = 10^{1.29}$$

Using the given value $$\text{Antilog}(1.29) = 19.5$$:

$$\frac{x}{4} = 19.5 \implies x = 19.5 \times 4 = 78\text{ mL}$$

Answer: 78