A current of 9.65 A is passed through 100 ml $N a C l$ solution containing sufficient $N a C l$ for 100 s. What will be the $p H$ of the resulting solution? Assume 100% current efficiency and constancy of the volume of solution.

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Solution

$9.65 A$ current for $100 s$
Charge flown $= 9.65 \times 100 = 965 C$
$\Rightarrow 96500 C = 1$ FARAD (F)
$965 C = 6.01 F$
$N a C l ⟶ {N a}^{+} (a q) + {C l}^{(-)} (a q)$
$H_{2} O ⇌ H^{+} (a q) + {O H}^{-} (a q)$
$\Rightarrow$ AT Anode,
${C l}^{(-)} ⟶ \frac{1}{2} {C l}_{2} + e^{(-)}$
${2 C l}^{-} ⟶ {C l}_{2} + {2 e}^{(-)}$
AT Cathode,
$H^{+} + e^{(-)} ⟶ \frac{1}{2} H_{2}$
${2 H}^{+} + 2 e^{(-)} ⟶ H_{2}$
According to Faraday's law
$2 F$ of electricity will retake $2$ mol of $H^{+}$
$0.01$ of electricity will retake $0.01$ mol of $H^{+}$
$H^{+}$ will be converted into $H_{2}$ gas.
This will increase concentration of ${O H}^{(-)}$ in solution.
After $100$ sec. $[{O H}^{-}] = 0.01$
Molarity of ${O H}^{+} = \frac{0.01}{\frac{100}{1000}} = 0.1 M$
$p O H = - l o g [{O H}^{-}] = - l o g [0.1] = 1$
$p H = 14 - p O H$
$= 14 - 1 = 13$

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A current of 9.65 A is passed through 100 ml NaCl solution containing sufficient NaCl for 100 s. What will be the pH of the resulting solution? Assume 100% current efficiency and constancy of the volume of solution.

A current of 9.65 A is passed through 100 ml $N a C l$ solution containing sufficient $N a C l$ for 100 s. What will be the $p H$ of the resulting solution? Assume 100% current efficiency and constancy of the volume of solution.