Question

The pH of $0.5L$ of $1.0MNaCl$ after the electrolysis for $965s$ using $5.0A$ current, is

• A. 1.0
• B. 12.7
• C. 13.0
• D. 1.30

Answer 1

Answer: (C)

$2NaCl+2H_{2}O\stackrel{\text{ electrolysis }}{\to }{H}_2+C{l}_2+2NaOH$
Weight of $NaCl$ present in $0.5L=0.5mol$
Charge $-965\times 5-4825C$
$\therefore$ Number of moles decomposed
$=\frac{1\times 4825}{96500}=0.05mol$
$\therefore$ Number of moles of $NaOH$ formed is also $0.05$.
$\therefore$ Molarity $=\frac{0.05\times 1000}{500}=0.1$
$pOH=1,pH=13$