1.

0.5 L of 1.0 M NaCl solution is electrolysed for 965 a using a current of 5 ampere. What will be the pH of the solution after the electrolysis?

Answer»

Solution :`NaCl+H_(2)O OVERSET("ELECTROLYSIS")to(1)/(2)H_(2)+(1)/(2)Cl_(2)+NaOH`
amount of NaCl present in 0.5 L of 1.0 M NaCl=0.5 mole
QUANTITY of electricity passed=`965xx5C=4825C`
1 mole of NaCl is decomposed by 96500C
`therefore4825`C will decompose `NaCl=(4825)/(96500)"mole"=0.05`mole
NaOH formed in the solution will also be 0.05 mole
volume of the solution`=0.5L`
`therefore`Molarity of NaOH in the solution`=(0.05)/(0.5)=0.1M=10^(-1)M`
`thereforepOH=1` or `pH=14-1=13`.


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