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Calculate the pH of 0.5 of 1.0 M NaCl so...

Calculate the pH of 0.5 of 1.0 M NaCl solution after electrolysis when a current of 5.0 ampere is passed for 965 seconds.

Text Solution

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`NaCl(aq)+H_(2)Ooverset("Electrolysis")rarr(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)+NaOH(aq)`
Amount of NaCl present in 0.5 L 1 M solution =0.5 mol
Quantity of charge passed (Q)`=(5A)xx(965 s)`=4825 As =4825 C
Now, 96500 C of charge decompose NaCl=1 mol.
4825 C of charge decompose NaCl `=((4825 C))/((96500 C))xx1 "mol"=0.05 "mol"`
Amount of NaOH formed in solution =0.05 mol
Volume of solution =0.5 L
Molarity of NaOH solution `=((0.05 "mol"))/((0.5 L))=0.1 " mol " L^(-1)=10^(-1) M`
pOH=1 or pH =14-1=13.
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Knowledge Check

  • pH of the NaCl solution after electrolysis will

    A
    increase
    B
    decrease
    C
    remain constant
    D
    can't be determined

  • The pH of 0.5L of 1.0"M NaCl" after the electrolysis for 965 s using 5.0 A current ( 100% efficiency ) is :

    A
    `1.00`
    B
    `13.00`
    C
    `12.70`
    D
    `1.30`

  • The electrochemical equivalent of silver is 0.001114 g . When an electric current of 0.5 ampere is passed through an aqueous silver nitrate solution for 200 seconds , the amount of silver deposited is

    A
    1.118 g
    B
    `0.1118 g`
    C
    `5.590` g
    D
    `0.5590 g`

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