1.

"50 cm"^(3) of 0.2" N HCl" is titrated against 0.1" N NaOH"The remaining titration adding "50 cm"^(3) of NaOH. The remaining titration is completed by adding 0.5 N KOH The volume of KOH required for completing the titration is

Answer»

A) `"12 cm"^(3)`
B) `"10 cm"^(3)`
C) `"25 cm"^(3)`
D) `10.5" cm"^(3)`

SOLUTION :Milli - equivqlents of `HCl=50xx0.2=10` milli-eq.
Milli-equivalents of `NaOH=50xx0.1=5` milli - eq.
Remaining milli - equivalents of acid = 5
REQUIRED volume of `KOH=0.5xxV_(1)=5`
`V_(1)=10cm^(3)`


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