100 " mL of " 0.2 M Kal`(OH)_(2)CO_(3)` solution is completely neutralised by a standard solution of `(M)/(4)H_(2)C_(2)O_(4)`. Which of the following is/are wrong?

Options

A. The volume of `H_(2)C_(2)O_(4)` required is 160 mL.
B. the volume of `H_(2)C_(2)O_(4)` required 80 mL.
C. The normality of `KAl` `(OH)_(2)CO_(3)` is 0.4 N
D. It is a redox reaction.

Answer

B, C, D

Explanation

n factor for `overset(+1+3)(Kal)(overset(-2)(OH))_(2)overset(-2)(CO_(3))` is the total positive or negative charge`=4`.
n factor for `H_(2)C_(2)O_(4)=2(2H^(o+)` ions)
It is an acid-base titration
`m" Eq of "Kal(OH)_(2)CO_(3)=m" Eq of "H_(2)C_(2)O_(4)`
`100mL xx0.2xx4=(1)/(4)xx2xxV mL`
`V_(H_(2)C_(2)O_(4))=160 mL`
`N_(KAl(OH)_(2)CO_(3))=0.1xx4=0.8N`