10 mL solution of `H_(2)SO_(4)` and `H_(2)C_(2)O_(4)` (oxalic acid), on titration with with 0.1 M KOH, required 20 " mL of " the base. 10 " mL of " the same solution on titration with `(M)/(300)K_(2)Cr_(2)O_(7)` required 50 " mL of " `K_(2)Cr_(2)O_(7)`.
Q. Strength of oxalic acid in the solution is:

Options

A. `4.5gL^(-1)`
B. `4.9gL^(-1)`
C. `2.25gL^(-1)`
D. `2.45gL^(-1)`

Answer

A

Explanation

In first titration, both `H_(2)SO_(4)` and `H_(2)C_(2)O_(4)` react with base, as acid-base titration n factor for both is 2 due to `2H^(o+)` per mol in each case. Let `x=m" Eq of "H_(2)C_(2)O_(4)` and `y=m" Eq of "H_(2)SO_(4)`
`thereforem" Eq of "H_(2)C_(2)O_(4)+m" Eq of "H_(2)SO_(4)=m" Eq of "KOH`
`thereforex+y=20xx0.1xx1` (n-factor)
`x+y=2` mEq ..(ii)
In seconds titration only `H_(2)C_(2)O_(4)` (being reducing agent)
Reacts with `K_(2)C_(2)O_(7)`.
`m" Eq of "H_(2)C_(2)O_(4)-=m" Eq of "K_(2)Cr_(2)O_(7)`
`(C_(2)O_(4)^(2-)to2CO_(2)+2e^(-))(6e^(-)+Cr_(2)O_(7)^(2-)to2Cr^(3+))`
`(x)-=m" Eq of "H_(2)C_(2)O_(4)-=50xx(1)/(300)xx6`
`thereforex-=1mEq` ..(ii)
From equations (i) and (ii) we get
`x=1 mEq, y=1 mEq`
Weight of `H_(2)C_(2)O_(4)=1xx10^(-3)xx45`
`(Ew of H_(2)C_(2)O_(4)=(90)/(2)=45)`
`=(0.045g)/(10mL)`
`=(0.045xx1000)/(10)gL^(-1)=4.5gL^(-1)`
Therefore, strength of `H_(2)C_(2)O_(4)=4.5gL^(-1)`