50 " mL of " an acidic solution of 0.255 M `K_2Cr_2O_7`, 30 " mL of " 0.4 M `K_2C_2O_4`, and 120 " mL of " 0.2 M `Fe^(2+)` are added together. Compute the molarities of `Fe^(3+)` ions and `Cr_2O_7^(2-)` ions in the final solution.

Explanation

(a). m" Eq of "oxidising agent `(K_2Cr_2O_7)`(n-factor`=6`)
`=50xx0.25xx6`
`(6e^(-)+Cr_2O_7^(2-)to2Cr^(3))=75mEq`
m" Eq of "reducing agent `(Fe^(2+)+K_2C_2O_4)`
`=(0.2xx120)+(2xx0.4xx30)`
`=24+24=48mEq`
`{:[(Fe^(2+)toFe^(3+)+e^(-)(n=1)),(C_2O_4^(2-)to2CO_2+2e^(-)(n=2)):}]`
`mEq-=mEq`
`75mEq-=48mEq`
Excess `m" Eq of "K_2Cr_2O_7=75-48=27mEq`
Total volume `=50+30+120=200mL`
Normality of `K_2Cr_2O_7=(mEq "left")/(V(mL))=(27)/(200)`
Molarity of `K_2Cr_2O_7=(N)/(n)=(27)/(200xx6)=0.0225M`
(b). m" Eq of "`Fe^(3+)` produced `=m" Eq of "Fe^(2+)` reacted
`=24 mEq`
Normality of `Fe^(3+)` produced `=(mEq)/(V mL)=(24)/(200)`
Molarity of `Fe^(3+)` produced `=(N)/(n)=(24)/(200xx1)=0.12M`