In an ore the only oxidizable material in `Sn^(2+)` This is titrated with a dichromate solution containing 2.5 of `K_(2)Cr_(2)O_(7)` in 0.50 litre A 0.40 g sample of the ore required 10.0 `cm^(3)` of titrant to reach equivalence point calculate the percentage of tin in the ore (k=39.1, Cr = 52 Sn =11.87)

Explanation

Step 1 To write balanced chemical equation f the redox reaction
`Sn^(2+)rarr Sn^(4+)+2 e^(-)[xx3`
`Cr_(2)O_(7)^(2-)+14 H^(+)+6 e^(-) rarr 2Cr^(3+)+7 H_(2)O`
`Cr_(2)O_(7)^(2-)+3 Sn^(2+)+14 H^(+) rarr 2Cr^(3+)+3 Sn^(4+)+7 H_(2)O`
Now 0.5 L of the solution cotains `K_(2)Cr_(2)O_(7)=2.5 g`
`therefore` 10 mL of solution will contain `K_(2)Cr_(2)O_(7)=(2.5)/(500)xx10 g=(2.5)/(500xx(210)/(294)` mole
From the balanced redox reaction
1 mole of `K_(2)Cr_(2)O_(7)` oxidises `Sn^(2+)=3` moles
`therefore (2.5)/(500)xx(10)/(294)` mole of `K_(2)Cr_(2)O_(7)` will oxidise `Sn^(2+)=3xx(2.5)/(500)xx(10)/(294) mol =3x(2.5)/(500)xx(10)/(294)xx118.7 g`
Step 2 To determine the % age of `Sn^(2+)` in the ore
Now 0.40 g of the or contain `Sn^(2+) =0.06 g`
`therefore % "of" Sn^(2+) "in the are" = 0.06/0.40xx100=15%`