In an ore, the only oxidizable material is `Sn^(2+)`. This ore is titrated with a dichromate solution containing `2.5g` of `K_(2)Cr_(2)O_(7)` in `0.5 “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 ore.
Redox changes are:
`Sn^(2+)rarrSn^(4+)+2e`
`6e+Cr_(2)^(6+)rarr2Cr^(3+)`
Since `Sn^(2+)` is oxidised by `K_(2)Cr_(2)O_(7)`.
`therefore` Meq.of `Sn^(2+)=”Meq.of” K_(2)Cr_(2)O_(7)`
used for tin `=NxxV_(“in ” mL) ( because N=(2.5)/((294.2)/(6)xx0.5))`
`= (2.5)/((294.2)/(6)xx0.50)xx10=1.0197`
`therefore (w_(Sn^(2+)))/(118//2)xx1000=1.0197`
`therefore w_(Sn^(2+))=0.06g`
`therefore % Sn=(0.06)/(0.4)xx100=15%`