In iodometric titrations, an oxidizing agent such as `KNnO_(4), K_(2)Cr_(2)O_(7),CuSO_(4),H_(2)O_(2)` is allowed to react in neutral medium or in acidic medium with excess of potassium iodide to liberate free iodine `Kl+` oxidizon agent `to l_(2)` Free iodine is titrated against stanard reducing agent usually with sodium thiosulphate i.e., `K_(2)Cr_(2)O_(7)+6Kl+7H_(2)SO_(4)toCr_(2)(SO_(4))_(3)+4K_(2)SO_(4)+7H_(2)O+l_(2)` `2CuSO_(4)+4Kl to Cu_(2)l_(2)+2K_(2)SO_(4)+l_(2)` `l_(2)+Na_(2)S_(2)O_(3)to 2Nal+Na_(2)S_(4)O_(6)` In iodometric titrations, starch solution is used as an indicator. Starch solution gives blue or violet colour with free iodine. At the end point, blue or violet colour disappear when iodine is completely changed to iodide. A `1.1g` sample of copper ore is dissovled and `Cu^(2+)` (aq.) is treated with `Kl.l_(2)` liberated required `12.12mL` of `0.1M Na_(2)S_(2)O_(3)` solution for titration. The `% Cu` in the ore in the ore is:A. `2.5%`B. `7%`C. `5.8%`D. `4.2%`

In iodometric titrations, an oxidizing agent such as `KNnO_(4), K_(2)C

1.	In iodometric titrations, an oxidizing agent such as `KNnO_(4), K_(2)Cr_(2)O_(7),CuSO_(4),H_(2)O_(2)` is allowed to react in neutral medium or in acidic medium with excess of potassium iodide to liberate free iodine `Kl+` oxidizon agent `to l_(2)` Free iodine is titrated against stanard reducing agent usually with sodium thiosulphate i.e., `K_(2)Cr_(2)O_(7)+6Kl+7H_(2)SO_(4)toCr_(2)(SO_(4))_(3)+4K_(2)SO_(4)+7H_(2)O+l_(2)` `2CuSO_(4)+4Kl to Cu_(2)l_(2)+2K_(2)SO_(4)+l_(2)` `l_(2)+Na_(2)S_(2)O_(3)to 2Nal+Na_(2)S_(4)O_(6)` In iodometric titrations, starch solution is used as an indicator. Starch solution gives blue or violet colour with free iodine. At the end point, blue or violet colour disappear when iodine is completely changed to iodide. A `1.1g` sample of copper ore is dissovled and `Cu^(2+)` (aq.) is treated with `Kl.l_(2)` liberated required `12.12mL` of `0.1M Na_(2)S_(2)O_(3)` solution for titration. The `% Cu` in the ore in the ore is:A. `2.5%`B. `7%`C. `5.8%`D. `4.2%`
Answer» Correct Answer - B meq. Of hypo `-=` meq. Of `Cu^(2+)-=` meq. Of `Cu` `12.12xx0.1=(w_(Cu))/(63.5//2)xx1000impliesw_(Cu)=0.076g` `:. %` purity `=0.076/1.1xx1000=6.99~~7%`

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