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Iodine titration can be iodemotric or io...

Iodine titration can be iodemotric or iodimetric depending on using iodine directly or indirectly is an oxidising agent in the redox titration.
a. Iodimetric titration in which a standard iodine solution is used as an oxidant and iodine is directly or indirectly titrated against a reducing agent. For example.
`2CuSO_(4)+4KJtoCu_(2)I_(2)+2K_(2)SO_(4)+I_(2)`
b. Iodimetric procedures are used for the datermination of strength of reducing agent such as thiosulphates, sulphites, arsenties adn stanous chloride etc. by titrating them against standard solution of iodine in a burette.
`2Na_(2)SO_(3)to2 NaI +Na_(2)S_(4)O_(6)`
Starch is used as indicator near the end point whilch form blue colour complex with `I_(3)^(-)`. The blue colour disappeams when there is not more of free `I_(2)` .
When 319.0 gm of `CuSO_(4)` in a solution is related with excess of 0.5 M KI solution, then librated iodine required 200 ml of `1.0M Na_(2)S_(2)O_(3)` for complete relation. The percentage purity of `CuSO_(4)` in the sample is

A

10%

B

20%

C

5%

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of determining the percentage purity of copper sulfate (CuSO₄) in the sample, we will follow these steps: ### Step 1: Determine the moles of Na₂S₂O₃ used We know that the volume of Na₂S₂O₃ solution used is 200 mL and its concentration is 1.0 M. \[ \text{Moles of Na₂S₂O₃} = \text{Volume (L)} \times \text{Concentration (M)} = 0.200 \, \text{L} \times 1.0 \, \text{mol/L} = 0.200 \, \text{mol} \] ### Step 2: Determine the moles of iodine (I₂) reacted From the reaction of Na₂S₂O₃ with iodine, the balanced equation is: \[ \text{I}_2 + 2 \text{Na}_2S_2O_3 \rightarrow 2 \text{NaI} + \text{Na}_2S_4O_6 \] From this equation, we see that 1 mole of I₂ reacts with 2 moles of Na₂S₂O₃. Therefore, the moles of I₂ can be calculated as follows: \[ \text{Moles of I₂} = \frac{0.200 \, \text{mol Na₂S₂O₃}}{2} = 0.100 \, \text{mol I₂} \] ### Step 3: Determine the moles of CuSO₄ reacted From the iodimetric reaction given: \[ 2 \text{CuSO₄} + 4 \text{KI} \rightarrow \text{Cu}_2\text{I}_2 + 2 \text{K}_2\text{SO₄} + \text{I}_2 \] From this equation, we see that 1 mole of I₂ reacts with 2 moles of CuSO₄. Therefore, the moles of CuSO₄ can be calculated as follows: \[ \text{Moles of CuSO₄} = 2 \times \text{Moles of I₂} = 2 \times 0.100 \, \text{mol} = 0.200 \, \text{mol} \] ### Step 4: Calculate the mass of CuSO₄ that reacted The molar mass of CuSO₄ is approximately 159.6 g/mol. Therefore, the mass of CuSO₄ that reacted is: \[ \text{Mass of CuSO₄} = \text{Moles} \times \text{Molar Mass} = 0.200 \, \text{mol} \times 159.6 \, \text{g/mol} = 31.92 \, \text{g} \] ### Step 5: Calculate the percentage purity of CuSO₄ The initial mass of the CuSO₄ sample is given as 319.0 g. The percentage purity can be calculated using the formula: \[ \text{Percentage Purity} = \left( \frac{\text{Mass of pure CuSO₄}}{\text{Total mass of sample}} \right) \times 100 \] Substituting the values: \[ \text{Percentage Purity} = \left( \frac{31.92 \, \text{g}}{319.0 \, \text{g}} \right) \times 100 \approx 10.00\% \] ### Final Answer The percentage purity of CuSO₄ in the sample is approximately **10%**. ---
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