Question

3.175g of copper is dissolved in an excess of h2so4 to prepare cuso4 solution.This solution is titrated with excess KI and the linerated iodine is titrated against 0.2M hypo (Na2S2O3).Volume of Na2s2O3 solution vonsumed is X ml.Find X/100​

Answer 1

Explanation:

3.175 of copper is dissolved in an excess of H2SO4 to prepare a CuSO4 soln. This solution is treated with excess of KI and the liberated iodine is titrated against .2M hypo. Vol of Na2S2O3 soln consumed is X mL. what is X/100

Answer 2

Answer:

Volume of 0.2M hypo solution(Na₂S₂O₃) consumed is 250mL.

Thus, value of X is 250 and, 250/100 = 2.5.

Explanation:

3.175 g of copper is dissolved in excess of H₂SO₄ solution to form CuSO₄, thus reaction taking place is:

Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O

Now, moles of Cu reacting:

$number\ of\ moles=\frac{given\ mass}{molar\ mass} =\frac{3.175}{63.5}=0.05$

From the reaction:

1 mole of copper is reacting to produce → 1 mole of CuSO₄

Thus, 0.05 mol of copper will react to produce → 0.05 mol of CuSO₄

Now, the CuSO₄ solution is titrated with excess KI and the reaction will take place as:

2CuSO₄ + 4KI → 2CuI + I₂ + 2K₂SO₄

Here, CuSO₄ reacts with excess KI and formation of Iodine(I₂) takes place.

Now, according to the reaction:

2 moles of CuSO₄ are reacting to produce → 1 mole of I₂

Thus, 1 mol of CuSO₄ will produce → 0.5 mole of I₂

and, 0.05 mol of CuSO₄ will produce → 0.5 × 0.05=0.025 mol of I₂

Now, further this iodine is titrated against 0.2M hypo solution(Na₂S₂O₃), and reaction will take place as:

I₂ + 2Na₂S₂O₃ → 2NaI + Na₂S₄O₆

From the reaction:

1 mol of I₂ is reacting with → 2 mol of Na₂S₂O₃

thus, 0.025 mol of I₂ will react with → 2×0.025 = 0.05 mol of Na₂S₂O₃

Also,

$Molarity=\frac{number\ of\ moles}{Volume\ of\ solution(in\ L)} \\\\or, Volume =\frac{number\ of\ moles}{Molarity}=\frac{0.05}{0.2} =0.25\ L\ or\ 250\ mL$

Thus, volume of 0.2M hypo solution(Na₂S₂O₃) consumed is 250mL

and, 250/100 = 2.5.