A3.452 g sample containing an unknown amount of a ce(iv) salt is dissolved in 250.0-ml of 1 m h2so4. a 25.00 ml aliquot is analyzed by adding ki and titrating the i3– that forms with s2o32–. the end point was reached following the addition of 13.02 ml of 0.03428 m na2s2o3. calculate the weight percent of ce4 in the sample. when should the indicator be added to this titration? (at the beginning or just before the end point? )

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻× = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻× = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)× = 0,0625 g of Ce(IV)

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = 1,812 wt%

I hope it helps!

the formula for the conversion of temperature from degree celsius to kelvin is:

the sample of water at is converted to kelvin temperature by using the above formula as:

on comparing the two samples of water at or we can say that at and at respectively. the difference between the temperatures of two samples of water is:

hence, one can say that the sample of water at is greater than the sample of water at by .