titration question help

tryer1998

I got 1.54x10^-6 moles for the answer for the first question, and 3.3x10^-6 mg for the second, but im not confident with these answers, so if anyone who knows what they are talking about can help, that would be great thanks

IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O Equation 24.1

1.

i. The student dissolves some of the iodised salt in a small volume of dilute hydrochloric acid. The student then transfers this solution to a 250 cm³ volumetric flask and makes up to the mark with distilled water.

25.0 cm³ of this solution is then added to a conical flask containing a solution of excess iodide ions. The iodine produced is titrated with a standard solution of 0.00100 mol dm⁻³ sodium thiosulfate, Na₂S₂O₃, using starch indicator.

The reaction equation is:

2Na₂S₂O₃(aq) + I₂(aq) → 2NaI(aq) + Na₂S₄O₆(aq) Equation 24.2

The student’s results give a mean titre of 15.40 cm³.

Calculate the amount (in mol) of IO₃⁻ in the 25.0 cm³ of iodate solution.

amount of IO₃⁻ = .......................................... mol [2]

ii Calculate the amount of IO₃⁻ in the original solution and hence the mass, in mg, of potassium iodate(V) in the iodised salt dissolved.
Give Your answer to an appropriate number of significant figures.

mass of KIO₃ = .......................................... mg

Reply 1

VioletPhillippo

17

Original post by tryer1998

I got 1.54x10^-6 moles for the answer for the first question, and 3.3x10^-6 mg for the second, but im not confident with these answers, so if anyone who knows what they are talking about can help, that would be great thanks

IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O Equation 24.1

1.

i. The student dissolves some of the iodised salt in a small volume of dilute hydrochloric acid. The student then transfers this solution to a 250 cm³ volumetric flask and makes up to the mark with distilled water.

25.0 cm³ of this solution is then added to a conical flask containing a solution of excess iodide ions. The iodine produced is titrated with a standard solution of 0.00100 mol dm⁻³ sodium thiosulfate, Na₂S₂O₃, using starch indicator.

The reaction equation is:

2Na₂S₂O₃(aq) + I₂(aq) → 2NaI(aq) + Na₂S₄O₆(aq) Equation 24.2

The student’s results give a mean titre of 15.40 cm³.

Calculate the amount (in mol) of IO₃⁻ in the 25.0 cm³ of iodate solution.

amount of IO₃⁻ = .......................................... mol [2]

ii Calculate the amount of IO₃⁻ in the original solution and hence the mass, in mg, of potassium iodate(V) in the iodised salt dissolved.
Give Your answer to an appropriate number of significant figures.

mass of KIO₃ = .......................................... mg

Hi,

I'm not sure if my answer is right either but this is the method that I used:

Moles sodium thiosulphate = CV = 0.001 x (15.4/1000) = 1.54 x 10^-5 mol.

From equation 24.2: 2 moles of thiosulphate react with 1 mole of iodine so the moles of iodine used are 1.54x10^-5/2 = 7.7 x 10^-6.

From equation 24.1: 3 moles of iodine were produced from 1 mole of the salt so the moles of the salt are moles of iodine/3 = 7.7×10^-6/3 = 2.57 x 10^-6.

For the 2nd part. If there are 2.57 x 10^-6 moles of the salt in 25cm^3 then there are 2.57 x 10^-6 x 10 moles in 250cm^3 = 2.57 x 10^-5.

m (KIO3)= n × Mr = 2.57 x 10^-5 × 214 = 5.49 × 10^-3 g x 1000 = 5.49 mg.

Hope this helps :smile: