A saturated aqueous solution of a slightly soluble salt, barium iodate [Ba(IO3)2

Question

A saturated aqueous solution of a slightly soluble salt, barium iodate [Ba(IO3)2], was prepared and titrated with sodium thiosulfate (according to the same procedure used in your experiment ). Based on the data tabulated below, calculated the indicated quantities in questions 1-5. Organize your calculations using the steps below, but first determine and write the proper chemical equation. 1. Calculate the number of moles of thiosulfate used in each titration n S2O3 ^2. = 2. Calculate the number of moles of iodate throated: n IO3 = 3. Calculate the molar concentration of iodate ions: [IO 3] 4. Calculate the average molar concentration of iodate ions: Avg [ IO3] = 5. Calculate Ksp for the dissolution of Ba(IO3)2 =

Explanation / Answer

Iodate and thiosulfate do not react directly, but an acid and potassium iodide are added to a solution of potassium iodate as follows:

KIO3 + 5KI + 3H2SO4 = 3I2 + 3K2SO4 + 3H2O

which can be represented by a ionic equation as:

IO3- + 5I- + 6H+ ---> 3I2 + 3H2O

Thiosulfate is titrated against the iodine formed in this solution:
I2 + 2Na2S2O3 = Na2S4O6 + 2NaI

Ionic equation:
I2 + 2S2O32- = S4O62- + 2I

so, from the above reactions, it can seen that 1mol IO3- reacts with 6mol S2O32-

1. Moles of thiosulphate = 0.01 moles/litre * 0.00065 L = 6*5 * 10-6 [1st trial]

= 0.01 moles/litre * 0.0133 litre = 0.000133 moles

= 0.01 moles/litre * 0.0265 litre = 0.000265 moles

2. 1mol IO3- reacts with 6mol S2O32-

so moles of iodate = 1/6 * moles of S2O32-

= 1/6 * 0.000133 moles (2nd titration)

= 2.1 * 10-5 moles

3.Molarity of iodate = moles of iodate / litres

= 2.1 * 10-5 moles / (0.0002 litre + 0.0133 litre = total volume)

= 0.00164 M

4. The above value is is the molar concentartion of iodate ions from 2nd titration. Similarly calculate for 1st and 3rd trials too. Add all the 3 resulties molarity values and divide by 3 to get average molarity

5. Ksp = [Ba2+][IO3-]2

For iodate, use value from question 4

For barium ion concentration, half the moles of iodate would be for barium ion since each Barium iodate gives 1 mole of Ba+2 ion and 2 moles of iodtate ions

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