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Question

Iodate (v) ions react with iodide ions according to the equation below . A .311 g sample of sodium iodate (V) is dissolved in water and made up to 250 mL. 25 mL portions are added to potassium iodide in dissolved in acid. The resultant iodine is titrated against sodium thiosulfate, the average titre being 12.5 mL. What is the molarity of the thiosulfate solution?

IO3 - (aq) + 5I - + 6H+ (aq) => 3I2 (aq) +3H2O (l)

please help, and answer this question. If you can explain too, tht wud be great. Sorry I'm just starting this chem stuff, I'm not lazy, lol.

thnx

Answer 1

0.311 g NaIO3 = 0.00157 mol

0.00157 mol IO3^-1 will produce 3x0.00157 mole I2

0.00471 mol I2

You started with (25/250) x 0.00157 mol or 0.000157 mol IO3^-1 and produced 0.000471 mol I2

I2 + 2S2O3^-2 --> 2I^-1 + S4O6^-2

So 0.000471 mol I2 requires 2x0.000471 mol S2O3^-2 or 0.000942 mol S2O3^2

12.5 mL x M = 0.942 millimol (0.000942 mol) S2O3^-2

M = 0.03768 Molar  (0.0377 M Na2S2O3)

Answer 2

First let's calculate how many moles of iodate you have. FW NaIO3 = 197.9

0.311 g NaIO3 x (1 mole NaIO3 / 197.9 g NaIO3) = 0.00157 moles NaIO3. When you dissolve NaIO3, you get one mole of Na+ and one mole of IO3-.

NaIO3(s) ==> Na+(aq) + IO3-(aq)

So 0.00157 moles of NaIO3 will produce 0.00157 moles of IO3- which is diluted to a final volume of 250 mL.

The amount of iodate taken for analysis is

0.00157 moles x (25 mL / 250 mL) = 0.000157 moles IO3-. How much iodine will that produce in the above reaction? According to the balanced equation above, one mole of IO3- produces three moles of I2. So

0.000157 moles IO3- x (3 moles I2 / 1 mole IO3-) = 0.000471 moles I2.

Now we need the reaction with thiosulfate:

2S2O3 2- + I2 ==> S4O6 2- + 2I-

We can see that there are two moles of thiosulfate (S2O3 2-) for every mole of I2.

0.000471 moles I2 x (2 moles S2O3 2- / 1 mole I2) = 0.000943 moles S2O3 2-

Molarity (M) = moles / L

moles = M x L

moles S2O3 2- = M S2O3 2- x L S2O3 2-

0.000943 = M S2O3 2- x 0.0125 L

0.0754 = M S2O32-

20 minutes ago

Answer 3

First let's calculate how many moles of iodate you have. FW NaIO3 = 197.9

0.311 g NaIO3 x (1 mole NaIO3 / 197.9 g NaIO3) = 0.00157 moles NaIO3. When you dissolve NaIO3, you get one mole of Na+ and one mole of IO3-.

NaIO3(s)  ==>  Na+(aq) + IO3-(aq)

So 0.00157 moles of NaIO3 will produce 0.00157 moles of IO3- which is diluted to a final volume of 250 mL.

The amount of iodate taken for analysis is

0.00157 moles x (25 mL / 250 mL) = 0.000157 moles IO3-. How much iodine will that produce in the above reaction? According to the balanced equation above, one mole of IO3- produces three moles of I2. So

0.000157 moles IO3- x (3 moles I2 / 1 mole IO3-) = 0.000471 moles I2.

Now we need the reaction with thiosulfate:

2S2O3 2-  +  I2  ==>  S4O6 2-  +  2I-

We can see that there are two moles of thiosulfate (S2O3 2-) for every mole of I2.

0.000471 moles I2 x (2 moles S2O3 2- / 1 mole I2) = 0.000943 moles S2O3 2-

Molarity (M) = moles / L

moles = M x L

moles S2O3 2- = M S2O3 2- x L S2O3 2-

0.000943 = M S2O3 2- x 0.0125 L

0.0754 = M S2O32-