Molarity of a compound in its solution is number of moles of the compound in 100

Question

Molarity of a compound in its solution is number of moles of the compound in 1000 mL of its solution.

In each trial 25 mL of 0.010 M KIO3 solution is used.

Molarity of KIO3 solution is 0.010 M

Volume of KIO3 solution used in each trial is, 25 mL

Number of moles of KIO3 in each trial is,

25 * 0.010 / 1000 = 2.50 * 10-4 mol

Thus, number of moles of KIO3in each trial is 2.50 * 10-4 mol.

According to the given balanced chemical equation,

1 mole of IO3- oxidizes 6 moles of thiosulfate (S2O32-)

2.50 * 10-4 mol of IO3- oxidizes, (2.50 * 10-4 mol) * 6 / 1 = 1.50 * 10-3 mol of thiosulfate.

So, number of moles of sodium thiosulfate added in each trial is 1.50 * 10-3 mol

25 mL of KIO3 solution is titrated against sodium thiosulfate solution. 25 mL of KIO3 solution contains 2.50 * 10-4 mol of KIO3. This 2.50 * 10-4 mol of KIO3 is completely consumed at equivalence point by oxidation of 1.50 * 10-3 mol of sodium thiosulfate. So, in each trial, the given volume of sodium thiosulfate solution contains 1.50 * 10-3 mol of sodium thiosulfate.

Trial 1

Number of moles of Na2S2O3 = 1.50 * 10-3 mol

Volume of Na2S2O3 = 14.92 mL

Molarity of Na2S2O3 = 1000 * (number of moles) / (volume in mL)

= 1000 * 1.50 * 10-3 / 14.92 = 0.100 M

So, by trial 1, Molarity of Na2S2O3 = 0.100 M

Trial 2

Number of moles of Na2S2O3 = 1.50 * 10-3 mol

Volume of Na2S2O3 = 14.80 mL

Molarity of Na2S2O3 = 1000 * (number of moles) / (volume in mL)

= 1000 * 1.50 * 10-3 / 14.8 = 0.101 M

So, by trial 2, Molarity of Na2S2O3 = 0.101 M

Thus, molarity of Na2S2O3 is 0.100 M by trial 1 and 0.101 M by trial 2.

Average molarity of Na2S2O3 is, (0.100 M + 0.101 M) / 2 = 0.1005 M

Average molarity of Na2S2O3 is 0.1005 M which can be approximated to 0.100 M

Average molarity of Na2S2O3 solution is 0.1005 M which can be approximated to 0.100 M.

Trial 1 Trial 2 Moles KIO3 used 2.50 * 10-4 mol 2.50 * 10-4 mol Mole Na2S2O3 added 1.50 * 10-3 mol 1.50 * 10-3 mol Volume Na2S2O3 used 14.92 mL 14.80 mL Molarity Na2S2O3 0.100 M 0.101 M

Explanation / Answer

Given procedure: Pipet 25 mL of the standard 0.010 M potassium iodate (KIO3) solution into a clean 125mL Erlenmeyer flask. Next, add about 10mL of 10% KI solution to the flask. Finally, add about 5mL of 1.0M sulfuric acid to the flask. Please help me calculate for the table below!

You can calculate the concentration of your diluted sodium thiosulfate solution using: IO3- (aq) + 6S2O3^2- + 6H+ --> I- + 3S4O6^2- +3H2O

Determination of the Concentration of the Potassium Iodate Standard Titration Data Trial 1 Trial 3 Trial 2 Sodium Thiosulfate Sodium ThiosulfateSodium Thioulfa 14.92 D. 00 14. 2 Calculations for determining the concentration of potassium iodate Final Buret Volume Initial Buret Volume Total Volume Used 15.0 0 Trial 1 Trial 2 Trial 3 3 use Mole Na2S203 Added Volume Na2S203 Used Molarity Na2S203 Average molarity of Na2S203 is M. 220

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