Calculate the concentration of an iodate solution prepared by dissolving 1.9853

Question

Calculate the concentration of an iodate solution prepared by dissolving 1.9853 g of KIO_3 and diluting to 500 mL with distilled water in a volumetric flask. A 25 mL aliquot of a 0.0195 M KIO_3 solution is added to a flask containing 2 g of KI and 10 mL of 0.5 M H_2SO_4. The resulting solution is titrated to a starch endpoint with 34.81 mL of the thiosulfate solution. Calculate the concentration of the thiosulfate solution. How will you know when the endpoint of the titration is reached? Identify the titrant, the primary standard, and the actual chemical species being titrated. Write the reaction for the formation of the species titrated. Write the reaction for the multi-step reaction sequence used to analyze Vitamin C. How will the volume of titrant used in the standardization compare with the titration volume when analyzing a Vitamin C sample? Why do you add sulfuric acid to the solution twice? Why do you add the sodium bicarbonate to the solution twice?

Explanation / Answer

Part A

#1 Concentration of KIO3 can be calculated as follows

M = 1.9853 g KIO3 x 1 mol/ 214 g KIO3 x 1/500 mL x 1000 mL/1 L

When we solve this, we are left with the unit mol/L which is concentration unit

So,M = 0.01855 M

KIO3 is a strong salt and dissociates as follows

KIO3 -----> K+ + IO3^-

concentration of IO3^- = 0.01855 mol KIO3/L * 1mol IO3^-/ 1 mol KIO3 ( ratio of KIO3 to IO3^- is 1:1)

Concentration of IO3^- = 0.01855M

#2 The balanced equation for the titration reaction is given as

5 I^- + IO3^- + 6H+ ----------> 3I2 + 3H2O............ ( K+is spectator ion)

We will find moles of each reactant to find the limiting reactant

---> 25 mL of 0.0195 M KIO3 is

25 mL x 1 L/1000 mL x 0.0195 mol/L = 0.0004875 mol KIO3 = 0.0004875 IO3^-

---> 2 g KI

2 g KI x 1mol/ 166 KI = 0.01204 mol KI = 0.01204 mol I-

----> 10 mL of 0.5M H2SO4

10 mL x 1 L/1000 mL x 0.5 mol H2SO4/L x 2 mol H+/ 1 mol H2SO4 = 0.01 mol

To find limiting reactant, we divide moles of each reactant by their stoichiometric coefficients

for IO3^- we have 0.0004875/ 1 = 0.0004875

for I- we have 0.01204/ 5 = 0.002408

for H+ we have 0.01/6 = 0.001667

Since moles of IO3^- are lowest, IO3^- is the limiting reactant.

We will find out amount of I2 that is formed in the reaction

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

0.0004875 mol IO3^- x 3 mol I2/ 1 mol IO3^- = 0.0014625 mol I2

I2 formed in this step reacts with thiosulfate solution as per the following reaction

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

0.0014625 mol I2 x 2 mol S2O3^-2 /1 mol I2 = 0.002925 mol

volume required for titration is 34.81 mL = 34.81 mL x 1 L/1000 mL = 0.03481 L

Molarity of thiosulfate is 0.002925 mol/ 0.03481 L = 0.0840 M

concentration of thiosulfate solution is 0.084 M

#3

when KI, KIO3 and H2SO4 are mixed, reaction occurs and I2 is liberated. When starch is added to this solution at this point, the solution turns blue in color because starch forms a blue color complex with I2 solution.

As we titrate this blue colored I2 solution with thiosulfate, I2 is reduced to I- and when all the I2 is reduced to I-, the solution loses its blue color which indicates end point of the titration

To answer part B, more details about the experiment are needed.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.