Redo the third probelm of the third problem of the Prelaboratory Assignment, sub

Question

Redo the third probelm of the third problem of the Prelaboratory Assignment, substituting NaIO3 (Sodium Iodate) for NaOCl. The new reaction yields I2, as the only product containing a halogen.

The third problem to redo is: A 10.0mL sample of aqueous NaOCl is treated with excess KI in an acidic solution. The quantity of iodine that is liberated is such that 27.46 mL of .0250 M Na2S2O3 solution must be added to cause the disappearance of the dark blue color due to the starch indicator. What is the molarity of the solution of NaOCl?

I would like a thorough explanation because I understand mostly all of this, but would still like to learn more. I always rate 5 stars for intelligent answers. Thank you so much, I really appreciate your knowledge and understanding.

Explanation / Answer

Hello,

The following questions pertain to Quantitative Analysis, which involves estimation of abundance and amount of a particular chemical species within a sample. This question of yours involves a central concept of Quantitative Analysis called Iodometry. Let us try to understand this.

Iodometry

This is an indirect method of estimation of iodine in a sample & the process involves the following two steps.

Step 1: In this step an oxidizing agent is made to react with solid KI (which serves as a reducing agent, since Iodine is in the form of I-1,and can only undergo oxidation rather than reduce itself). In this redox reaction, I-1 changes to I2. I2 thus liberated is absorbed in starch solution to form a starch iodine complex, which imparts blue colour to the solution.

Step 2: In this step, the liberated iodine is made to react with sodium thiosulphate solution (Na2S2O3) of known concentration & known volume. The amount of Na2S2O3 utilized for reaction with I2 can tell us how much iodine was liberated and which in turn can tell us how much how iodide is present in the sample of KI or how much oxidizing agent was used in the reaction with KI.

Now coming to your question:

A 10.0mL sample of aqueous NaIO3 is treated with excess KI in an acidic solution. The quantity of iodine that is liberated is such that 27.46 mL of 0.0250 M Na2S2O3 solution must be added to cause the disappearance of the dark blue color due to the starch indicator. What is the molarity of the solution of NalO3?

We’ll divide the question into two parts:

Part 1: Reaction of NaIO3 & KI

Part 2: Reaction of liberated I2 & Na2S2O3

The reactions involved in this question are as follows:

Part 1: Reaction of NaIO3 & KI

[since the reaction isn’t given & also the nature of acid isn’t given so I’m taking the liberty of including only those species who’s oxidation state changes in the process. The rest of the ions, viz. Na+ & K+ behave as spectator ions and their presence or absence doesn’t create any difference to the stoichiometry of the reaction]

6H+(aq) + IO3-1(aq) + 5I-1(aq) à 3I2 (aq) + H2O (l)

Part 2: Reaction of liberated I2 & Na2S2O3

[Na+ is the spectator ion, so I’m ignoring it in the equation]

I2(aq) + 2S2O3-2(aq) ? 2I-1(aq) + S4O62-(aq)

This question can be solved using the equivalents method:

To use this method, you’ll have to remember that for any reaction the number of equivalents of all the combining species is the same as the number of equivalents of all the products formed.

Also, number of equivalents = number of moles x n-factor

number of moles = molarity x volume

For an oxidation or reduction reaction the n-factor is equal to the number of moles of electrons lost or gained respectively, per mole of reactant.

In this reaction:

Part 1:

a) IO3-1 has chlorine in +5 oxidation state, which undergoes reduction to I2 in which the oxidation state is 0. So, for this change, 5 moles of electrons are lost for every mole of IO3-1 during its conversion to I2, so the n-factor is equal to 5.

b) I- has iodine in -1 oxidation state, which undergoes oxidation to I2 which has iodine in 0 oxidation state. So, for this change, 1 mole of electrons are lost for every mole of I- during its conversion to I2, so the n-factor is equal to 1.

Part2:

a) I2 has iodine in 0 oxidation state, which undergoes reduction to l-1 in which the oxidation state is -1. So, for this change, 2 moles of electrons are gained for every mole of I2 during its conversion to l-1, so the n-factor is equal to 2.

b) S2O32- has sulphur in +2 oxidation state, which undergoes oxidation to S4O62- which has sulphur in +2.5 oxidation state. So, for this change, 1 mole of electrons are lost for every mole of S2O32- during its conversion to S4O62-, so the n-factor is equal to 1.

Next,

Let the molarity of IO3-1 be a Molar

Millimoles of IO3-1 = Molarity x Volume = a x 10 ml = 10a millimoles

Milliequivalents of IO3-1 = 10a millimoles x 5 = 50a milliequivalents

Milliequivalents of I-1 = Millimoles x n-factor = 10a x 1 = 10a milliequivalents (since n-factor of IO3-1 & I-1 are in the ratio 5 & 1 respectively)

The reason why I’m including I-1 in this assessment is because even though I-1 is in excess, I2 in the product also comes from it. Iodate and iodide react by comproportionation reaction and it would be incorrect to assume that all the I2 in the product comes only from iodate. This is the rectification that I’ve incorporated.

Millimoles of Na2S2O3 required for completely consuming the liberated I2 = 27.46 mL x 0.0250 M = 0.6865 millimoles

Milliequivalents of Na2S2O3 required for completely consuming the liberated I2 = millimoles x n-factor = 0.6865 x 1 = 0.6865 milliequivalents

Since the number of milliequivalents of both the I2 & Na2S2O3 has to equal in the reaction & milliequivalents of I2 liberated is equal to sum of the milliequivalents of IO3-1 & I-1 involved in the reaction , so we can write

Milliequivalents of IO3-1 + Milliequivalents of I-1 = Milliequivalents of I2 liberated = Milliequivalents of Na2S2O3 required

Or, 50a + 10a milliequivalents = 0.6865 milliequivalents

Or,60a = 0.6865 milliequivalents

Or, a = 0.01143 M

Thus, the molarity of NaIO3 solution used for this reaction using equivalent method turns out to be 0.01143 M

Verification of Result

We can verify the result using the conventional Mole Method.

Millimoles of IO3-1 = Molarity x Volume = a x 10 ml = 10a millimoles

Millimoles of I-1 = Millimoles x n-factor = 10a x 5 = 50a millimoles (since n-factor of IO3-1 & I-1 are in the ratio 5:1 & moles of IO3-1 & I-1 are in the ratio 1:5)

Now if we look at the respective half reaction:

10 electrons + 2IO3-1 = I2

2I- = I2 + 2 electrons

Millimoles of I2 liberated from IO3-1 = 5a millimoles (since for every two moles of IO3-1 involved one mole of I2 is produced)

Millimoles of I2 liberated from I-1 = 25a millimoles (since for every two moles of I-1 involved one mole of I2 is produced)

So total millimoles of I2 produced = Millimoles of I2 liberated from IO3-1 + Millimoles of I2 liberated from I-1 = 30a millimoles

Millimoles of S2O32- utilized in the reaction = 27.46 mL x 0.0250 M = 0.6865 millimoles

further we can see that moles of I2 liberated : moles of S2O32- is in the ratio 1:2

So, we can write,

(Moles of I2)/1 = (Moles of S2O32-)/2

Or, 30a millimoles = 0.6865/2

Or, a = 0.01143 M

Thus, the molarity of NaIO3 solution used for this reaction using the mole method also turns out to be 0.01143 M

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