What would the molarity of Sodium Thiosulfate solution be? Lab steps attached...

Question

What would the molarity of Sodium Thiosulfate solution be? Lab steps attached... for our results we had an average of 24.18 ml for the titration.

Lab:

            The solution that you make this week will be used next week in the determination of dissolved oxygen in water samples. It is important that this solution be made up correctly. Be sure to check with you instructor when the lab requests you to do so.

Because sodium thiosulfate is a hydrated compound, it is impossible to determine the exact amount of sodium and thiosulfate ions in a solution by weighing a particular mass since this mass contains an amount of water that will vary with humidity, exposure to air and many other variables. Consequently, you will only be able to make your sodium thiosulfate solution to approximately the desired molarity. In order to determine the exact concentration you will perform a titration using the sodium thiosulfate solution you have made with a compound which doesn't absorb water and thus can be weighed exactly. Potassium iodate, a free flowing solid that can be weighed out very accurately will be used in the standardization. From the volume of sodium thiosulfate solution necessary to titrate a potassium iodate solution of known concentration, the concentration of the sodium thiosulfate solution can be determined. The standardization is based on these reactions:

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

            2 Na2S2O3 + I2 (aq) ---à 2Na+ + S4O6-2 + 2Na+ + 2I-        (2)

Starch is a convenient indicator to use in titrations involving iodine; the presence of iodine and starch form a deeply colored blue-purple complex which gives clear endpoint.

Solution Preparation (work with a partner)

Weigh the amount of sodium thiosulfate necessary to make 500 ml of 0.025 M solution (this is the answer to your pre-lab question 3) using a top loading balance.

Put your sodium thiosulfate in a 400 mL beaker and add approximately 300 ml of DI water to it.

Pour this solution into a 500 ml volumetric flask and dilute it to make 500 ml using DI water.

Pour this solution into a 500 ml bottle provided on the side bench and label this with your names and section, and formula of the sodium thiosulfate.

Add about 0.4 grams of NaOH to your solution to retard bacterial growth.

Standardization of the solution

*The potassium iodate must be weighed on the analytical balance (4 decimal points)

Weigh out approximately 0.24 grams of dry potassium iodate. Record the exact mass you have in your lab notebook.

Dissolve the potassium iodate in 50 ml DI water in a small beaker and pour this into a 250 ml volumetric flask. Rinse the beaker twice with 10 ml portions of water and add these to the volumetric flask and fill to the mark

Transfer a 25.00 ml aliquot (using a volumetric pipet) to a 250 ml Erlenmeyer flask and add 0.5 g (weighed on the normal top loading balances) of KI and 2 ml (use a graduated cylinder) of 1 M H2SO4.

Rinse a buret several times with 5 ml portions of the sodium thiosulfate solution and then fill the buret.

Record the initial volume in the buret. This is a quantitative experiment so you should read and record all buret readings to two figures past the decimal place (for example 25.39 ml or 0.00 ml)

Titrate the iodate solution with the thiosulfate solution constantly swirling the flask while the solution is being added.

When the solution turns pale yellow, dilute to approximately 150 ml with distilled water, add 2 ml of 1% freshly prepared starch solution and continue the titration until he color changes from blue to colorless for the first time.

Record the final buret reading.

Repeat this titration on two more aliquots of potassium iodate solution (starting at the step where you use the volumetric pipet to remove 25 ml and place it in the 250 ml flask with the KI and H2SO4). Record initial and final buret readings for each titration. You will probably need to refill the buret.

Explanation / Answer

In this iodometric titration, the iodate ions react with iodide ions and acidic protons to liberate iodine. This liberated iodine is then titrated with sodium thiosulfate. The balanced reactions are given as:

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

I2(aq) + 2S2O3-2(aq) ------> 2I(aq) + S4O6-2(aq)

From the above equation we know that 1 mol of I2 is consumed per 2 mol of S2O3-2

And 3 mol of I2 are formed per mol of IO3

Therefore, 6 mol of S2O3-2 are consumed per mol of IO3

Number of moles of IO3 required = 6 X mol of S2O3-2

Number of molesof S2O3-2 = concentration of S2O3-2 solution X volume of S2O3-2 solution

Number of moles of IO3 required = 6 X concentration of S2O3-2 solution X volume of S2O3-2 solution

concentration of S2O3-2 solution = Number of moles of IO3 required / 6 X volume of S2O3-2 solution

= Mass of IO3 / Molar mass of IO3 X 6 X volume of S2O3-2 solution

= 0.24 g / 214 g mol-1 X 6 X 24.18 X 10-3 L

= 0.0077 mol L-1

The concentration of S2O3-2 solution is 0.0077 mol L-1 or 0.0077 M

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