Titration. [Na2S2O3] = 0.0499 M 2I2 ( aq ) + 4Na2S2O3 ( aq ) 2Na2S4O6 (aq) + 4Na

Question

Titration.

[Na2S2O3] = 0.0499 M

                                           2I2 (aq) + 4Na2S2O3 (aq) 2Na2S4O6 (aq) + 4NaI (aq)

VH2SO4 added

(mL)

VNa2S2O3 used (mL)

molNa2S2O3

molI2

How do I find molNa2S2O3?

How do I find molI2?

Analysis.

Knowing:

2I (aq) I2 (aq) + 2e

2H2O () + 2e H2 (g) + 2OH (aq)

1. Are the calculated moles of hydrogen gas, iodine, and electrons consistent with the anode and cathode halfreactions for each trial? Explain.

VH2SO4 added

(mL)

VNa2S2O3 used (mL)

molNa2S2O3

molI2

Explanation / Answer

Solution :-

Molarity of Na2S2O3 = 0.0499 M

Using the molarity and volume of the Na2S2O3 we can calculate its moles for each trial

Moles= molarity * volume in liter

Moles of Na2S2O3 in trial 1 = 0.0499 mol per L * 0.02324 L = 0.00116 mol

Moles of Na2S2O3 in trial 2 = 0.0499 mol per L * 0.02620 L = 0.001307 mol

Moles of Na2S2O3 in trial 3 = 0.0499 mol per L * 0.0256 L = 0.001277 mol

Now using the mole ratio of the Na2S2O3 and I2 we can calculate the moles of I2 for each trial

Mole ratio of the Na2S2O3 to I2 is 4 : 2

Therefore

Moles of I2 in trial 1 = 0.00116 mol Na2S2O3*2 mol I2/4 mol Na2S2O3 = 0.00058 mol I2

Moles of I2 in trial 2 = 0.001307 mol Na2S2O3*2 mol I2/4 mol Na2S2O3 = 0.0006535 mol I2

Moles of I2 in trial 3 = 0.001277 mol Na2S2O3*2 mol I2/4 mol Na2S2O3 = 0.0006385 mol I2

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