Assume that a 2.130 g sample of bleach was weighed out as outlined in the proced

Question

Assume that a 2.130 g sample of bleach was weighed out as outlined in the procedure. When a sodium thiosulfate solution containing 6.829g of Na2S2O3 x 5H2O in 250.0ml was used to titrate the above bleach sample, it required 11.21ml to reach the strach endpoint.

what was the molarity of the sodium thiosulfate pentahyrdate solution?

what is the number of moles of Na2S2O3 in the 11.21 ml?

what is the number of moles of I2 that react with the Na2S2O3?

how many moles of NaOCl are in the bleach sample? How many grams?

what is the percent by mass of NaOCl in the blach?

Explanation / Answer

MW of Na2S2O3, 5H2O = 248.18 g/mol

So 6.829 g Na2S2O3, 5H2O corresponds to (6.829 / 248.18) moles = 0.0275 moles

So concentration of Na2S2O3, 5H2O solution = 0.0275 moles / 0.25 L = 0.11 moles / L or 0.11 M

Number of moles of Na2S2O3 in 11.21 mL = molarity x volume (L) = 0.11 x 11.21 x 10-3 = 0.00123 moles

1 mole of I2 + 2 moles of Na2S2O3 = 2 NaI + NaS4O6

So, 0.00123 moles of Na2S2O3 reacts with 0.00123 / 2 = 0.000615 moles of I2

1 mole of ClO-1 + 2 I-1 + 2 H+ = Cl-1 + 1 mole of I2 + H2O

So, 0.000615 moles of I2 was produced from 0.000615 moles of NaOCl (bleech).

Grams of NaOCl = MW of NaOCl x number of moles = 74.44 x 0.000615 = 0.04578 g

% mass of NaOCl in bleech = (0.04578 g NaOCl / 2.130 g of bleech) x 100 = 2.15%

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