Problem 2: (3 pts) Calculate the molarity of the solution of NaOCl if a 10.0-mL

Question

Problem 2: (3 pts) Calculate the molarity of the solution of NaOCl if a 10.0-mL sample of aqueous NaOCI is treated with excess KI in an acidic solution, then titrated with 25.00 mL of a 0.0250 M Na2S203 solution. Problem 3: (2 pts total) Assume the solution from Problem 2 was from a concentrate of bleach, where 3.0 mL of bleach was previously diluted to 30.0 mL final volume. What is: a) The molarity of the concentrate? b) % Composition of the concentrate assuming density of 1.0 g/mL for the bleach.

Explanation / Answer

Ans 2 :

The reaction is given as :

3NaOCl + Na2S2O3 = 3NaCl + Na2S2O6

Number of moles of Na2S2O3 = 0.025 x 0.0250 = 0.000625 moles

So number of moles of NaOCl = 3 x 0.000625 = 0.001875 moles

Molarity of NaOCl = 0.001875 / 0.010

= 0.1875 M

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