a.) What is the molarity of a solution prepared by dissolving 47.1 grams of Ca(O

Question

a.) What is the molarity of a solution prepared by dissolving 47.1 grams of Ca(OH)2 in a total solution volume of 400.0 mL?

b.) What volume of the solution in part (a) is required to prepare 5.00 L of 0.100 M Ca(OH)2?

c.) The dilute solution in part (b) is used in a titration experiment to neutralize 35.00 mL of an HClO solution. 24.48mL of 0.100 M Ca(OH)2 are required to fully neutralize the hypochlorous acid. Give the balanced chemical equation and calculate the concentration of the hypochlorous acid solution.

I understand how to find the molarity but I dont understand the rest of it. For molarity i got 1.59

Explanation / Answer

a)

mass of Ca(OH)2 = 47.1 g

moles of Ca(OH)2 = 47.1 / 74.09 = 0.6357

volume = 400 mL = 0.4 L

Molarity = moles / volume

             = 0.6357 / 0.4

Molarity = 1.59 M

b)

C1 = 1.59 M

V1 = ??

V2 = 5.0 L

C2 = 0.100 M

C1 V1 = C2 V2

1.59 x V1 = 0.1 x 5

V1 = 0.3145

volume of solution required = 314 mL

c)

2 HClO   +   Ca(OH)2   -------------> Ca(OCl)2 + 2 H2O

moles of Ca(OH)2 = 24.48 x 0.1 / 1000 = 2.448 x 10^-3

2 mol HClO   ----------> 1 mol Ca(OH)2

?? mol HClO   -----------> 2.448 x 10^-3 mol

moles of HClO = 4.896 x 10^-3

moles = molarity x volume

4.896 x 10^-3 = Molarity x 0.035

Molarity of HClO = 0.140 M

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