A sample of solid Ca(OH)2 was stirred in water at a certain temperature until the solution...

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A sample of solid Ca(OH)2 was stirred in water at a certaintemperature until the solution contained as much dissolved Ca(OH)2as it could hold. A 78.7-mL sample of this solution was withdrawnand titrated with 0.0696 M HBr. It required 68.0 mL of the acidsolution for neutralization.
(a) What was the molarity of the Ca(OH)2 solution?
_M
(b) What is the solubility of Ca(OH)₂ in water, atthe experimental temperature, in grams of Ca(OH)₂ per100 mL of solution?

_g/100mL

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4.2 Ratings (687 Votes)

Ca(OH)2 + 2HBr -----> CaBr2 + H2O

1 moleÂ Â Â Â Â Â 2 moles

Ca(OH)2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â HBr

molarity M1 =Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â molarity M2 = 0.0696M

volume V1 = 78.7 mlÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â volume V2 = 68ml

no of moles n1 = 1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â no of moles n2 = 2

M1V1/n1Â Â Â Â = M2V2/n2

M1Â Â = M2V2n1/n2V1

Â Â Â Â Â = 0.0696*68*1/2*78.7 = 0.032 M

solubility of Ca(OH)2 = 0.032moles/L

molar mass of Ca(OH)2Â Â = 74g/mole

solubility of Ca(OH)2 = 0.032*74/1000ml

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.2368 g/100ml

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