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QUESTION

# Ksp=2.4x10-5 for calcium sulfate What is the molar solubility of calcium sulfate in pure water? What is the mass solubility of calcium sulfate in pure water, expressed in g/L

Molar solubility: 4.9 * 10^(-3)"M"

Calcium sulfate, "CaSO"_4, will not dissociate completely in aqueous solution to form calcium cations, "Ca"^(2+), and sulfate anions, "SO"_4^(2-).

What actually goes on is that the solid calcium sulfate will be in equilibrium with the amounts of ions that do dissolve. The position of this equilibrium, i.e. how much of the solid will actually dissolve to form ions in aqueous solution, is determined by the magnitude of the , K_(sp).

The smaller the value of the K_(sp), the less soluble will an ionic compound be.

To find the molar solubility of the calcium sulfate, use an ICE table

"CaSO"_text(4(s]) " "rightleftharpoons" " "Ca"_text((aq])^(2+) " "+" " "SO"_text(4(aq])^(2-)

color(purple)("I")" " " " " " - " " " " " " " " " " "0" " " " " " " " " "0 color(purple)("C")" " " " " " - " " " " " " " "(+s)" " " " "(+s) color(purple)("E")" " " " " " - " " " " " " " " " "s" " " " " " " "s

By definition, K_(sp) is equal to

K_(sp) = ["Ca"^(2+)] * ["SO"_4^(2-)]

K_(sp) = s * s = s^2

This means that the molar solubility of calcium sulfate is

s = sqrt(K_(sp)) = sqrt(2.4 * 10^(-5)) = color(green)(4.9 * 10^(-3)"M")

To get the mass solubility of calcium sulfate, use the compound's molar mass

4.9 * 10^(-3)color(red)(cancel(color(black)("moles")))/"L" * "136.141 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.67 g/L")

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