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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Boxxxed on March 16, 2011, 11:42:32 PM

Title: Simple Solubility Problem
Post by: Boxxxed on March 16, 2011, 11:42:32 PM

25 mg of RaSO4 is stirred in 100 ml of water. How much dissolves? ksp=4.2x10^-11

RaSO4 ---> Ra + SO4

25mg = 0.025 g

moles = 0.025/322.09g/mol = 7.76x10^-5

7.76x10^-5 / 0.1 L = 7.76x10^-4 M

Q = (Ra)(SO4) = (7.76x10^-4)(7.76x10^-4) = 6.02x10^-7 M

Equilibrium Concentrations = Sqroot ksp = 6.48x10^-6 M can dissolve

n=c*v
6.48x10^-6 M * 0.1 L = 6.48x10^-7

6.48x10^-7 moles * 322.06 = 0.000209 grams = 0.21 mg dissolves.


If I wanted to figure out how much precipitates I just subtract the dissolved from the total?

7.76x10^-5 M - 6.48x10^-6 M = 7.11x10^-5 M * Volume = 7.11x10^-6 moles precipitated * molar mass = 0.00229g = 2.3 mg precipitated.

Why doesn't this value = 25-0.21 = 24.79 mg?

Edit, figured it out. Typo with exponent.

Title: Re: Simple Solubility Problem
Post by: Nobby on March 17, 2011, 12:00:50 PM

I would calculate in this way.

K = c(Ra²⁺) * c(SO₄^2-) = 4,2 *10^-11 mol²/l²

c(Ra²⁺) = c(SO₄^2-)

c(Ra²⁺) = SQR(4,2 *10^-11 mol²/l²) = 6,48 *10^-6 mol/l

In 100 ml we have then 6,48 *10^-7 mol

Calculated with the molecular weight of Radium sulfate 322,06 g/mol we get 2,09 *10^-4 g = 0,21 mg

25 mg is used what means 24,79 mg left.