AP Chemistry

Consider the following reaction.
CaSO4(s) Ca2+(aq) + SO42-(aq)
At 25°C the equilibrium constant is Kc = 2.4 10-5 for this reaction.
(a) If excess CaSO4(s) is mixed with water at 25°C to produce a saturated solution of CaSO4, what are the equilibrium concentrations of Ca2+ and SO42-?
(b) If the resulting solution has a volume of 5.5 L, what is the minimum mass of CaSO4(s) needed to achieve equilibrium?

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  1. Claire, you and your alter ego, Olivia, need to learn how to make an arrow. ==> or --> or >>>.

    .........CaSO4(s) ==> Ca^2+ + SO4^2-
    I........sat'd........0.........0
    C........sat'd........x.........x
    E........sat'd.........x.........x

    Kc = (Ca^2+)(SO4^2-) = 2.4E-5
    Substitute from the ICE chart and solve for x = (Ca^2+) = (SO4^2-) in units of mols/L.

    After you have part A probably you can do part B yourself but post your work here if you need additional help.

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  2. WASSAP HOMESKILLETSi need help.. so does anyone know how much we need to know about equilibrium expressions for the AP gov and pol exam?? because my friend told me his exam last year asked him about chemical equilibrium on his exam and im afraid i might be asked this question and theni will not get a 5 on the exam and will not get into college and then wont have a job and be begging on the streets and living in a hobo shelter...

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  3. A:
    CaSO4(s) is not used in the equilibrium concentration expression because it is a solid. So, the concentrations of Ca^2+(aq) and SO4^2-(aq) are equal. The equilibrium constant, Kc, is 2.4 x 10^-5. The equation I used is 2.4 x 10^-5 = x^2, x being the concentrations of Ca^2+(aq) and SO4^2-(aq). So solve for x by finding the square root of 2.4 x 10^-5 which I found to be 4.9 x 10^-3 rounding to two significant figures. That value is the concentration for both products.
    B:
    To find the minimum mass I used the concentration found in part A: 4.9 x 10^-5 and converted it to grams of CaSO4(s) using its molar mass. Here is my conversion factor:
    4.9 x 10^-3 mol/ 1 L X 5.5 L X 136.14 g/1 mol = 3.7 g (Rounded to two significant figures.)

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