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?

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.)

(a) Well, it looks like quite a party for CaSO4! So, let's see. Since the solution is saturated with CaSO4, it means that CaSO4 is at its maximum solubility at that temperature. That being said, the concentration of Ca2+ and SO42- in the saturated solution will be determined by the equilibrium constant.

Since the equilibrium constant (Kc) is given as 2.4 × 10-5, we can set up an equation using the stoichiometry of the reaction:

Kc = [Ca2+][SO42-]/[CaSO4]
2.4 × 10-5 = [Ca2+][SO42-]/[CaSO4]

Now, since we have excess CaSO4, its concentration will remain constant and can be represented by [CaSO4]. Let's say it's x (mol/L).

Substituting these values into the equation, we get:
2.4 × 10-5 = ([Ca2+][SO42-])/x

Now, since x is a constant, we can rewrite the equation as:
2.4 × 10-5 * x = [Ca2+][SO42-]

So, the equilibrium concentrations of Ca2+ and SO42- will be 2.4 × 10-5 * x.

(b) Now, let's calculate the minimum mass of CaSO4 needed to achieve equilibrium. To do this, we need to first determine the moles of CaSO4 required.

We know the volume of the solution is 5.5 L, so we can use the given concentration of CaSO4 (which is also the same as its solubility) to calculate the moles:
x (mol/L) * 5.5 L = moles of CaSO4

Once we have the moles, we can calculate the minimum mass of CaSO4 using its molar mass. So, the minimum mass of CaSO4 needed would be:
moles of CaSO4 * molar mass of CaSO4 = minimum mass of CaSO4

Voila! That's how you calculate the equilibrium concentrations of Ca2+ and SO42- and the minimum mass of CaSO4 needed to achieve equilibrium. Now, let's get this party started!

To answer these questions, we need to use the given equilibrium constant (Kc) and apply the concept of equilibrium expressions, as well as the stoichiometry of the reaction.

(a) To find the equilibrium concentrations of Ca2+ and SO42-, we can assume x moles of CaSO4 dissolve in water to form x moles of Ca2+ and x moles of SO42-. Since the stoichiometry of the reaction is 1:1 between CaSO4 and each ion, the equilibrium concentrations of Ca2+ and SO42- will also be x.

Using the equilibrium constant expression:
Kc = [Ca2+][SO42-]

Substituting the equilibrium concentrations with x:
2.4 x 10^-5 = x * x

Simplifying:
2.4 x 10^-5 = x^2

Solving for x by taking the square root (ignoring the negative root, as concentrations cannot be negative):
x = √(2.4 x 10^-5) = 1.55 x 10^-3

Therefore, the equilibrium concentrations of Ca2+ and SO42- are both 1.55 x 10^-3 M.

(b) To find the minimum mass of CaSO4 needed to achieve equilibrium, we can use the equation:

moles of CaSO4 = volume of solution (in liters) * concentration of CaSO4

Since the resulting solution has a volume of 5.5 L and the concentration of CaSO4 is 1.55 x 10^-3 M, we can calculate the moles of CaSO4 required:

moles of CaSO4 = 5.5 L * 1.55 x 10^-3 M = 8.53 x 10^-3 moles

The molar mass of CaSO4 is 136.14 g/mol, so we can now calculate the minimum mass of CaSO4:

mass of CaSO4 = moles of CaSO4 * molar mass of CaSO4 = 8.53 x 10^-3 moles * 136.14 g/mol ≈ 1.16 g

Hence, the minimum mass of CaSO4 needed to achieve equilibrium in a 5.5 L solution is approximately 1.16 grams.

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.

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...