I have no idea how to go about this question, please help!

A clumsy chemist accidentally pours 100 mL of a 0.004 M Na2SO4 solution into a beaker containing 500 mL of 0.050 M CaCl2. Will a precipitate be formed? Explain your reasoning and/or show all calculations?

If a ppt forms it must be because Ksp is exceeded. The only slightly soluble material is the possibility of CaSO4. Therefore, you must determine Qsp for the solution and see if Ksp is exceeded.

(Na2SO4) in final solution is 0.004 x (100/600) = approx 6.67E-4
(CaCl2) in final solution is 0.050 x (500/600) = approx 0.042

So Qsp = (Ca^2+)(SO4^2-) = apprx 2.8E-5. Compare this Qsp with Ksp for CaSO4. If Ksp is exceeded (Ksp smaller than Qsp) a ppt will occur; otherwise, no ppt. You may want to go through the calculations again if it's close to make sure of the approximations. Bt the way, I have assumed that the solutions are additive in volume; i.e., 100 mL + 500 mL = 600 mL.

To determine if a precipitate will be formed, we need to compare the ion concentrations of the resulting solution with the solubility product constant (Ksp) for the possible precipitate compound.

First, let's write out the balanced chemical equation for the reaction between Na2SO4 and CaCl2:

Na2SO4 (aq) + CaCl2 (aq) -> 2NaCl (aq) + CaSO4 (s)

We can observe that NaCl remains in the solution as it is soluble, while CaSO4 may precipitate since it is less soluble.

Now, we need to calculate the ion concentrations in the resulting solution after the two solutions are mixed.

For the Na2SO4 solution:
Volume = 100 mL = 0.1 L
Molarity (M) = 0.004 M

For the CaCl2 solution:
Volume = 500 mL = 0.5 L
Molarity (M) = 0.050 M

From the balanced equation, we know that 1 mole of Na2SO4 produces 1 mole of CaSO4. Thus, the ratio between the ions is 1:1. Therefore, the concentration of Ca2+ ions after mixing will be the same as the original CaCl2 solution.

The concentration of Ca2+ ions is:

Concentration of Ca2+ = 0.050 M

Now, let's calculate the concentration of SO4^2- ions in the resulting solution using the conservation of moles.

Moles of Na2SO4 = Molarity x Volume = 0.004 M x 0.1 L = 0.0004 moles

Since the ratio between Na2SO4 and CaSO4 is 1:1, we have 0.0004 moles of CaSO4 ions.

The volume of the resulting solution is the sum of the volumes of the two solutions:

Volume = 100 mL + 500 mL = 600 mL = 0.6 L

Concentration of SO4^2- = Moles of SO4^2- / Volume
Concentration of SO4^2- = 0.0004 moles / 0.6 L = 0.00067 M

Now, we have the concentrations of Ca2+ and SO4^2- ions in the resulting solution. We can compare these concentrations with the solubility product constant (Ksp) of CaSO4, which is 4.93 x 10^-5 according to the literature.

Ksp = [Ca2+][SO4^2-]

Substituting the values, we get:

4.93 x 10^-5 = (0.050 M)(0.00067 M)

Simplifying, we find:

4.93 x 10^-5 = 3.32 x 10^-6

Since 3.32 x 10^-6 is less than the solubility product constant (4.93 x 10^-5), a precipitate of CaSO4 will form in the resulting solution.

Therefore, based on the calculations, we can conclude that a precipitate will be formed when the 0.004 M Na2SO4 solution is mixed with the 0.050 M CaCl2 solution.