If 2.00 mL of 0.600 M NaOH are added to 1.000 L of 0.850 M CaCl2, what is the value of the reaction quotient and will precipitation occur?

I've been trying this problem over and over and haven't been able to get it. I know precipitation will not occur from my previous answer and that part was correct but not my reaction quotient.

Ksp for Ca(OH)2 is 5.02 x 10^-8.

The way to predict whether a precipitation will occur is to calculate Qsp and compare to Ksp.

Qsp = (Ca2+)(OH-)^2. Concentration of OH- will be given by (M1V1/V2) = 0.0012

Qsp = 0.850 * (.0012)^2 = 1.22*10^-6

My homework says this is not correct...

To determine the value of the reaction quotient (Qsp) and whether precipitation will occur in this scenario, you need to calculate the concentrations of Ca2+ and OH- ions after the reaction between NaOH and CaCl2.

First, let's calculate the amount (in moles) of NaOH added:
0.600 M NaOH = 0.600 moles/L * 0.002 L = 0.0012 moles

Since NaOH is a strong base and completely dissociates in water, it will yield 0.0012 moles of OH- ions.

Now, let's calculate the concentrations of Ca2+ and OH- ions after the reaction:

Ca2+ concentration after the reaction:
Initial concentration = 0.850 M CaCl2
Volume = 1.000 L CaCl2
Therefore, the concentration of Ca2+ ions remains unchanged.

OH- concentration after the reaction:
Initial concentration = 0.0012 moles / 1.002 L (total volume after adding NaOH)
Concentration = 0.0012 moles / 1.002 L = 0.0011988 M

Now, let's calculate the value of the reaction quotient (Qsp):
Qsp = [Ca2+][OH-]^2
Substituting the values:
Qsp = (0.850 M)(0.0011988 M)^2
Qsp = 1.207 x 10^-6

Comparing Qsp to the Ksp value for Ca(OH)2 (5.02 x 10^-8), we find that Qsp > Ksp. This means that the solution is already saturated with Ca(OH)2 and precipitation will occur.

Please note that the value you calculated for Qsp in your question was close to the correct answer, but the slight difference may be due to rounding errors during calculations.