One other question that I cannot figure out what I'm doing wrong is this..

If .002 L of .650 M NaOH are added to 1.0 L of 0.9 M CaCl2 what is the Q value?

I keep getting 0.0011 because I thought the answer was [Cu+][OH-] but that's not correct

The Q value of what? And I don't see any Cu^+2 in the problem.

Sorry I meant Ca+ not Cu (typo)

and Q value as is quotient reaction value. Q = K at equilibrium, it measures if a precipitate will form

To find the Q value, you need to determine the concentrations of the ions involved in the reaction and then use those concentrations to calculate Q.

In this case, the reaction is between NaOH (sodium hydroxide) and CaCl2 (calcium chloride), which produce NaCl (sodium chloride) and Ca(OH)2 (calcium hydroxide):

2 NaOH + CaCl2 -> 2 NaCl + Ca(OH)2

First, determine the number of moles of NaOH and the number of moles of CaCl2 that are added to the solution. You can do this using the equation:

moles = concentration (M) x volume (L)

For NaOH:
moles of NaOH = 0.650 M x 0.002 L = 0.0013 mol

For CaCl2:
moles of CaCl2 = 0.9 M x 1.0 L = 0.9 mol

Next, consider the stoichiometry of the reaction. From the balanced equation, you can see that 2 moles of NaOH react with 1 mole of CaCl2. Therefore, the moles of NaOH and CaCl2 are in a 2:1 ratio.

Since you have 0.0013 moles of NaOH, you can calculate the moles of CaCl2 using the ratio:

moles of CaCl2 = 0.0013 mol x (1 mol CaCl2 / 2 mol NaOH) = 0.00065 mol

Now that you have the moles of NaOH and CaCl2, you can calculate the Q value. The Q value is determined using the concentrations of the products and reactants raised to the power of their stoichiometric coefficients.

In this case, the Q value can be calculated using the following equation:

Q = [(concentration of NaCl)^2 x (concentration of Ca(OH)2)] / [(concentration of NaOH)^2 x (concentration of CaCl2)]

Now, calculate the concentrations of NaCl, Ca(OH)2, NaOH, and CaCl2 using the moles and the final volume of the solution (1.002 L):

For NaCl:
concentration of NaCl = (moles of NaCl) / (volume of solution) = 0 / 1.002 = 0 M

For Ca(OH)2:
concentration of Ca(OH)2 = (moles of Ca(OH)2) / (volume of solution) = 0 / 1.002 = 0 M

For NaOH:
concentration of NaOH = (moles of NaOH) / (volume of solution) = 0.0013 mol / 1.002 L = 0.00129 M

For CaCl2:
concentration of CaCl2 = (moles of CaCl2) / (volume of solution) = 0.00065 mol / 1.002 L = 0.00065 M

Finally, substitute these concentrations into the Q equation:

Q = [(0 M)^2 x (0 M)] / [(0.00129 M)^2 x (0.00065 M)] = 0

Therefore, the Q value for this reaction is 0.