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.
Why don't you post your work and let us find the error.
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...
I don' see anything wrong with this. I calculated OH using 0.6 x 0.002/1.002 and came up with a slightly smaller number but it rounds to your value and if I square that and multiply by Ca (I used 0.850 x 1/1.002) = 0.848 but no matter how I slice it the answer always rounds to 1.22E-6. If the only thing you are keying in is the answer of 1.22E-6 then I think the database must be wrong. Do they want unit? Technically there are no units although I see a lot of prof using them.
To determine the reaction quotient, Q, you need to write the balanced chemical equation for the reaction between NaOH and CaCl2. The balanced equation is:
2NaOH + CaCl2 -> 2NaCl + Ca(OH)2
Looking at the equation, we can see that two moles of NaOH react with one mole of CaCl2 to produce two moles of NaCl and one mole of Ca(OH)2.
Next, we need to calculate the concentrations of the reactants and products at equilibrium. To do this, we will use the given volumes and concentrations.
Given:
Volume of NaOH (V1) = 2.00 mL = 2.00 x 10^-3 L
Concentration of NaOH (C1) = 0.600 M
Volume of CaCl2 (V2) = 1.000 L
Concentration of CaCl2 (C2) = 0.850 M
First, we need to convert the volume of NaOH to liters:
V1 = 2.00 x 10^-3 L
Next, we can calculate the moles of NaOH using the formula:
moles = volume (L) x concentration (M)
moles of NaOH = V1 x C1 = (2.00 x 10^-3 L) x (0.600 M) = 1.20 x 10^-3 mol
Since two moles of NaOH react with one mole of CaCl2, we divide the moles of NaOH by 2 to find the moles of CaCl2:
moles of CaCl2 = 1.20 x 10^-3 mol / 2 = 6.00 x 10^-4 mol
Now we can calculate the concentrations of NaCl and Ca(OH)2 at equilibrium. The volume and concentration of CaCl2 do not change, so the concentration of CaCl2 at equilibrium remains 0.850 M.
For NaCl, we multiply the number of moles (1.20 x 10^-3 mol) by 2 and divide by the total volume in liters (V1 + V2):
Concentration of NaCl = (2 x (1.20 x 10^-3 mol)) / (2.00 x 10^-3 L + 1.000 L) = 2.40 x 10^-3 mol / 1.002 L = 2.39 x 10^-3 M
For Ca(OH)2, we multiply the number of moles (6.00 x 10^-4 mol) by 1 and divide by the total volume in liters (V1 + V2):
Concentration of Ca(OH)2 = (1 x (6.00 x 10^-4 mol)) / (2.00 x 10^-3 L + 1.000 L) = 6.00 x 10^-4 mol / 1.002 L = 5.99 x 10^-4 M
Finally, we can calculate the reaction quotient, Q, by multiplying the concentrations of the products and dividing by the concentrations of the reactants:
Q = (C(NaCl))^2 x C(Ca(OH)2) / (C(NaOH))^2 x C(CaCl2)
Q = (2.39 x 10^-3 M)^2 x (5.99 x 10^-4 M) / (0.600 M)^2 x (0.850 M)
Q = 2.28 x 10^-8
If the value of the reaction quotient, Q, is less than the solubility product constant, Ksp, precipitation will not occur. However, to determine whether precipitation will occur, we need to compare Q and Ksp for Ca(OH)2.