The Ksp for BaCrO4 is 1.2X10-10. Will BaCrO4 precipitate when 10 mL of 1 X10-5M Ba(NO3)2 is mixed with 10 mL of 1X10-3M K2CrO4?

I am not sure where to start

You want to calculate Qsp and compare that with Ksp.

Ba(NO3)2 + K2CrO4
You are given concns before mixing.
After mixing each is diluted by factor of 2. New concns are
(Ba^2+) = 1E-5M x (10 mL/20 mL) = ?
(CrO4^2-) = 1E-3 x (10 mL/20 mL) = ?

Plug those new concns and calculate Qsp. If Qsp > Ksp there will be a ppt; if Qsp < Ksp there will be no ppt.

Well, let's put on our detective hats and solve this precipitating mystery!

Step 1: Write the balanced chemical equation. In this case, it would be:
Ba(NO3)2 + K2CrO4 → BaCrO4 + 2 KNO3

Step 2: Calculate the concentration of ions in the mixture.
For Ba(NO3)2:
Ba2+ concentration = 10 mL × 1 × 10^-5 M = 1 × 10^-7 mol/L

For K2CrO4:
CrO42- concentration = 10 mL × 1 × 10^-3 M = 1 × 10^-5 mol/L

Step 3: Use the solubility product constant (Ksp) to determine if precipitation occurs.
Ksp = [Ba^2+][CrO42-]
Ksp = (1 × 10^-7)(1 × 10^-5) = 1 × 10^-12

Step 4: Compare the calculated Ksp value with the given Ksp value (1.2 × 10^-10). Since the calculated Ksp value (1 × 10^-12) is smaller than the given Ksp value, precipitation will not occur.

So, my detective friend, BaCrO4 will not be donning its precipitation robe in this case. It remains happily dissolved in solution, spreading its chromate cheer!

To determine if BaCrO4 will precipitate, we need to compare the product of the ion concentrations (Q) with the solubility product constant (Ksp). Here's how to calculate it step by step:

Step 1: Write the balanced equation for the dissociation of BaCrO4:
BaCrO4(s) ↔ Ba2+(aq) + CrO4^2-(aq)

Step 2: Calculate the initial concentrations of Ba2+ and CrO4^2- ions. Both solutions have equal volumes of 10 mL (0.01 L), so we can assume that the final volume is 20 mL (0.02 L).

For Ba(NO3)2 solution:
Initial concentration of Ba2+ = 1 X 10^-5 M
Therefore, [Ba2+] = 1 X 10^-5 M

For K2CrO4 solution:
Initial concentration of CrO4^2- = 1 X 10^-3 M
Therefore, [CrO4^2-] = 1 X 10^-3 M

Step 3: Calculate the values of Q, which is the product of the ion concentrations:
Q = [Ba2+] x [CrO4^2-]
= (1 X 10^-5 M) x (1 X 10^-3 M)
= 1 X 10^-8

Step 4: Compare the value of Q with the solubility product constant (Ksp):
Ksp = 1.2 X 10^-10

If Q > Ksp, then the product (BaCrO4) will precipitate. If Q < Ksp, then it will remain dissolved.

In this case, since Q (1 X 10^-8) is less than Ksp (1.2 X 10^-10), BaCrO4 will not precipitate. It will remain dissolved in the solution.

To determine if BaCrO4 will precipitate when the solutions are mixed, we need to compare the ion product (Q) to the solubility product (Ksp) for this compound.

1. First, let's write the balanced equation for the dissociation of BaCrO4:
BaCrO4(s) ⇌ Ba2+(aq) + CrO42-(aq)

2. The solubility product expression (Ksp) is written by multiplying the concentrations of the dissociated ions raised to the power of their stoichiometric coefficients. In this case, it is:
Ksp = [Ba2+][CrO42-]

3. We are given the Ksp value as 1.2x10^-10. Since the Ksp expression only contains the concentrations of the ions, we need to find their concentrations in the final solution.

4. To do this, we can use the concept of stoichiometry and the given initial concentrations of Ba(NO3)2 and K2CrO4.

5. The total volume of the final solution is given as 10 mL + 10 mL = 20 mL = 0.02 L.

6. First, let's calculate the number of moles of Ba2+ ions present in the 10 mL of 1x10^-5 M Ba(NO3)2.
Moles of Ba2+ = concentration × volume = 1x10^-5 M × 0.01 L = 1x10^-7 mol.

7. Next, let's calculate the number of moles of CrO42- ions present in the 10 mL of 1x10^-3 M K2CrO4.
Moles of CrO42- = concentration × volume = 1x10^-3 M × 0.01 L = 1x10^-5 mol.

8. Since both solutions have the same volume, the final concentrations of Ba2+ and CrO42- ions will be the same in the mixed solution.

9. Therefore, the concentrations of Ba2+ and CrO42- ions in the mixed solution will be:
[Ba2+] = [CrO42-] = moles/volume = (1x10^-7 mol + 1x10^-5 mol)/0.02 L = 5x10^-6 M.

10. Now we can calculate the ion product, Q, using these concentrations:
Q = [Ba2+][CrO42-] = (5x10^-6 M) × (5x10^-6 M) = 2.5x10^-11.

11. To determine if BaCrO4 will precipitate, we compare Q to Ksp. If Q is greater than Ksp, the compound will precipitate. If Q is less than Ksp, no precipitate will form.

12. In this case, since Q (2.5x10^-11) is less than Ksp (1.2x10^-10), BaCrO4 will not precipitate.

Therefore, BaCrO4 will not precipitate when 10 mL of 1x10^-5 M Ba(NO3)2 is mixed with 10 mL of 1x10^-3 M K2CrO4.