Ca(OH)2 has a Ksp of 6.5×10−6. If 50 mL of the solution from part A (0.370 g Ca(OH)2 added to 500 mL of water) is added to each of the beakers shown here, in which beakers, if any, will a precipitate form?

Question

Ca(OH)2 has a Ksp of 6.5×10−6. If 50  mL of the solution from part A (0.370 g Ca(OH)2 added to 500 mL of water) is added to each of the beakers shown here, in which beakers, if any, will a precipitate form?

1. 50 mL 1.0 M HCl   
2. 50 mL 1.0 M NaCl   
3. 50 mL 1.0 M CaCl2   
4. 50 mL 0.10 M CaCl2

DrBob222 · Accepted Answer

.........................Ca(OH)2(s) ==> Ca^2+ + 2OH^-
Ksp = (Ca^2+)(OH^-)^2 = 6.5E-6
To answer if a ppt will occur when two solutions are mixed you plug in (Ca^2+) and (OH^-)^2 in the mixed solution. If the product is larger than Ksp you get a ppt. If the product is smaller than Ksp no ppt will be formed. Here is how you do part c (#3) of the problem.
Concn of part A solution =?
I rounded the molar mass Ca(OH)2 to 74
mols = grams/molar mass = 0.370/74 = 0.005 moles.
Molarity = M = moles/L of solution = 0.005 moles/0.500 L = 0.01 M Ca(OH)2. So you know (Ca^2+) = 0.01 M and (OH^-) = 0.02 M in that solution. All of the Ca(OH)2 solid has been dissolved.

You mix 50 mL of this with 50 mL of 1.0 M CaCl2 so that the total (Ca^2+) = 1.0 + 0.01 = 1.01 M. In the newly mixed solution you have 50 mL of the above solution with 50 mL of 1.0 M CaCl2. So what are the final concentrations?
(Ca^2+) from the part A + added CaCl2 has been diluted from 50 mL to 100 mL.
1.01 M x (50 mL/100 mL) = 0.505 M.
(OH^-) = 0.02 x (50 mL/100 mL) = 0.01M
Qsp= (Ca^2+)(OH^-)^2 = (0.505)(0.01)^2 = 5.05E-5 and this is larger than 6.5E-6 so ppt will occur.
#4 is done the same way.
You can go throught the calculations for #1 and #2 but that's extra work. You should realize for #1 that you're adding an acid to a base so they neutralize each other and you're forming CaCl2 by adding the acid so you should realize Ca(OH)2 can't ppt. For #2, you're adding NaCl and that has nothing in it to ppt with Ca(OH)2 so a ppt can't occur. Post your work if you get stuck.

mina · Answer

So in the end, beaker 3 with 1.0 M CaCl2 will form a precipitate with 0.370 g Ca(OH)2 added to 500 mL of water because the calculated Qsp of Ca(OH)2 is greater than Ksp of Ca(OH)2. Thank you DrBob!

DrBob222 · Answer

yes