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
.........................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.
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!
yes
To determine if a precipitate will form in each of the beakers, we need to compare the ion product (IP) of Ca(OH)2, known as Ksp, with the ion product (IP) of the potential precipitate formed with the added reagent.
First, let's write the balanced equation for the dissolution of Ca(OH)2:
Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH-(aq)
The Ksp expression for Ca(OH)2 is:
Ksp = [Ca2+][OH-]^2
Given that the Ksp of Ca(OH)2 is 6.5×10−6, we can calculate the concentration of Ca2+ and OH- ions in the solution.
Since Ca(OH)2 is sparingly soluble, we assume that both Ca2+ and OH- ions come from the dissociation of Ca(OH)2.
Now, let's examine each beaker:
1. 50 mL 1.0 M HCl
Since HCl is a strong acid and fully dissociates, we can assume that it will provide a high concentration of H+ ions. However, H+ ions will react with OH- ions to form water molecules, reducing the concentration of OH- ions. Therefore, no precipitate will form in this scenario.
2. 50 mL 1.0 M NaCl
NaCl is a salt that fully dissociates into Na+ and Cl- ions. Neither of these ions will react with Ca2+ or OH- ions, so no precipitate will form.
3. 50 mL 1.0 M CaCl2
CaCl2 dissociates into Ca2+ and 2Cl- ions. The concentration of Ca2+ ions from CaCl2 is higher than the concentration of Ca2+ ions from Ca(OH)2. Since Ca2+ ions and OH- ions react to form Ca(OH)2, a precipitate will form in this scenario.
4. 50 mL 0.10 M CaCl2
The concentration of Ca2+ ions from CaCl2 is lower than in the previous scenario. Therefore, the concentration of Ca2+ ions from Ca(OH)2 will exceed the solubility product, resulting in the formation of a precipitate.
In summary, a precipitate will form in beakers 3 and 4 when 50 mL of the Ca(OH)2 solution is added.