Question

The solubility constant of calcium fluoride CaF2 is 4.0x10^-11 will 10.0mL
Will of 0.01M cacl2 and 20.0 mL of 0.01M HF to added sufficient water to make a 1.0L solution. Would CaF2 precipitate because of this?

Yes
No
Yes if temperature is increased
There is insufficient data.

Answer 1

Can someone tell me whether this is the correct logic towards approaching this problem?

\[CaF _{2} --> Ca ^{2+} +2F ^{-}\]

\[Ksp = [x][2x]^{2}\]

Answer 2

yeah that's right I guess !

Answer 3

I'm not sure if it is approaching or not but its correct

Answer 4

What's the volume of CaCl2 in the addition of HF?

Answer 5

They did not give it

Answer 6

\[10.0 mL/1000*CaCl _{2} * (0.01) M = 1*10^-4 mol CaCl _{2} \]

\[20.0mL/1000*(0.01mol/L) HF = 2x10\]

Answer 7

2x10^-4 mol of HF

Answer 8

need to figure out the concentrations of both ions

Answer 9

but they said that the volume was diluted to 1L.

Answer 10

HF --> H + F

CaCl2 --> Ca + 2Cl

1 mol of Ca ions.
1 mol of F ions from those equations. i'm guessing that this is the next step.

so we do

1x10^-4 mol/1L = 10^-4 M Ca ions
2x10^-4 mol/1L = 2x10^-4 M F ions

Answer 11

\[ Ksp=[x][2x]2\]

[1x10^-4M][2(2x10^-4)]^2 = Qsp

(1.6x10^-7)(1x10^-4) = 1.6x10^-11

Answer 12

\[Qsp = 1.6x10^-11 \]

Answer 13

Ksp = 4.1x10^-11

Answer 14

So its gonna be a no

Answer 15

\[Qsp < Ksp \] so no it won't.

Answer 16

Yeah, this was a long problem but I think what you had to do was to write out the reactions and keep track of how many ions you had total for each.

Answer 17

1. you multiply the Molarity of each by the volume to get the number of moles.
2. you find the new concentration of each compound (new volume) for 1L
3. then you figure out how many total moles of ions you have, in this case for each of them it's one.
4. after that you set up the Ksp for CaF2 which was [x][2x]^2
5. plug in and solve.

Answer 18

You also realize that Qsp < Ksp which means that no precipitate will form.

if you raised the temperature you'll change the value of your Ksp and if you don't add any more of those compounds nothing will happen as a result.

it can't be yes.

Ksp is an equilibrium value, so with all equilibrium values they will be temperature dependent.

Answer 19

But the general idea behind this question was that you didn't really have enough of both substance to really cause precipitate to form.

for that Qsp >Ksp for that to happen.

Answer 20

Qsp is the [ ] of ions at any point while Ksp is the equilibrium value.