Solubility Product Constant

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[chrisf]

10.0mL of 0.10M Pb(NO₃)₂ is mixed with 10.0mL of 0.10M NaCl. After equilibrium is attained, it is determined that the concentration of Cl^- in solution is 0.0218M. Calculate the K_sp of PbCl₂.

Pb(NO₃)₂ + 2NaCl -> PbCl₂ + 2NaNO₃

(0.10mol Pb(NO₃)₂ / 1.0L) * 0.010L = 0.001mol Pb(NO₃)₂

(0.10mol NaCl / 1.0L) * 0.010L = 0.001mol NaCl

	Pb(NO3)2	2NaCl	PbCl2	2NaNO3
Initial	0.001mol	0.001mol	0.000mol	0.000mol

And... I have no idea where to go after that.

I realize K_sp = [Pb⁺][2Cl^-] but I'm at a loss where to go.

A sample of solid CaF₂ is placed into a test tube of water and allowed to reach equilibrium. Calculate the equilibrium concentrations of Ca²⁺ and F^-, given the K_sp of CaF₂ = 3.9 x 10^-11

Yay. I figured out this one this morning. I was WAY overcomplicating it.

3.9x10^-11 = [Ca²⁺][F^-]²
3.9x10^-11 = (x)(2x)²
3.9x10^-11 = 4x³
x = 2.14x10^-4

[Ca²⁺] = x = 2.14x10^-4
[F^-] = 2x = 2(2.14x10^-4) = 4.27x10^-4

[spock]

Let's start by getting rid of the spectator ions:
                                Pb+2(aq)           +   2Cl-(aq)   -->   PbCl2 (s)
Then let's do an ICE diagram (moles of Cl- at equil = .028X.020L= .00056mol)
 Initial                          0.001mol             0.001mol
 Change                           -x                     -2x
 Equil                            .001 -x               .00056

Moles of Cl- reacted = .001 - .00056 = .00044mol = 2x

Moles of Pb reacted = .00044/2 = .00022 mol Cl- (aq)
Moles of Pb remaining in solution at equil = .001 - .00044 = .00056 mol Pb+2 (aq)

Ksp = [Pb+2] [Cl-]^2  =  (.00056) (.00022)^2

[chrisf]

Doing it that way comes out with K_sp ~ 2.71x10^-11 which is pretty far off from 1.7x10^-5 (which is approx what it should be).

Hmm, I'll have to think about this one some more.

[chrisf]

I think I figured it out. 

Pb(NO₃)₂ + 2NaCl -> PbCl₂ + 2NaNO₃

Moles of NaCl = 0.10mol/1L * 0.010L = 0.001mol NaCl or 0.001mol Na⁺ and 0.001mol Cl^-

Moles of Cl^- at equilibrium = 0.0218mol/1L * 0.020L = 0.000436mol Cl^-

Moles of Cl^- reacted = 0.001mol Cl^- - 0.000436mol Cl^- = 0.000564mol Cl^-

Moles of Pb reacted = 0.000436mol Pb⁺/2 = 0.000218mol Cl^- <-- Should this be 0.000564mol/2 instead?

Moles of Pb left in solution = 0.001mol Pb⁺ - 0.000218mol = 0.000782mol Pb⁺

K_sp = [Pb⁺][Cl^-]²

[Cl^-] = 0.000436mol/0.020L = 0.0218M
[Pb²⁺] = 0.00782mol/0.020L = 0.0391M

K_sp = (0.0391)(0.0218)² = 1.86x10^-5

That's close enough to 1.7x10^-5 I think (hope).

[spock]

Your right, I forgot to convert the moles of Pb+2 and Cl- back into molarity by dividing by the volume.

Good job!

[Kagome]

hi chrisf, A sample of solid CaF2 is placed into a test tube of water and allowed to reach equilibrium. Calculate the equilibrium concentrations of Ca2+ and F-, given the Ksp of CaF2 = 3.9 x 10-11

looked very similar to my homework but the only thing is mine is asking for CAF2 not them separate so how would i find the answer to it?

[chrisf]

hi chrisf, A sample of solid CaF2 is placed into a test tube of water and allowed to reach equilibrium. Calculate the equilibrium concentrations of Ca2+ and F-, given the Ksp of CaF2 = 3.9 x 10-11

looked very similar to my homework but the only thing is mine is asking for CAF2 not them separate so how would i find the answer to it?

CaF2 is a solid. You can't have a concentration of a solid.

[Kagome]

The mineral fluorite is composed of CAF2. The molar solubility of calcium fluoride in water is 2.1X106-4
moles/L. what is the Ksp of CaF2?

thats what my equation

[chrisf]

K_sp = [Ca²⁺][F^-]²

K_sp = (2.1X10^-4)(4.2X10^-4)²

K_sp = 3.70x10^-11

[Kagome]

hi yes! thats how I thought you did that one thank you!
ok I hope you dont mind answering one more question for me
what ratio of benzoate ion to benzoic acid would be required to prepare a buffer with a pH of 7.20?
Ka(C6H5COOH)=6.5X10-4

ok what I did so far was find H+ since they gave you the pH and I got 6.30X10-8
I found OH by dividing (1X10-14)/(6.30X10-8) to get 1.58X-7
so I was woundering if Im suppose to conugate base/acid(ka)??

[chrisf]

I'll try to get back to you on this later today or tomorrow, but first write out your complete reaction.

[spock]

The system you are describing is a buffer made up of benzoic acid (a weak acid = HBenz), and its conjugate base (benzoate ion = Benz-)

Using the symbols above for benzoic acid and benzoate ion, here is the equation:

               HBenz   <===>  H+    +  Benz-

The Ka expression for the above equilibrium is
                 Ka  =   [H+]  [Benz-] / [HBenz]

Rearranging to solve for [H+] we get
                  [H+] =  Ka ([HBenz]/[Benz-]

What this tells us is that the [H+] (and thus the pH is dependent on two things:
                   1)  the Ka of the acid  (which is fixed)
                   2)  the ratio of the concentration of the benzoic acid to benzoate ion (which can be varied.

Since we know the [H+] (you found that to be 6.30 x 10-8 above), you can look up the Ka value for benzoic acid, and then calculate the ratio:
                    [Benz-]/[HBenz]  =   Ka/ [H+]

[Kagome]

hi spock, this was homework, and my teacher let us know we would be given ka, and it wasn't on my sheet. since I got H+ what do I do next? do I find OH- too?
and Im looking at your equation of  [Benz-]/[HBenz]  =   Ka/ [H+]
couldn't I just use 6.30X10-8 (for the H+ HBenz) and the OH-Benz to get the ka??

[spock]

You are given the Ka in the problem:  6.5 x 10-5   (I think you have a typo as I check the Ka in a textbook and found it listed as 6.5 x10-5 rather than 6.5 x 10-4)

You found the [H+] to be 6.30 x 10-8:

Since the ratio of [Benz-]/[HBenz] = Ka/[H+] it must be equal to

                       6.5 x 10-5/6.30 x 10-8 = 1031

That seems like a very high ratio for an effective buffer.  Buffers are usually pretty close to a 1:1 ratio (say .5M Benz- to .5M HBenz).  In this case you would need something like .5M Benz to .00485 M HBenz.

You should be aware that [H+] is not equal to [HBenz] in a buffer.  Nor is [OH-]=[Benz-].  So, while you could find the [OH-], it is not useful in finding the Ka value in this case.

[chrisf]

I think I figured it out. 

Pb(NO₃)₂ + 2NaCl -> PbCl₂ + 2NaNO₃

Moles of NaCl = 0.10mol/1L * 0.010L = 0.001mol NaCl or 0.001mol Na⁺ and 0.001mol Cl^-

Moles of Cl^- at equilibrium = 0.0218mol/1L * 0.020L = 0.000436mol Cl^-

Moles of Cl^- reacted = 0.001mol Cl^- - 0.000436mol Cl^- = 0.000564mol Cl^-

Moles of Pb reacted = 0.000564mol Pb⁺/2 = 0.000282mol Pb²⁺

Moles of Pb left in solution = 0.001mol Pb⁺ - 0.000282mol = 0.000718mol Pb⁺

K_sp = [Pb⁺][Cl^-]²

[Cl^-] = 0.000436mol/0.020L = 0.0218M
[Pb²⁺] = 0.00718mol/0.020L = 0.0359M

K_sp = (0.0359)(0.0218)² = 1.70x10^-5

I was looking over this problem again yesterday and I noticed a problem. Now it's exactly right.