-Suppose K= 4.5x10^-3 at a certain temperature for the reaction :
PCl5 <--> PCl3+ Cl3, what must be the concentration of Cl2 under these conditions?

-Chromate ion, Cro4^-2, is used as a qualitative test for lead(II) ion, forming a birght yello precipitate of lead chromate, PbCrO4. the Ksp for PbCrO4 is 2.8x10^-13. calculate the solubility in grams per liter

Can someone please walk me through these. i tried doing these but couldn't figure it out. thanks!

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1. Suppose K= 4.5x10^-3 at a certain temperature for the reaction :
PCl5 <--> PCl3+ Cl3, what must be the concentration of Cl2 under these conditions?
Is this Kc or Kp? I assume Kc. The easiest way to solve this is to set up an ICE chart, substitute the unknowns into the Kc expression, and solve for the unknown. How far can you get with that?

-Chromate ion, Cro4^-2, is used as a qualitative test for lead(II) ion, forming a birght yello precipitate of lead chromate, PbCrO4. the Ksp for PbCrO4 is 2.8x10^-13. calculate the solubility in grams per liter

PbCrO4(s) ==> Pb^+2 + CrO4^-2
If you let S = solubility of PbCrO4, then S = (Pb^+2) and S = CrO4^-2
Substitute S for those ions in the Ksp expression and solve for S. That will give you solubility in moles/L (M). Convert that to grams for g/L.

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2. Uhh. what is an ICE chart?

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3. The PbCrO4 solution I showed you is an abbreviated ICE chart. But here is how you set it up.
PCl5 ==> PCl3 + Cl2

When you do this, write
ICE off to the left, then under each of the reactants and products, fill in the chart. With these boards, spacing is a very real problem and I can't do that with horizontal lines but I can do it vertically. You should look at my vertical chart and make a horizontal one. It makes a lot of sense that way.
ICE, by the way, stands for initial, change, equilibrium. But I can't go further until I know if this is Kc or Kp. Also, there must be a beginning concn of something or a beginning pressure of something.

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4. i'm still not understanding the first question. and there's nothing saying if its a Kc or Kp it just says K. and there is no beginning pressure.

but for the second question i got 5.3x10^-7. i haven't converted that to grams per liter yet.

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5. S for PbCrO4 = 5.29 x 10^-7 M (which rounds to 5.3; however, I would leave that 5.2915 in the calculator as is and multiply by the molar mass of PbCrO4, THEN round to two significant figures.) Sometimes that makes a difference; most of the time it doesn't.

Since the first question asks for the concn of Cl2 I assume that K is Kc. But we must start with something. I can show you how if we assume a starting concn of PCl5 of 1 mol/L.
PCl5 ==> PCl3 + Cl2

Initial:
PCl5 = 1 M
PCl3 = 0 M
Cl2 = 0 M

change:
PCl3 = +x
Cl2 = +x
PCl5 = -x

equilibrium:
just add intial to change to arrive at equilibrium. You should get this.
PCl5 = 1-x
Cl2 = x
PCl3 = x

Now substitute that into the Kc expression.
Kc = 4.5 x 10^-3 = (x)(x)/(1-x) and solve for x.
x will give you PCl3 and Cl2 and 1-x will give you PCl5, all of these AT EQUILIBRIUM although the problem doesn't say that.

Here is what I was trying to describe. Ignore the ..... I've used those for spacing.
.....PCl5 ==> PCl3 + Cl2
...I..1.00 M...0 M....0 M
...C...-x M....+x M...+x M
...E..1-x M.... +x M ..+x M

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6. ok i think i might understand. my packet doesn't mention ICE, so i guess that might be a problem. but thanks for your help!

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