-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!
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.
Uhh. what is an ICE chart?
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.
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.
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
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!
Sure, I can walk you through these problems. Let's start with the first one:
1. For the reaction PCl5 <--> PCl3 + Cl2, the equilibrium constant K is given as K = [PCl3][Cl2] / [PCl5].
To find the concentration of Cl2, we need to know the concentrations of PCl3 and PCl5. However, the problem statement does not provide these concentrations. Therefore, we cannot directly solve for the concentration of Cl2.
If the initial concentrations of PCl3 and PCl5 are given, we could use stoichiometry or an ICE (initial, change, equilibrium) table to determine the equilibrium concentrations. However, without any initial concentrations, we cannot proceed with a numerical calculation.
Moving on to the second problem:
2. The solubility product constant (Ksp) for a salt is an equilibrium constant that tells you how much of a solid salt will dissolve in a solvent to form ions. The equation for the dissolution of lead chromate (PbCrO4) can be written as:
PbCrO4 ⇌ Pb^2+ + CrO4^2-
The Ksp expression for this reaction is Ksp = [Pb^2+][CrO4^2-].
The problem asks us to calculate the solubility of PbCrO4 in grams per liter. Solubility is usually expressed in terms of molarity, so we need to convert grams per liter to molarity. To do this, we need the molar mass of PbCrO4, which is 323.2 g/mol.
Let's assume that the solubility of PbCrO4 is "x" mol/L. Then, at equilibrium, the concentrations of Pb^2+ and CrO4^2- are also "x" mol/L. Therefore, we can substitute these values into the Ksp expression:
Ksp = [Pb^2+][CrO4^2-] = (x)(x) = x^2.
Given that Ksp = 2.8 x 10^-13, we can set up the following equation:
2.8 x 10^-13 = x^2.
To solve for x, take the square root of both sides:
√(2.8 x 10^-13) = x.
This will give you the value of x, which represents the molar solubility of PbCrO4. Finally, to convert the molar solubility to grams per liter, multiply by the molar mass of PbCrO4:
Solubility (in grams per liter) = x (mol/L) × molar mass (g/mol).
I hope this helps! Let me know if you have any further questions.