General chem help needed?
Problem #2. What is the molar solubility of PbS? Ksp=2.5 x 10 to the -27
problem #3. The solubility of ZX3 (as Z+3 + 3X-) in water at 298 K is 4.0 x 10 to the -4 M. Ksp for ZX3 is?
problem #4 What is the maximum concentration of chloride ions that can be in a solution containing 0.100 M Pb2+ without precipitating any lead ions? Ksp for lead chloride is 1.0 x 10 to the -4.
If you can please help me. I need to understand how to do it. As always my huge thank to you.
Answer:
Problem 1: Ok, given lead iodide's, PbI2, Ksp value with nothing else, therefore write the equilibrium formula for PbI2 out.
PbI2(s) <--> Pb²+(aq) + 2Iˉ (aq)
--s- - - - - - - - s- - - - - - - - 2s
(Ignore the dashes, spaces won't allow me to align my s)
Since the ratio is 1:1:2, we know that Ksp=[Pb²][Iˉ]²=(s)(2s)²
(Notice how only 's' gets doubled in size and not Iˉ. I don't know how to explain it, but you just have to remember)
Ksp= s(2s)² (use arithmetic)
1.39 x10^-8 = 4s³
s= 1.515 x10^-3 M
So refer to the equilibrium formula above, when s= 1.515 x10^-8, iodide will be twice the amount. So multiply it by 2 and you get [Iˉ] =3.03 x10^-3 (Answer B.)
Problem 2: Similar to the first question, write the equilibrium formula so you can somewhat have a visual on what is going on.
PbS(s) <--> Pb²+(aq) + S²ˉ(aq)
---s- - - - - - - - s- - - - - - - s
Ksp= [Pb²+][S²ˉ] = s²
s² = 2.5 x10^-27
s= 5.0 x10^-14 = [PbS]
Problem 3: Sorry I don't really understand what you mean by 'ZX3 (as Z+3 + 3X-)'
However, if you're able to understand the questions above, you should be able to do fine on this one.
Problem 4: When it says 'without precipitating', then it is implying that it is just at the verge of, so using the given Ksp value is appropriate.
PbCl2(s) <--> Pb²+(aq) + 2Clˉ(aq)
---s- - - - - - - - - s- - - - - - - - 2s-------
Ksp= 1.0 x10^-4 = [Pb²+][Clˉ]²
√(1.0 x10^-4 / [Pb²+]) = [Clˉ]
√(1.0 x10^-4 / 0.100M) = [Clˉ]
[Clˉ] = 0.032M
These questions are very similar to one another. Once you understand the concept of this unit, it will be a piece of cake. Good luck!
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: