gen chem problem

[topdent1]

[SIZE=+1]The Ksp for PbCl2 is 1.6 * 10–5. Which maximum amount of NaCl(s) may be added to 1.0 L of 0.010 Pb(NO3)2 solution without causing precipitate to occur?[/SIZE]
[SIZE=+1] A. 1.6 * 10-5 mol[/SIZE]
[SIZE=+1] B. 3.2 * 10-5 mol[/SIZE]
[SIZE=+1] C. 1.6 * 10-3 mol[/SIZE]
[SIZE=+1] D. 3.2 * 10-3 mol[/SIZE]
[SIZE=+1] E. 1.6 * 10-2 mol

Can someone give me a detailed solution please?
[/SIZE]

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[harrygt]

[SIZE=+1]The Ksp for PbCl2 is 1.6 * 10–5. Which maximum amount of NaCl(s) may be added to 1.0 L of 0.010 Pb(NO3)2 solution without causing precipitate to occur?[/SIZE]
[SIZE=+1]A. 1.6 * 10-5 mol[/SIZE]
[SIZE=+1]B. 3.2 * 10-5 mol[/SIZE]
[SIZE=+1]C. 1.6 * 10-3 mol[/SIZE]
[SIZE=+1]D. 3.2 * 10-3 mol[/SIZE]
[SIZE=+1]E. 1.6 * 10-2 mol[/SIZE]

[SIZE=+1]Can someone give me a detailed solution please?[/SIZE]

Ksp of PbCl2 = [Pb+2] [Cl-]^2
in order to have no solid precipitate, the above equation should be equal or less than KSP. in the equal sitauation, we have the maximum amount that can be added without causing precipitation.
we have 0.01 molar Pb(NO3)2 which is 0.01 molar with respect to Pb2+ too.

so we have: 0.01 * [Cl-]^2 = 1.6 * 10^-5

[Cl-]^2 = .16 * 10^-2
[Cl-] = .4 * 10^-1 = 0.04 molar with respect to Cl-. It means we can add up to 0.04 mole [4.0 * 10^-2] to the solution.
I don't see this in the multiple choices. Someone correct me if I'm wrong.

 

[Mstoothlady2012]

[SIZE=+1]The Ksp for PbCl2 is 1.6 * 10–5. Which maximum amount of NaCl(s) may be added to 1.0 L of 0.010 Pb(NO3)2 solution without causing precipitate to occur?[/SIZE]
[SIZE=+1] A. 1.6 * 10-5 mol[/SIZE]
[SIZE=+1] B. 3.2 * 10-5 mol[/SIZE]
[SIZE=+1] C. 1.6 * 10-3 mol[/SIZE]
[SIZE=+1] D. 3.2 * 10-3 mol[/SIZE]
[SIZE=+1] E. 1.6 * 10-2 mol

Can someone give me a detailed solution please?
[/SIZE]

Pbcl2 ---> Pb+2 + 2Cl-


Ksp = [Pb+2] [Cl-]
[1.6 x 10 ^ -5] = (x) (2x)^2
[1.6 x 10^-5] = 4x^3
x= 1.59 x 10^-2 = E

I am not sure if this is right or not, haven't done one of these in long time.

 

[harrygt]

Pbcl2 ---> Pb+2 + 2Cl-


Ksp = [Pb+2] [Cl-]
[1.6 x 10 ^ -5] = (x) (2x)^2
[1.6 x 10^-5] = 4x^3
x= 1.59 x 10^-2 = E

I am not sure if this is right or not, haven't done one of these in long time.

we already have X by having the conc. of the Pb(NO3)2 solution. We are not dissolving any PbCl2 solid. If so, then your calculations would be legitimate.

 

[DentalDeac]

I'm really confused by this problem. Topdent where did you get the problem from and do you have the correct answer?

 

[osimsDDS]

Pbcl2 ---> Pb+2 + 2Cl-


Ksp = [Pb+2] [Cl-]
[1.6 x 10 ^ -5] = (x) (2x)^2
[1.6 x 10^-5] = 4x^3
x= 1.59 x 10^-2 = E

I am not sure if this is right or not, haven't done one of these in long time.

I think you are correct, we have a solution of Pb(NO3)2 and you are adding NaCl....a double displacement reaction occurs where Pb and Cl react to form a precipitate in solution (chloride ion with heavy metal = precipitate) so we are trying to find out how much to add without it precipitating as we know the solubility product of the precipitate is 1.6x10^-5. You are correct this is how you do it... the answer is E as you mentioned...

 

[DentalDeac]

we already have X by having the conc. of the Pb(NO3)2 solution. We are not dissolving any PbCl2 solid. If so, then your calculations would be legitimate.

So does the concentration of the Pb(NO3)2 not matter?

 

[harrygt]

So does the concentration of the Pb(NO3)2 not matter?

Yes, it does matter. The concentration of Pb(NO3)2 is representing the concentration of Pb2+ in the solution, and this concentration is not actually going to change due to adding a small amount of NaCl.
we should set up Ksp = 0.01 * [Cl-]^2 which will let us find the concentration of Cl- as I did. However, I did not find the answer in any of the choices.
What is the source of the question?

 

[DentalDeac]

Yes, it does matter. The concentration of Pb(NO3)2 is representing the concentration of Pb2+ in the solution, and this concentration is not actually going to change due to adding a small amount of NaCl.
we should set up Ksp = 0.01 * [Cl-]^2 which will let us find the concentration of Cl- as I did. However, I did not find the answer in any of the choices.
What is the source of the question?

Right, I did it the same way you did and thats why I'm so confused.

 

[DentalDeac]

I think you are correct, we have a solution of Pb(NO3)2 and you are adding NaCl....a double displacement reaction occurs where Pb and Cl react to form a precipitate in solution (chloride ion with heavy metal = precipitate) so we are trying to find out how much to add without it precipitating as we know the solubility product of the precipitate is 1.6x10^-5. You are correct this is how you do it... the answer is E as you mentioned...

osims don't you have to take into account the concentration of the Pb(NO3)2 that it is being added to?

 

[harrygt]

Right, I did it the same way you did and thats why I'm so confused.

Ok then, I'm becoming pretty certain that the correct answer is not in the choices. Usually, whenever I don't get the answer to a question, I read that question about 3 subsequent times again, and check my reasoning and calculations very well. I can't see anything wrong with my calculations though. Anyone's critic on my calculations and way of solving this problem would be appreciated.

 

[DentalDeac]

Ok then, I'm becoming pretty certain that the correct answer is not in the choices. Usually, whenever I don't get the answer to a question, I read that question about 3 subsequent times again, and check my reasoning and calculations very well. I can't see anything wrong with my calculations though. Anyone's critic on my calculations and way of solving this problem would be appreciated.

I'm talking to a friend on the phone right now and we are both trying to figure this problem out. Very frustrating. I emailed my Kaplan teacher this question and I'll see what his answer is.

 

[topdent1]

I got the question from this website, question number 34 i believe. http://www.geocities.com/crazytwo2000/

Also, the correct answer is E.

 

[DentalDeac]

I don't trust that geocities website. I still think the answer is 0.04 mol

 

[harrygt]

I don't trust that geocities website. I still think the answer is 0.04 mol

Yup, the answer is 0.04. I will bet my Name on it.

 

[DentalDeac]

Yup, the answer is 0.04.

I just wasted an hour of studying trying to figure this stupid problem out only to discover that it came from a stupid geocities site. The guy who posted those problems probably doesn't know what he's doing, I wouldn't use that as a study source

 

[harrygt]

I just wasted an hour of studying trying to figure this stupid problem out only to discover that it came from a stupid geocities site. The guy who posted those problems probably doesn't know what he's doing, I wouldn't use that as a study source

Calm down man. You have not lost anything; just gained more confidence for next time you approaach these problems. You can find wrong stuff all over the net, even sometimes in wiki.

 

[DentalDeac]

Calm down man. You have not lost anything; just gained more confidence for next time you approaach these problems. You can find wrong stuff all over the net, even sometimes in wiki.

Haha true. That is a good point, because before reading this thread, I didn't really know how to approach a Ksp problem, but now I feel like I'm an expert on the topic.

So I guess in a way this was a good exercise.

Now on to Acids and Bases, specifically Ka, pKa, and pH

 

[harrygt]

Haha true. That is a good point, because before reading this thread, I didn't really know how to approach a Ksp problem, but now I feel like I'm an expert on the topic.

So I guess in a way this was a good exercise.

Now on to Acids and Bases, specifically Ka, pKa, and pH

Good that you got it. Ksp problems are very important to know. The same applies to Ka, pKa, and PH.

 

[doc toothache]

Ksp of PbCl2 = [Pb+2] [Cl-]^2
in order to have no solid precipitate, the above equation should be equal or less than KSP. in the equal sitauation, we have the maximum amount that can be added without causing precipitation.
we have 0.01 molar Pb(NO3)2 which is 0.01 molar with respect to Pb2+ too.

so we have: 0.01 * [Cl-]^2 = 1.6 * 10^-5

[Cl-]^2 = .16 * 10^-2
[Cl-] = .4 * 10^-1 = 0.04 molar with respect to Cl-. It means we can add up to 0.04 mole [4.0 * 10^-2] to the solution.
I don't see this in the multiple choices. Someone correct me if I'm wrong.

You are on the right track. However since

PbCl2 = (Pb+2) + 2 Cl-;

KSp= [Pb+2][2 Cl-]^2
Since the molar concentration calculated is twice that of actual molar concentration of NaCl, the max that can be added is 2.0 x 10^-2. We can double check the numbers by
The Ksp becomes [1 x 10^-2][2 x (2 x 10^-2)]^2= 1.6 x 10^-5

 

[harrygt]

You are on the right track. However since

PbCl2 = (Pb+2) + 2 Cl-;

KSp= [Pb+2][2 Cl-]^2
Since the molar concentration calculated is twice that of actual molar concentration of NaCl, the max that can be added is 2.0 x 10^-2. We can double check the numbers by
The Ksp becomes [1 x 10^-2][2 x2 x 10^-2]^2= 1.6 x 10^-5

Sorry, but you are on the wrong track. The Cl is coming from outside, not from dissociation of the PbCl2. It is not important that each PbCl2 will give 2 moles of Cl, except for writting down the Ksp formula. once you dump for example 5 moles of Cl- in the solution, you have 5 moles, not 10 moles.
Actually when you are double checking your number, you are contradicting yourself. if you have 2 x 2 x 10^-2, then that means you have 0.04 moles of Cl- in solution. How can you get 0.04 moles of Cl- in a solution of Pb(NO3)2 that has no other source of Cl-, by only adding 0.02 moles.
Give it a second thought.

 

[Mstoothlady2012]

Ksp of PbCl2 = [Pb+2] [Cl-]^2
in order to have no solid precipitate, the above equation should be equal or less than KSP. in the equal sitauation, we have the maximum amount that can be added without causing precipitation.
we have 0.01 molar Pb(NO3)2 which is 0.01 molar with respect to Pb2+ too.

so we have: 0.01 * [Cl-]^2 = 1.6 * 10^-5

[Cl-]^2 = .16 * 10^-2
[Cl-] = .4 * 10^-1 = 0.04 molar with respect to Cl-. It means we can add up to 0.04 mole [4.0 * 10^-2] to the solution.
I don't see this in the multiple choices. Someone correct me if I'm wrong.

Harry I think you have set up your equation wrong. It should be [2Cl]^2

 

[harrygt]

Harry I think you have set up your equation wrong. It should be [2Cl]^2

Nope, this question is different I'm pretty sure with my answer. tell me, how much Cl- would you have if you add 1 moles of Cl- to this solution that has neither Cl-, nor any source of Cl-? Would you have 1 mole or two moles after adding one mole? You will have one mole!
You would put the 2 over there if some Pb(Cl)2 was added to the solution, which dissociates to 2 Cl-. However, in this case, the amount of Cl- that you add, is going to be the amount of Cl- that you have. That amount will go right in the paranthesis, without any irrelevent 2 multipliyed by it.
Think of it this way. Based on what you say, if I add 0.04 moles of Cl- to the solution, you are gonna put 0.08 in the paranthesis. But why? do we have 0.08 moles in the solution after adding 0.04 moles? Of course not!

 

[Mstoothlady2012]

Nope, this question is different I'm pretty sure with my answer. tell me, how much Cl- would you have if you add 1 moles of Cl- to this solution that has neither Cl-, nor any source of Cl-? Would you have 1 mole or two moles after adding one mole? You will have one mole!
You would put the 2 over there if some Pb(Cl)2 was added to the solution, which dissociates to 2 Cl-. However, in this case, the amount of Cl- that you add, is going to be the amount of Cl- that you have. That amount will go right in the paranthesis, without any irrelevent 2 multipliyed by it.
Think of it this way. Based on what you say, if I add 0.04 moles of Cl- to the solution, you are gonna put 0.08 in the paranthesis. But why? do we have 0.08 moles in the solution after adding 0.04 moles? Of course not!

I see what you are saying, but hey I always learned to balance the equation & then use Ksp. So I am just following that....m not a genius in this topic so i ain't gonna argue wid u....

 

[harrygt]

I see what you are saying, but hey I always learned to balance the equation & then use Ksp. So I am just following that....m not a genius in this topic so i ain't gonna argue wid u....

You always balance when you start with with putting a reactant which is going to give us some products. We still balance this one to find the Ksp expression [and the exponents], but using the coefficients in the rest of the problem is irrelevant.
if we put .1 mole of Mg(OH)2 in water to make a 1 liter solution, we will get .2 moles of OH- as the balanced equation tells.
This case is different though. Imagine dumping .2 moles of OH- in water to make a one liter solution. Then you are gonna have .2 moles of OH-.
In this question, there is no Cl- present at the begining, nor any PbCl2 dissociating. once you put .04 moles in the solution, you should take that as the number getting in [Cl-] paranthesis which is equal to the conc. of Cl-. Just look at what the doc tooch guy put in this [Cl-] paranthesis. He put 2 * 2 * 10^-2. we are supposed to put the overall conc of Cl- in parantheis which is 0.04 and that guy is doing the same thing as I did. But he wrongly assumes that it is double the amound of Cl-. Well, then I say how did it get doubled if we added only 0.02? See the contradiction?
I hope you get this.

 

[Mstoothlady2012]

You always balance when you start with with putting a reactant which is going to give us some products. We still balance this one to find the Ksp expression [and the exponents], but using the coefficients in the rest of the problem is irrelevant.
if we put .1 mole of Mg(OH)2 in water to make a 1 liter solution, we will get .2 moles of OH- as the balanced equation tells.
This case is different though. Imagine dumping .2 moles of OH- in water to make a one liter solution. Then you are gonna have .2 moles of OH-.
In this question, there is no Cl- present at the begining, nor any PbCl2 dissociating. once you put .04 moles in the solution, you should take that as the number getting in [Cl-] paranthesis which is equal to the conc. of Cl-. Just look at what the doc tooch guy put in this [Cl-] paranthesis. He put 2 * 2 * 10^-2. we are supposed to put the overall conc of Cl- in parantheis which is 0.04 and that guy is doing the same thing as I did. But he wrongly assumes that it is double the amound of Cl-. Well, then I say how did it get doubled if we added only 0.02? See the contradiction?
I hope you get this.

If PbCl2 is not dissociating, then why are you using PbCl2 dissociating constant value to find the answer?

 

[harrygt]

If PbCl2 is not dissociating, then why are you using PbCl2 dissociating constant value to find the answer?

because we are going to have Pb2+ and Cl- in this solution, which are capabale of precipitating into PbCl2. That is why we have to use Ksp of PbCl2 to find how much Pb and Cl we can have in this solution withough making them precipitate.. There is no PbCl2 in the solution to dissociate.

 

[doc toothache]

Harrygt is correct. The answer is 0.04M.

 

[Danny289]

Pbcl2 ---> Pb+2 + 2Cl-


Ksp = [Pb+2] [Cl-]
[1.6 x 10 ^ -5] = (x) (2x)^2
[1.6 x 10^-5] = 4x^3
x= 1.59 x 10^-2 = E

I am not sure if this is right or not, haven't done one of these in long time.

you are right . the second part of this question (1.0 L of 0.010 Pb(NO3)2) just teling us we have enogh Pb2+ in our solution and we don't need to use that in our calculation!

 

[harrygt]

you are right . the second part of this question (1.0 L of 0.010 Pb(NO3)2) just teling us we have enogh Pb2+ in our solution and we don't need to use that in our calculation!

Danny, so you mean there is no difference in the amount of Cl- we can add to 1) a 0.01 molar Pb(N03)2 solution 2) a 0.00001 molar Pb(NO3)2 solution and 2) a 1 molar Pb(NO3)2 solution, without causing any solid to precipitate?

KSP = [Pb2+] (which is already known) * [Cl-]^2 when [Cl-] is the amount we are going to add.
It's that simple. Check my previous explanations.

 

[Glycogen]

I have done this specific problem million times to get to where you are harry, but still I get 1.6*10^-2 !
If we have .01 Pb and also we know what is our ksp for this solution after completion which is 1.6*10^-5 then then we have to set cl as our X.
So we say that PbCl2 --> Pb+2 + 2 Cl
then we have 1.6 *10^-5=(.01)(2X^2)
and the result will be 2*10^-2 or 1.6*10^-2
In case Dr.tooth he set up his ksp wrong and therefore he got .04 because he has (.01)(2(2*10^-2)^2)
this should be our X I think.
I'm not sure about the whole thing but this sound completely reasonable,I have gone over your reasoning but I really don't get your point.
Thanks.

 

[Glycogen]

By the way I've tried to convince myself that your reasoning is right since you are the king of chemistry!!!!
But,as I said I can not follow your reasoning!!!!

 

[250rsavage]

I got the same answer as Harry. I not 100% sure why, but in my notes, doing PPT problems like this one PbCl2 ->Pb+2Cl Ksp =1.6 *10-5 and with a conc. of 0.010 Pb(NO3)2[SIZE=+1]
[/SIZE]
1.6*10-5=[10-2][Cl-]2
1.6*10-3=[Cl-]2
Cl- = .04

 

[Danny289]

Danny, so you mean there is no difference in the amount of Cl- we can add to 1) a 0.01 molar Pb(N03)2 solution 2) a 0.00001 molar Pb(NO3)2 solution and 2) a 1 molar Pb(NO3)2 solution, without causing any solid to precipitate?

KSP = [Pb2+] (which is already known) * [Cl-]^2 when [Cl-] is the amount we are going to add.
It's that simple. Check my previous explanations.

No, I mean not matter how much Pb+2 we have in soultion! we control Pb+2 + 2Cl-===>Pb(Cl)2 just with adding Cl-. the precipitation is related to "Cl-" in this case based on "Ksp" formula not pb+2 concentration. if you have even 1 mole pb+2in solution but not having enough Cl- to preicipiate with your cation. part of pb+2 will react with Cl- ( whatever you are looking for) and the rest of that will be in soultion in pb+2 form.

IMO Mstoothlady2012 did correct. ignore Pb+2 in this case and answer will be 1.6 X10^-2

 

[DRHOYA]

This is from my understanding, and how I did the question.

The Pb(NO3) or whatever is telling us the concentration of Pb....which I think we all agree on. The KSP is given.......ok.

Setting up the equation we have: PbCl2 -----> Pb + 2Cl. We can all agree this is the balanced equation right?

Ok, so setting up this dissociation goes (as far as I have learned it):
PbCl2 -----> Pb + 2Cl
- 0.01 x

KSP = [Pb] * [2x]squared
1.6*10^-5 = [0.01] * [4x squared]
0.0016 = [4x squared]
4 * 10^-4 = [x squared]
x = .02

Crap........and my solution is not an answer choice....sigh....

somone please correct me.

 

[Mstoothlady2012]

This is from my understanding, and how I did the question.

The Pb(NO3) or whatever is telling us the concentration of Pb....which I think we all agree on. The KSP is given.......ok.

Setting up the equation we have: PbCl2 -----> Pb + 2Cl. We can all agree this is the balanced equation right?

Ok, so setting up this dissociation goes (as far as I have learned it):
PbCl2 -----> Pb + 2Cl
- 0.01 x

KSP = [Pb] * [2x]squared
1.6*10^-5 = [0.01] * [4x squared]
0.0016 = [4x squared]
4 * 10^-4 = [x squared]
x = .02

Crap........and my solution is not an answer choice....sigh....

somone please correct me.

x = 0.02 = 2 x 10 ^ -2 = E

 

[klutzy1987]

That is exactly how it is supposed to be done. The answer is 2*10^-2. The correct answer is not there however Choice E is preety close to the correct answer so i would go with that choice. I was about to type the whole long explanation out but I realized that DRHOYA already did a wonderful job of showing how it is supposed to be done.

 

[DRHOYA]

x = 0.02 = 2 x 10 ^ -2 = E

Ok yeah. But is it ok so assume at since it is 2 * 10^-2, that the correct answer is E which is 1.6 * 10^-2?? This scares me, b/c if I did the question right, how come the "exact" answer is not there.

btw, is the way I solved the question correct?? thanksss

 

[Mstoothlady2012]

Ok yeah. But is it ok so assume at since it is 2 * 10^-2, that the correct answer is E which is 1.6 * 10^-2?? This scares me, b/c if I did the question right, how come the "exact" answer is not there.

btw, is the way I solved the question correct?? thanksss

Honestly, I dont know anymore after all these arguments in this thread!! Look at the way I did, this is how I have learned to do it in past. Now I have forgotten my "chemistry concepts". So all I can do is follow the steps & do the math. What harry is doing is conceptual, he might be right I don't know. The way you are doing, I think you are suppose to assume that PbCl2 will disscoiate proprotionally, so we assume x for both. Ofcourse we take into account the exponents though!

 

[DRHOYA]

Honestly, I dont know anymore after all these arguments in this thread!! Look at the way I did, this is how I have learned to do it in past. Now I have forgotten my "chemistry concepts". So all I can do is follow the steps & do the math. What harry is doing is conceptual, he might be right I don't know. The way you are doing, I think you are suppose to assume that PbCl2 will disscoiate proprotionally, so we assume x for both. Ofcourse we take into account the exponents though!

Ok. Yeah same here, I havn't taken this kind of chem since freshmen year, so its not exactly right on my mind. Anyway, about the dissociate aspect, can we not assume the concentration of the Pb(NO3) thing is whats given for Pb in the balanced dissociated equation?? Looking at the way you did it makes perfect sense, however I'm just wondering where the 0.01M Pb(NO3) plays in.

 

[Mstoothlady2012]

Ok. Yeah same here, I havn't taken this kind of chem since freshmen year, so its not exactly right on my mind. Anyway, about the dissociate aspect, can we not assume the concentration of the Pb(NO3) thing is whats given for Pb in the balanced dissociated equation?? Looking at the way you did it makes perfect sense, however I'm just wondering where the 0.01M Pb(NO3) plays in.

yea, you are right! We have to use that given conc somewhere. I am kind of leaning towards Harry's explanation. I read all of his reply again, but I still kind of don't und why he didn't multiply cl by 2 but he did square the cl. Now your answer makes more sense to me than mine. Lol I am so confused.

 

[DRHOYA]

yea, you are right! We have to use that given conc somewhere. I am kind of leaning towards Harry's explanation. I read all of his reply again, but I still kind of don't und why he didn't multiply cl by 2 but he did square the cl. Now your answer makes more sense to me than mine. Lol I am so confused.

Yeah, I feel somewhat confident in the way I did the question, but you are right. Reading Harry's explanation, idk.

Harry, where are ya man?

We need some help on this one.

 

[DentalDeac]

Guys, I emailed my Kaplan instructor this question and had him look at the problem. He came up with the same answer that Harry and I came up with, .04

 

[harrygt]

No, I mean not matter how much Pb+2 we have in soultion! we control Pb+2 + 2Cl-===>Pb(Cl)2 just with adding Cl-(?????). the precipitation is related to "Cl-" in this case based on "Ksp" formula not pb+2 concentration. if you have even 1 mole pb+2in solution but not having enough Cl- to preicipiate with your cation. part of pb+2 will react with Cl- ( whatever you are looking for) and the rest of that will be in soultion in pb+2 form.

IMO Mstoothlady2012 did correct. ignore Pb+2 in this case and answer will be 1.6 X10^-2

Hey bro, in an 2 way reaction which is trying to reach an equilibrium, increasing the concentration of any of the reactants and products will shift the reaction to a specific way, so you are making a big mistake here by saying that the Pb2+ concentration does not matter. NO WAY!

 

[DentalDeac]

Also, everyone needs to realize that this question came from a GEOCITIES website. Not the most credible source. The answer is .04

 

[harrygt]

Yeah, I feel somewhat confident in the way I did the question, but you are right. Reading Harry's explanation, idk.

Harry, where are ya man?

We need some help on this one.

Hey man, sorry, I had not been in front of my pc for the whole day. I had a very good Gchem teacher who made us do a lot of these problems. I got the 0.04 answer with confidence, the first time I solved it. But then Misstoothy and many others put me in doubt, so I checked my gchem notebook up, looking for similar problems [we did tons of these KSP problems in class, lab and SI last year]. I found about 4 of these types of questions, and none of them had the 2 mulitplied. All were done the same way I did at my first attempt. Read my previous arguments from dental deac on this thread carefully from the begining, and I hope you will get it. I really tried my best to clear this one up. I'm 100% sure about the answer, and I will bet my name on it

 

[Danny289]

Hey bro, in an 2 way reaction which is trying to reach an equilibrium, increasing the concentration of any of the reactants and products will shift the reaction to a specific way, so you are making a big mistake here by saying that the Pb2+ concentration does not matter. NO WAY!

Ok bro. yes you are right about equilibrium we both know the "Le Chatelier" principal and we know it applys in equilibrium.
but the " Le Chatelier" principal is not applyiny here for simple reason as you explained in your respond to this question earlier and we talked about that in PM you go for Q<ksp : "Ion Product" that means the situation before equilibrium and Le Chatelier principal. actualy the question asking about that point.
lets simply the problem : the question asking about the situation Na+, Cl- Pb+2 and No3- in solution in the same time, and giving 0.01 mole Pb(No3)2 telling us we don't have saturated Pb(No3)2 in our solution
maybe I am wrong. just I go by logic not just formula.

 

[Mstoothlady2012]

Hey man, sorry, I had not been in front of my pc for the whole day. I had a very good Gchem teacher who made us do a lot of these problems. I got the 0.04 answer with confidence, the first time I solved it. But then Misstoothy and many others put me in doubt, so I checked my gchem notebook up, looking for similar problems [we did tons of these KSP problems in class, lab and SI last year]. I found about 4 of these types of questions, and none of them had the 2 mulitplied. All were done the same way I did at my first attempt. Read my previous arguments from dental deac on this thread carefully from the begining, and I hope you will get it. I really tried my best to clear this one up. I'm 100% sure about the answer, and I will bet my name on it

I am sorry but I still don't understand why you are squaring Cl but not multiplying by its coefficient. Explain it one more time plz? If I don't get it again, I promise I will leave you alone

 

[DRHOYA]

Hey man, sorry, I had not been in front of my pc for the whole day. I had a very good Gchem teacher who made us do a lot of these problems. I got the 0.04 answer with confidence, the first time I solved it. But then Misstoothy and many others put me in doubt, so I checked my gchem notebook up, looking for similar problems [we did tons of these KSP problems in class, lab and SI last year]. I found about 4 of these types of questions, and none of them had the 2 mulitplied. All were done the same way I did at my first attempt. Read my previous arguments from dental deac on this thread carefully from the begining, and I hope you will get it. I really tried my best to clear this one up. I'm 100% sure about the answer, and I will bet my name on it

Oooooooooooo ok I think I see it. I carefully read through your explanations. So, since PbCl2 is not actually dissociating itself. You have Pb(NO3) with a given concentration, and they are telling you that you are going to add NaCl as the source of Cl-. So when you write the equation......you have [Pb] (given) and [Cl-] from NaCl that you are "going to put in".......ok. I hope that made sense. I think thats what it is....and thats why you are not having [2Cl]. But if thats the case.....you still have to square it??

 

[harrygt]

I am sorry but I still don't understand why you are squaring Cl but not multiplying by its coefficient. Explain it one more time plz? If I don't get it again, I promise I will leave you alone

You have to square Cl conc, because that is how you find the ion product [product of each ion conc that has been reaised to the coefficient of the balanced reaction.]
You have to put the concentration of each ion in its bracket. if we had x moles of PbCl2 dissovling in the solution, we would have 2x Cl-, and therefore, we would put 2x in the brackets of Cl- and raise it to 2. For this case, the reasaon is simply obvious. You get 2x Cl- for each x mole disssociated PbCl2, so 2x is the Cl- conc.

this question is different though. There is no x moles of PbCl2 dissolving to give us 2x moles of Cl-. As mentioned in the prev paragraph, you should put the real conc of Cl- in the brackets. if we add x moles of Cl-, how many moles of Cl- would there be in the solution? Of course X!
Look at it this way. What you are doing [the wrong way] is that you put x Cl- in the solution, but then you put 2x in the brackets. This means based on your way of solving the problem, the Cl- conc is getting doubled somehow. You are adding X mole, and then declaring the conc of the Cl- to be 2X. This is not right.