Solubility Question (Mole Dissolve)

[DrCheese]

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I stumble upon a solubility question and theres no answer for it. I just want to confirm if my choice is correct. Thanks!

How many moles of AgIO3 (Ksp = 3.1x10^-8) will dissolve in one liter of a 10^-5M solution of NaIO3?

A SqRt(3.1x10^-8)
B SqRt(3.1x10^-8) (10^-5)
C SqRt(3.1x10^-8) / (10^-5)
D (10^-5) - SqRt(3.1x10^-8)
E SqRt(3.1x10^-8) - (10^-5)

I am thinking it is C. The solubility of AgIO3 is SqRt(3.1x10^-8) since 3.1x10^-5 =(Ag)(IO3). I divide by that amount by the number of IO3 supplied by NaIO3.

 

[nze82]

I stumble upon a solubility question and theres no answer for it. I just want to confirm if my choice is correct. Thanks!

How many moles of AgIO3 (Ksp = 3.1x10^-8) will dissolve in one liter of a 10^-5M solution of NaIO3?

A SqRt(3.1x10^-8)
B SqRt(3.1x10^-8) (10^-5)
C SqRt(3.1x10^-8) / (10^-5)
D (10^-5) - SqRt(3.1x10^-8)
E SqRt(3.1x10^-8) - (10^-5)

I am thinking it is C. The solubility of AgIO3 is SqRt(3.1x10^-8) since 3.1x10^-5 =(Ag)(IO3). I divide by that amount by the number of IO3 supplied by NaIO3.

Correct!

 

[DrCheese]

Cool thanks!!

 

[kponenation]

Can someone explain to me how you solve this problem?

I thought you do it like 3.1E-8 = x (10E-5) so x = 3.1E-4..

 

[nze82]

Can someone explain to me how you solve this problem?

I thought you do it like 3.1E-8 = x (10E-5) so x = 3.1E-4..

This is how I've always solved these:

AgIO3 <-->Ag+ + IO3-

KSP = [Ag+][IO3-] = (x)(x) = x^2

Remember that KSP is a measure of the solubility of a particular compound at a specific temperature, and for this particular compound:

KSP = x^2

Let's see what happens, when AgIO3 dissolves in presence of NaIO3:

AgIO3 <-->Ag+ + IO3-

NaIO3 <-->Ag+ + IO3-

As you can see, due to the common ion effect, the solubility of AgIO3 decreases, under such circumstances. This new solubility is what you've calculated above, and it turns out to be equal to 3.1E-4.
However, as we showed above, any measure of solubility for this particular compound is equal to x^2. Therefore, to find the molar solubility (x) for this compound we must proceed as follows:

3.1E-4 = x^2

x = Molar solubility = Sqrt (3.1E-4)

 

[DrCheese]

Can someone explain to me how you solve this problem?

I thought you do it like 3.1E-8 = x (10E-5) so x = 3.1E-4..

You are not finding the solubility of AgIO3 after you add in NaIO3.

The question is asking how many can you put in NaIO3 before it reaches equilibrium or precipitates. Thus you want to find out the solubility of IO3 when dissolved in water that will be your IO3 max solubility, which is 3.1E-8 = x^2 or SqRt(3.1E-8).

Because NaIO3 donates IO3, you want to know how many will go into AgIO3. You are given E-5M of NaIO3 in 1 liter. So how many NaIO3 can you add before it equal to max solubility. The equation is SqRt(3.1E-8)/E-5.

(I guess nze82 answered your question lol..)

 

[faerielynx]

I stumble upon a solubility question and theres no answer for it. I just want to confirm if my choice is correct. Thanks!

How many moles of AgIO3 (Ksp = 3.1x10^-8) will dissolve in one liter of a 10^-5M solution of NaIO3?

A SqRt(3.1x10^-8)
B SqRt(3.1x10^-8) (10^-5)
C SqRt(3.1x10^-8) / (10^-5)
D (10^-5) - SqRt(3.1x10^-8)
E SqRt(3.1x10^-8) - (10^-5)

I am thinking it is C. The solubility of AgIO3 is SqRt(3.1x10^-8) since 3.1x10^-5 =(Ag)(IO3). I divide by that amount by the number of IO3 supplied by NaIO3.

its actually 'E'... its a tricky kaplan ques

 

[nze82]

its actually 'E'... its a tricky kaplan ques

Hmmmm...is it? How's that possible? Could you explain it please? tnx
I personally don't see how E could be the correct answer. I might be wrong though.

 

[Aceofspades]

The answer is 3.1E-3 and there is something wrong with the OP's question. It has no answer represented in the multiple choice questions. All the answer choices have a Sqrt() function, but there will be no Sqrt() function if you assume the Ksp to be so small that only a neglibile amount dissolves. If you consider the Ksp to be large enough that a considerable amount dissolves then you have no choice but to use the quadratic formula. The DAT won't require you to use the quadratic formula on a gen chem question, or any question for that matter.

Case 1: Ksp is very small and amount dissolved is negligible (this is always the case for the DAT, and any gen chem class actually)

First you write out the equation

AgIO3(s) -> Ag+(aq) + IO3-(aq)

Now write out the Ksp equation. Remember Ksp is just another equilibrium constant like K and is calculated through [products]/[reactants], but in this case the reactant is a pure solid and can b e ignored

Ksp = [products]

Since they told you the Ksp is 3.1E-8 you can just replace Ksp with that value. The concentration of Ag+ and IO3- are unkown, but equal to each other because when one molecule of AgIO3 dissolves it produces an equal amount of both. However, note that you have an additional source of IO3-(aq) brought in by the addition of 1 liter of a 10^-5M solution of NaIO3. This will add 1E-5 to the IO3- side

3.1E-8 = [x][x + 1E-5]

Now, x (the amount that dissolves) is considered very small (small Ksp) so x + 1E-5 is very close to 1E-5. Therefore

3.1E-8 = [x][1E-5]
3.1E-3=[x]

So there we have it, 3.1E-3 moles will dissolve in 1 liter of this solution.

Case 2: Amount dissolved is not negligible (would never happen on the DAT, but let us entertain the idea anyway)

AgIO3(s) -> Ag+(aq) + IO3-(aq)

Ksp = [products]

3.1E-8 = [x][x + 1E-5]
3.1E-8 = x^2 + xE-5
0 = x^2 + xE-5 - 3.1E-8

Now we must use the quadratic formula

a = 1
b = 1E-5
c = -3.1E-8

x = (-1E-5 +- Sqrt(1E-10 - -1.24E-7))/2
x = (-1E-5 +- Sqrt(1.241E-7))/2... this does not even remotely match any of the answer choices

Solving...
x = (-1E-5 +- 3.523E-4)/2
x = 1.716E-4 and x = -1.812E-4

Of these, you will obviously take the positive one and x = 1.716E-4. All this is a mute point because the DAT will never require you to go through this second case.

 

[kponenation]

The answer is 3.1E-3 and there is something wrong with the OP's question. It has no answer represented in the multiple choice questions. All the answer choices have a Sqrt() function, but there will be no Sqrt() function if you assume the Ksp to be so small that only a neglibile amount dissolves. If you consider the Ksp to be large enough that a considerable amount dissolves then you have no choice but to use the quadratic formula. The DAT won't require you to use the quadratic formula on a gen chem question, or any question for that matter.

Case 1: Ksp is very small and amount dissolved is negligible (this is always the case for the DAT, and any gen chem class actually)

First you write out the equation

AgIO3(s) -> Ag+(aq) + IO3-(aq)

Now write out the Ksp equation. Remember Ksp is just another equilibrium constant like K and is calculated through [products]/[reactants], but in this case the reactant is a pure solid and can b e ignored

Ksp = [products]

Since they told you the Ksp is 3.1E-8 you can just replace Ksp with that value. The concentration of Ag+ and IO3- are unkown, but equal to each other because when one molecule of AgIO3 dissolves it produces an equal amount of both. However, note that you have an additional source of IO3-(aq) brought in by the addition of 1 liter of a 10^-5M solution of NaIO3. This will add 1E-5 to the IO3- side

3.1E-8 = [x][x + 1E-5]

Now, x (the amount that dissolves) is considered very small (small Ksp) so x + 1E-5 is very close to 1E-5. Therefore

3.1E-8 = [x][1E-5]
3.1E-3=[x]

So there we have it, 3.1E-3 moles will dissolve in 1 liter of this solution.

Case 2: Amount dissolved is not negligible (would never happen on the DAT, but let us entertain the idea anyway)

AgIO3(s) -> Ag+(aq) + IO3-(aq)

Ksp = [products]

3.1E-8 = [x][x + 1E-5]
3.1E-8 = x^2 + xE-5
0 = x^2 + xE-5 - 3.1E-8

Now we must use the quadratic formula

a = 1
b = 1E-5
c = -3.1E-8

x = (-1E-5 +- Sqrt(1E-10 - -1.24E-7))/2
x = (-1E-5 +- Sqrt(1.241E-7))/2... this does not even remotely match any of the answer choices

Solving...
x = (-1E-5 +- 3.523E-4)/2
x = 1.716E-4 and x = -1.812E-4

Of these, you will obviously take the positive one and x = 1.716E-4. All this is a mute point because the DAT will never require you to go through this second case.

So according to this my answer of 3.1E-4 is correct not your 3.1E-3 because it's 10E-5 not 1E-5. Right?

 

[Aceofspades]

So according to this my answer of 3.1E-4 is correct not your 3.1E-3 because it's 10E-5 not 1E-5. Right?

No, the answer is 3.1 x 10^-3 NOT 3.1 x 10^-4. E means "x 10^", so 1E-5 is 1 * 10^-5 which is just 10^-5. 10E-5 is actually 10 * 10^-5 which is 10^-4 and that is incorrect. Your setup skipped some steps but was otherwise correct.

 

[kponenation]

Oh crap... I thought it was 10E-5.. I mis-read it... yeah it will be 3.1E-3.
So this is correct??? that's what I thought at first but looks like others think differ.

 

[Aceofspades]

Oh crap... I thought it was 10E-5.. I mis-read it... yeah it will be 3.1E-3.
So this is correct??? that's what I thought at first but looks like others think differ.

Haha, the others are wrong so don't bet your DAT score on it. Get used to it, it happens on SDN. Luckily there are people here to correct it. Frankly all the other answers are absurd. Ksp is just another equilibrium constant. Everyone should know [products]/[reactants]. Ksp questions are very easy to solve once you realize this.

 

[MNova]

The answer is 3.1E-3 and there is something wrong with the OP's question. It has no answer represented in the multiple choice questions. All the answer choices have a Sqrt() function, but there will be no Sqrt() function if you assume the Ksp to be so small that only a neglibile amount dissolves. If you consider the Ksp to be large enough that a considerable amount dissolves then you have no choice but to use the quadratic formula. The DAT won't require you to use the quadratic formula on a gen chem question, or any question for that matter.

Case 1: Ksp is very small and amount dissolved is negligible (this is always the case for the DAT, and any gen chem class actually)

First you write out the equation

AgIO3(s) -> Ag+(aq) + IO3-(aq)

Now write out the Ksp equation. Remember Ksp is just another equilibrium constant like K and is calculated through [products]/[reactants], but in this case the reactant is a pure solid and can b e ignored

Ksp = [products]

Since they told you the Ksp is 3.1E-8 you can just replace Ksp with that value. The concentration of Ag+ and IO3- are unkown, but equal to each other because when one molecule of AgIO3 dissolves it produces an equal amount of both. However, note that you have an additional source of IO3-(aq) brought in by the addition of 1 liter of a 10^-5M solution of NaIO3. This will add 1E-5 to the IO3- side

3.1E-8 = [x][x + 1E-5]

Now, x (the amount that dissolves) is considered very small (small Ksp) so x + 1E-5 is very close to 1E-5. Therefore

3.1E-8 = [x][1E-5]
3.1E-3=[x]

So there we have it, 3.1E-3 moles will dissolve in 1 liter of this solution.

Case 2: Amount dissolved is not negligible (would never happen on the DAT, but let us entertain the idea anyway)

AgIO3(s) -> Ag+(aq) + IO3-(aq)

Ksp = [products]

3.1E-8 = [x][x + 1E-5]
3.1E-8 = x^2 + xE-5
0 = x^2 + xE-5 - 3.1E-8

Now we must use the quadratic formula

a = 1
b = 1E-5
c = -3.1E-8

x = (-1E-5 +- Sqrt(1E-10 - -1.24E-7))/2
x = (-1E-5 +- Sqrt(1.241E-7))/2... this does not even remotely match any of the answer choices

Solving...
x = (-1E-5 +- 3.523E-4)/2
x = 1.716E-4 and x = -1.812E-4

Of these, you will obviously take the positive one and x = 1.716E-4. All this is a mute point because the DAT will never require you to go through this second case.

Your "Case 1" above was exactly how I approached the problem. The fact that it's not an answer choice though concerns me!

 

[doc toothache]

No, the answer is 3.1 x 10^-3 NOT 3.1 x 10^-4. E means "x 10^", so 1E-5 is 1 * 10^-5 which is just 10^-5. 10E-5 is actually 10 * 10^-5 which is 10^-4 and that is incorrect. Your setup skipped some steps but was otherwise correct.

If the answer is 3.1 x 10^-3 it suggests that the solubility of AgIO3 is increased by the presence of NaIO3, when in reality the solubility of a compound decreases in the presence of a common ion and it becomes significant when the concentration is high. An exception would be the effect of pH on a compound which contains an anion of a weak acid (e.g. CaCO3). When the concentration of the common ion is negligible the solubility remains unchanged and is simply calculated from the Ksp; when it is not negligible it is subtracted from the max concentration calculated from the Ksp. In this case the answer should be E.

 

[Innulja]

If the answer is 3.1 x 10^-3 it suggests that the solubility of AgIO3 is increased by the presence of NaIO3, when in reality the solubility of a compound decreases in the presence of a common ion and it becomes significant when the concentration is high. An exception would be the effect of pH on a compound which contains an anion of a weak acid (e.g. CaCO3). When the concentration of the common ion is negligible the solubility remains unchanged and is simply calculated from the Ksp; when it is not negligible it is subtracted from the max concentration calculated from the Ksp. In this case the answer should be E.

If the answer choice is E you get a negative number which is not possible!!!

 

[nze82]

OK! Now I really want to know the correct answer to this question!!

 

[doc toothache]

E SqRt(3.1x10^-8) - (10^-5)

If the answer choice is E you get a negative number which is not possible!!!

Could be that one of our calculators is off.

(3.1 x 10^-8)^1/2= 1.76 x 10^-4=0.000176
10^-5= 1 x 10^-4= 0.0001
then 0.000171-0.0001=0.000076=(7.6 x 10)^-5

Can you make the corrections or show how you came up with a negative #?

 

[kponenation]

I'm with nze82.. I really want to know THE answer...

 

[GiTsticker]

Could be that one of our calculators is off.

(3.1 x 10^-8)^1/2= 1.76 x 10^-4=0.000176
10^-5= 1 x 10^-4= 0.0001
then 0.000171-0.0001=0.000076=(7.6 x 10)^-5

Can you make the corrections or show how you came up with a negative #?

Just a minor correction, but 10^-5 is the same as 1 x 10^-5, not 1 x 10^-4

PS: I'm with the 3.1x10^-3 crowd. That's how Kaplan says to do it, and that's how my textbook says to do it.
On second thought...3.1x10^-3 IS larger than 10^-5, meaning it is significant enough that we need to use the quadratic equation to figure it out. hmm

Ace of Spade's case 2 would be the appropriate way to do this problem. Which means we don't have to know it!

 

[ethanyoo726]

I think the answer choices are wrong.

AceofSpades Case 1 is correct. That answer is in Molarity, but it's in 1 liter of solution, so you know that it's just 3.1 x 10^-3 moles.

 

[doc toothache]

More a case of rusty cortex than calculator problem.

 

[Innulja]

Could be that one of our calculators is off.

(3.1 x 10^-8)^1/2= 1.76 x 10^-4=0.000176
10^-5= 1 x 10^-4= 0.0001
then 0.000171-0.0001=0.000076=(7.6 x 10)^-5

Can you make the corrections or show how you came up with a negative #?

If you have the Destroyer look at question #98, I think it will clear things out.
Maybe your reasoning is correct, however that is not how you do this type of questions.