# Ksp with significant anion hydrolysis

• Qube
In summary: Ksp. ...Ksp is defined, calculated and determined using equilibrium concentrations of ions, not using solubility. But we can still go from solubility to equilibrium concentration of ions, right? As shown above?Yes, we can go from solubility to equilibrium concentration of ions, but that does not mean that solubility and equilibrium concentration of ions are the same thing. Solubility is a measure of the maximum amount of a substance that can dissolve in a solvent, while equilibrium concentration of ions is a measure of the concentration of ions at equilibrium. Ksp is determined using equilibrium concentrations of ions, not solubility.
Qube
Gold Member
When calculating a Ksp for a salt with a significantly basic anion, e.g. silver cyanide, AgCN, given the experimental molar solubility value (s), should one write the Ksp expression as:

$K_{sp} = [Ag^{+}][CN^{-}]^{x}$, where x is the fraction of the concentration cyanide ion at equilibrium with respect to s ...

or as

$K_{sp} = [Ag^{+}][CN^{-}]^{x}[HCN]^{1-x}$?

I'm thinking it's the latter because Ksp is defined as the ion concentration product and obviously when silver cyanide dissolves in water solution we get not only cyanide ion in water but also hydrocyanic acid. The first expression would be incomplete. The complete dissolution equation of AgCN is:

$AgCN \leftrightharpoons Ag^{+} + (1/9)CN^{-} + (8/9)HCN$

The coefficients represent the hydrolysis of the cyanide ion factored in with the ionization of hydrocyanic acid to make more cyanide ion. Hydrolysis of the cyanide ion with molar solubility given by Wikipedia to be s = 0.00023 M is about 91%. However, about 1.7% of the HCN which is formed through cyanide ion hydrolysis ionizes to form cyanide ion again. So call "net" hydrolysis about 88%. There is approximately 1 cyanide ion to every 9 hydrocyanic acid molecules in solution. So should the Ksp therefore be:

$K_{sp} = [Ag^{+}][CN^{-}]^{1/9}[HCN]^{8/9}$?

Which, because of mass balance:

$s = [Ag^{+}] = [CN^{-}] + [HCN]$

is equivalent to:

$K_{sp} = [s/9]^{1/9}[8s/9]^{8/9}$?

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As you mentioned, Ksp is ion concentration product, so it just [Ag+][CN-].

Now, equilibrium in the solution is described by several equations, starting with three equilibria:

$$K_{sp} = [Ag^+][CN^-]$$

$$K_w = [H^+][OH^-]$$

$$K_a = \frac {[H^+][CN^-]}{[HCN]}$$

ending with mass balance and charge balance:

$$[Ag^+] = [HCN] + [CN^-]$$

$$[Ag^+] + [H^+] = [CN^-] + [OH^-]$$

(actually there is more to it, as Ag+ gets complexed by CN-, creating Ag(CN)2-, but let's not go there).

Molar solubility equals either [Ag+] or [HCN]+[CN-].

1 person
Thanks. But that will give me a new Ksp which is dramatically (more than just a few magnitudes) different from the Ksp of AgCN in literature. Why the big difference?

Qube said:
Thanks. But that will give me a new Ksp which is dramatically (more than just a few magnitudes) different from the Ksp of AgCN in literature.

No, it won't.

Hard to comment not knowing where you got this information from.

1 person
Borek said:
No, it won't.

Hard to comment not knowing where you got this information from.

These are the values I've been using:

$K_{a} HCN = 6.2*10^-10$

As we can see, the conjugate acid of the cyanide ion, $NC^{-}$, is fairly weak. So we can expect the cyanide ion itself to be a rather good base. At at such a low molar solubility value, we expect hydrolysis of the cyanide ion to be rather extensive. I get about 88% after factoring in ionization of the hydrocyanic acid.

so $K_{b} NC- = Kw / K_{a} HCN$

The s value for silver cyanide was taken from Wikipedia.

$s = 0.000023 g/100 mL$

http://en.wikipedia.org/wiki/Silver_cyanide

And this value can be readily found (values vary somewhat across sources but all of the sources have the molar solubility value pegged at $10^{-16}$ to $10^{-17}$. I'm using the value below:

$Ksp AgCN = 2.2 * 10^{-16}$.

So:

Revised Ksp (with consideration of anion hydrolysis):

$K_{sp} = [Ag^{+}][NC^{-}] = [s/9]^{1/9} = 9.2 * 10^{-4}$.

Plus it's clear that …

$K_{sp} = [Ag^{+}][NC^{-}] = [s/9]^{1/9}$ is probably dramatically different from $K_{sp} = [Ag^{+}][NC^{-}] =$.

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You have calculated something you call "revised Ksp", it has nothing to do with Ksp as is defined in a standard way, and you are surprised its value is different from the standard one?

Ksp is defined, calculated and determined using equilibrium concentrations of ions, not using solubility.

Yes, the equilibrium concentration of ions is what I plugged into my Ksp; I used the measured molar solubility value from Wikipedia as the empirical value on the EQ concentration of ions. Because we know that in solution

$s = [Ag^{+}] + [Ag(CN)_{2}^{-}] = [NC^{-}] + [HCN]$

And that $K_{sp} = [Ag^{+}]_{equilibrium}[NC^{-}]_{equilibrium}$

And $[Ag^{+}]_{equilibrium} =![NC^{-}]_{equilibrium}$

And from Wikipedia we have the molar solubility of AgCN. We can take it as a decent measure of $[Ag^{+}]$.

$s = [Ag^{+}]_{equilibrium} = [NC^{-}]_{equilibrium} + [HCN]_{equilibrium}$

And $[NC^{-}]_{equilibrium}$ is easily ascertained using the $K_{b}$ of the cyanide ion.

Borek said:
Ksp is defined, calculated and determined using equilibrium concentrations of ions, not using solubility.

But we can still go from solubility to equilibrium concentration of ions, right? As shown above?
I think the issue we have here is that we are lacking a good, formal definition of the solubility product. Does the solubility product really exclude molecular products of salt disassociation? Just about every textbook I've consulted has no say on this specific matter, but then just about every textbook ignores anion hydrolysis, and for those which do brave into the messy chemistry of anion hydrolysis, do not say anything about this issue.

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Qube said:
we are lacking a good, formal definition of the solubility product

We don't lack anything, you just made it up (probably because you are mixing solubility and Ksp). I told what the definition is, and apparently all books you consulted defined it the same way. They don't speak about other ions, because other ions are NOT part of Ksp.

Does the solubility product really exclude molecular products of salt disassociation?

Yes. They are necessary when you want to calculate solubility, but they are not required for Ksp. Ksp is NOT a measure of solubility, it is an equilibrium constant of the dissolution reaction. When the dissolution is followed by other reaction, they increase solubility, but they are not part of Ksp.

Borek said:
We don't lack anything, you just made it up (probably because you are mixing solubility and Ksp). I told what the definition is, and apparently all books you consulted defined it the same way. They don't speak about other ions, because other ions are NOT part of Ksp.

What do you mean other ions are not part of Ksp? What if I have calcium carbonate; would I not factor in the equilibrium concentrations of not only carbonate ion but also hydrogen carbonate ion?

Yes. They are necessary when you want to calculate solubility, but they are not required for Ksp. Ksp is NOT a measure of solubility, it is an equilibrium constant of the dissolution reaction. When the dissolution is followed by other reaction, they increase solubility, but they are not part of Ksp.

Did I calculate Ksp correctly - was the revised Ksp calculated correctly and do you see where I'm going with it?

Qube said:
What do you mean other ions are not part of Ksp? What if I have calcium carbonate;

I mean exactly what I said. For calcium carbonate

$$K_{sp} = [Ca^{2+}] [CO_3^{2-}]$$

That's by unambiguous definition that you are stubbornly refusing to accept.

would I not factor in the equilibrium concentrations of not only carbonate ion but also hydrogen carbonate ion?

No, you would not. Hydrogen carbonate doesn't matter for Ksp, it matters for solubility - but solubility and Ksp are different things.

Let's get back to AgCN. Let's say we keep pH buffered at pH that equals pKa of HCN - if so, [CN-]=[HCN].

$$K_{sp} = [Ag^+][CN^-]$$

mass balance is

$$[Ag^+] = [HCN] + [CN^-]$$

as we know [CN-]=[HCN], we can write

$$[Ag^+] = 2[CN^-]$$

(that works only for this particular pH!)

We have two equations in two unknowns now. Solve for [Ag+]

$$[Ag^+] = \frac {\sqrt{K_{sp}}}{2}$$

and you will get the AgCN solubility in this particular pH. No need to redefine Ksp, calculated solubility takes presence of both HCN and CN- into account.

Borek said:
I mean exactly what I said. For calcium carbonate

$$K_{sp} = [Ca^{2+}] [CO_3^{2-}]$$

That's by unambiguous definition that you are stubbornly refusing to accept.

Right, and for that system,

$[Ca^{2+}]_{equilibrium} =![CO_3^{2-}]_{equilibrium}$

unless the system is buffered at a certain pH, of course, and so we cannot say

$K_{sp} = [Ca^{2+}] [CO_3^{2-}] =$, where s is the equilibrium concentration of each ion, or square root of Ksp, or molar solubility.

Right?

No, you would not. Hydrogen carbonate doesn't matter for Ksp, it matters for solubility - but solubility and Ksp are different things.

Let's get back to AgCN. Let's say we keep pH buffered at pH that equals pKa of HCN - if so, [CN-]=[HCN].

$$K_{sp} = [Ag^+][CN^-]$$

mass balance is

$$[Ag^+] = [HCN] + [CN^-]$$

as we know [CN-]=[HCN], we can write

$$[Ag^+] = 2[CN^-]$$

(that works only for this particular pH!)

We have two equations in two unknowns now. Solve for [Ag+]

$$[Ag^+] = \frac {\sqrt{K_{sp}}}{2}$$

and you will get the AgCN solubility in this particular pH. No need to redefine Ksp, calculated solubility takes presence of both HCN and CN- into account.

This is all fine, but we don't have a 1:1 base to conjugate acid buffer solution of silver cyanide. So we cannot define cyanide ion concentration as half the silver ion concentration. Now what? Does my method apply? I don't see why not ... after all I am using the equilibrium concentrations of each ion.

I am starting to think you are misunderstanding one of the concepts used, I just don't know which one yet.

Solubility - it is the amount of the substance that dissolves. It doesn't tell anything about equilibrium concentrations (although it puts some restrictions on their values).

Equilibrium concentration - concentration of a specific ion observed in the solution, once it is at equilibrium. Doesn't matter where the ion comes from.

Ksp - is expressed using equilibrium concentrations.

Qube said:
$K_{sp} = [Ca^{2+}] [CO_3^{2-}] =$, where s is the equilibrium concentration of each ion, or square root of Ksp, or molar solubility.

Right?

Wrong. Molar solubility is NOT the equilibrium concentration. Furthermore, neither is guaranteed to be equal to square root of Ksp. You already wrote that [Ca2+]≠[CO32-], which precludes this result.

This is all fine, but we don't have a 1:1 base to conjugate acid buffer solution of silver cyanide. So we cannot define cyanide ion concentration as half the silver ion concentration. Now what?

If so, we do what I explained much earlier - we describe whole system with a full set of equations, describing every mass balance, charge balance, and every equilibrium, and we solve the whole system.

Borek said:
I am starting to think you are misunderstanding one of the concepts used, I just don't know which one yet.
Solubility - it is the amount of the substance that dissolves. It doesn't tell anything about equilibrium concentrations (although it puts some restrictions on their values).
Equilibrium concentration - concentration of a specific ion observed in the solution, once it is at equilibrium. Doesn't matter where the ion comes from.
Ksp - is expressed using equilibrium concentrations.Molar solubility is NOT the equilibrium concentration. Furthermore, neither is guaranteed to be equal to square root of Ksp. You already wrote that [Ca2+]≠[CO32-], which precludes this result.
Well, I wrote molar solubility is the concentration of calcium ion. And equal to the sum of all carbonate derived species.

Thank you for the insight, I probably did foul up somewhere and I just have to see where.

Okay, this is what my teacher did, and indeed he did use molar solubility to arrive at Ksp. "Offending" parts in orange boxes. Borek, can you clarify?

http://i.minus.com/jFjOBJDj83U4M.png

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Sorry, I have serious problems following what he wrote.

First part - about molar solubility - looks perfectly correct.

He is right you can't neglect hydrolysis of PO43-. It doesn't stop equation

$$K_{sp} = [Ag^+]^3[PO_4^{3-}]$$

from being right, you just can't plug concentration calculated from the solubility data directly into this definition.

Molar solubility s is

$$s = \frac {[Ag^+]}3 = [H_3PO_4] + [H_2PO_4^-] + [HPO_4^{2-}] + [PO_4^{3-}]$$

so to properly calculate Ksp from the solubility you should calculate what part of the phosphate is in the form of PO43-. Not trivial, but doable (assuming you know all dissociation constants).

If I read his print correctly that's exactly what he did, just neglecting further hydrolysis steps.

Borek said:
Sorry, I have serious problems following what he wrote.

First part - about molar solubility - looks perfectly correct.

He is right you can't neglect hydrolysis of PO43-. It doesn't stop equation

$$K_{sp} = [Ag^+]^3[PO_4^{3-}]$$

from being right, you just can't plug concentration calculated from the solubility data directly into this definition.

Okay, I did what you suggested, and I got 1.65*10^-13. I plugged in the amount of phosphate ion at equilibrium after considering phosphate ion hydrolysis (which is ~99% using molar solubility of silver phosphate as the initial concentration of phosphate ion).

This answer is several orders of magnitude off from 1) reference values and 2) several more magnitudes off from his calculated value of Ksp for silver phosphate. Here is the remainder of his work in which he actually finds a new Ksp.

http://i.minus.com/j4fBLdAe9gEB5.png

Also why did he drop the phosphate term from the new Ksp? True, hydrolysis is significant, but still, shouldn't that term be included? The amount of phosphate remaining as he calculated is in the order of something times ten to the negative eighth power; including that term would most definitely have a big impact on the Ksp.

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Qube said:
Okay, I did what you suggested, and I got 1.65*10^-13.

No way is it correct. We start with molar solubility in the 10- range, [Ag+]3 alone is 10-15, and that's before even taking PO43- into consideration.

http://i.minus.com/j4fBLdAe9gEB5.png

Also why did he drop the phosphate term from the new Ksp? True, hydrolysis is significant, but still, shouldn't that term be included? The amount of phosphate remaining as he calculated is in the order of something times ten to the negative eighth power; including that term would most definitely have a big impact on the Ksp.

I disagree with his statement "this is Ksp". This is dissolution equilibrium, but not the solubility product. Ksp has a well defined meaning, mentioned many times in this very thread.

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Borek said:
No way is it correct. We start with molar solubility in the 10- range, [Ag+]3 alone is 10-15, and that's before even taking PO43- into consideration.

Good point. I think I forgot to cube the silver ion concentration. Cubing it and account for the amount of phosphate left upon hydrolysis gives me a Ksp of 4.0 * 10^-23, which not that far off from my prof's calculated value.

$K_{sp} = [Ag^{+}]^{3}[PO_{4}^{3-}] = [1.55 x 10^{-5}]^{3}[1.55 x 10^{-5} - 1.5489 x 10^{-5}] = 4.0 x 10^{-23}$
Borek said:
I disagree with his statement "this is Ksp". This is dissolution equilibrium, but not the solubility product. Ksp has a well defined meaning, mentioned many times in this very thread.

What do you think the correct Ksp for silver phosphate is? This webpage gives it as something to the negative 17th power.

http://www.ktf-split.hr/periodni/en/abc/kpt.html

Wikipedia gives a similar value.

http://en.wikipedia.org/wiki/Silver_phosphate

Using your definition of Ksp and my teacher's definition of Ksp we both get something to the negative 23rd power. The values however still differ significantly.

So who's correct?

Did the literature really not take into account hydrolysis of the phosphate ion?

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Borek said:
disagree with his statement "this is Ksp". This is dissolution equilibrium, but not the solubility product. Ksp has a well defined meaning, mentioned many times in this very thread.

If I understand you correctly, you are saying that Ksp is the ion concentration product, where the ions are the constituent ions of the solid salt. Because hydroxide ion is NOT a constituent ion of silver phosphate, it should be dropped from the Ksp (but can be included in the dissociation equilibrium).

TBH - no idea. I see pKsp around 17 here (in my reference books) as well, and it clearly doesn't make sense.

These values are often repeated and copied from older sources, so it is enough that someone at some moment printed wrong value and it is repeated since then. The best approach would be to search the literature, but I have no access.

Qube said:
If I understand you correctly, you are saying that Ksp is the ion concentration product, where the ions are the constituent ions of the solid salt. Because hydroxide ion is NOT a constituent ion of silver phosphate, it should be dropped from the Ksp (but can be included in the dissociation equilibrium).

Yes.

There is no problem with defining and using any equilibrium you want, some of them are quite convenient to work with. Just don't try to call them by the names already used for something else.

Borek said:
Yes.

There is no problem with defining and using any equilibrium you want, some of them are quite convenient to work with. Just don't try to call them by the names already used for something else.

Okay, I'll perhaps try raising this with my professor. I guess I'll see how this goes. And upon consulting Daniel C Harris' book, Quantitative Chemical Analysis, he shares a similar definition of Ksp as being the ion product concentration (ions being the constituent solid's ions).

Daniel C. Harris said:
The solubility product is the equilibrium constant for the reaction in which a solid salt dissolves to give its constituent ions in solution.

So if that is so, would this also be incorrect? See the bottom of this screenshot of my prof's information sheet.

http://i.minus.com/jOxQFZ67LXICS.png

Frankly I can't see this going down well with my professor as the idea that Ksp = ion product concentration (ions being both those directly from the salt and those created upon hydrolysis) seems pretty damn ingrained in his mind .

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On the other hand he hasn't been entirely self-consistent; this is a picture of his own textbook.

So what's your take? Do I have a confused professor? Does this issue merit a chat with him? I think it does.

Qube said:
So if that is so, would this also be incorrect? See the bottom of this screenshot of my prof's information sheet.

http://i.minus.com/jOxQFZ67LXICS.png

That's OK with me. List contains Ksp for all salts, and some other constant for two salts - but it is clearly explained what the other constants are. As I wrote earlier, you can define and use any equilibrium constant you want, as long you are clear about what it is and you don't try to mislabel it.

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Borek said:
That's OK with me. List contains Ksp for all salts, and some other constant for two salts - but it is clearly explained what the other constants are. As I e wrote earlier, you can define and use any equilibrium constant you want, as long you are clear about what it is and you don't try to mislabel it.

Well he does give those two equilibrium constants but he takes them as Ksp anyway in class. Plus on the info sheet note that he lists the other two EQ constants under the heading of Ksp without any further elaboration of the fact that they resemble (actually are) dissociation equilibrium constants. Plus as you can see on the quiz key he mistakes a dissociation equilibrium for a solubility product. I think I should go see him about drawing the line between dissociation equilibrium constants and solubility product constants and how the latter involves "constituent ions" as he himself wrote in his own textbook.

I'll let you know how it goes. I believe the quiz key needs correction. Thank you for clarifying what Ksp means exactly :). Words are tricky sometimes. At least for silver phosphate the two EQ constants do not differ by a ton (as in a few magnitudes), but they do still differ by a fair bit. Percent difference is about 40% between your (Borek's) Ksp and my prof's "Ksp".

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There is no such thing as "my Ksp". I can list the value and give the reference (book title), but in no way I am claiming this is the right/wrong value.

If anything, I doubt the Ksp around 17, as it doesn't look OK when compared with the molar solubility, which is relatively easy to determine. But it doesn't have to be correct either.

1 person
Borek said:
There is no such thing as "my Ksp". I can list the value and give the reference (book title), but in no way I am claiming this is the right/wrong value.

If anything, I doubt the Ksp around 17, as it doesn't look OK when compared with the molar solubility, which is relatively easy to determine. But it doesn't have to be correct either.

I'm not referring to the numerical value; I'm referring to the method by which one arrives at Ksp. You used the concentrations of the salt's constituent ions; my prof uses the products of dissolution (which includes the products of salt anion hydrolysis). My issues are:

1) These are two different methods2) The literature and the prof's book seem to indicate that only the former method is the correct way of ascertaining Ksp (the former being: use the concentration of constituent ions, not products of ion/solvent interactions). 3) Using the two methods yields different answers too. In silver phosphate's case, the difference is about 40%.

I think these are the points I will raise with him the next time I see him.

What do you think?

I have nothing new to add.

1 person
Cool! I'll let you know what he thinks. This is an interesting problem.

Okay, I just got back from asking. He says:

1) The below method of calculating Ksp is wrong or at the very least fraught with uncertainty.

$K_{sp}=[Ag^{+}]^{3}[PO_{4}^{3-}]$

2) Why? Wikipedia molar solubility data gives us three sig figs. The amount of phosphate ion that is hydrolyzed is 100% to 3 sig figs! We can of course go to 4 sig figs and calculate the concentration of phosphate ion left over, but the data only support 3 sig figs (not 4!)

3) So what's your take, Borek? He told me that if Daniel C. Harris was with him in the room then Harris would completely agree with him on the statement that Ksp does not necessarily involve the salt's constituent ions.

Earlier he told me that if Linus Pauling were alive, Pauling would agree with him too (about a different issue, and I don't see why Pauling wouldn't have agreed with him).

4) My take: I agree with my chem instructor.

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I never suggested using Ksp in the form given to calculate solubility directly.

Argument about sig figs is completely off. We calculate concentration of PO43- left not by subtraction, but by multiplication (he did it himself!), so knowing total concentration of all forms of phosphates with 3 sig figs we can calculate concentrations of all forms of phosphates with the same accuracy (assuming we know all three Ka values with accuracy high enough). What he says is "we know concentration is 1.0 M, and as only 1.0% dissociated, we can't calculate concentration of the dissociated part, because we have not enough sig figs". Really? What about 0.010 M?

To some extent this is just semantics - looks like he doesn't differentiate between "dissolution equilibrium" and "solubility product". I do. Fact that you were confused seems to suggest his approach can be misguiding.

Edit: I just checked the IUPAC definition (as published in the orange book). It says

The product of the ion activities raised to appropriate powers of an ionic solute in its saturated solution expressed with due reference to the dissociation equilibria involved and the ions present.

so (unless my English fails me) it looks like it allows using the "solubility product" name for other formulas than just simple ions concentrations product. Didn't know. But if so, it requires listing the reaction itself, otherwise it is ambiguous.

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Borek said:
I never suggested using Ksp in the form given to calculate solubility directly.

No, we're going from solubility to Ksp, not the other way around. Solubility values are taken from Wikipedia.

Argument about sig figs is completely off. We calculate concentration of PO43- left not by subtraction, but by multiplication (he did it himself!), so knowing total concentration of all forms of phosphates with 3 sig figs we can calculate concentrations of all forms of phosphates with the same accuracy (assuming we know all three Ka values with accuracy high enough). What he says is "we know concentration is 1.0 M, and as only 1.0% dissociated, we can't calculate concentration of the dissociated part, because we have not enough sig figs". Really? What about 0.010 M?

Huh? Well, the Ka values we were working with only had 2 sig figs. So right from the beginning that limited us in how many sig figs we can extract for phosphate ion hydrolysis. Most (make that all) Ka values I've seen only have two sig figs - i.e. #.# * 10#. Plus the amount of phosphate ion we calculated to have been converted through hydrolysis is 100% to 2 or 3 sig figs.

I'm not sure if I follow the second part of your quote above.

To some extent this is just semantics - looks like he doesn't differentiate between "dissolution equilibrium" and "solubility product". I do. Fact that you were confused seems to suggest his approach can be misguiding.

Edit: I just checked the IUPAC definition (as published in the orange book). It says
so (unless my English fails me) it looks like it allows using the "solubility product" name for other formulas than just simple ions concentrations product. Didn't know. But if so, it requires listing the reaction itself, otherwise it is ambiguous.

Okay, looks like he's in the clear then, because he definitely listed the reactions before doing any calculations.

Qube said:
No, we're going from solubility to Ksp, not the other way around. Solubility values are taken from Wikipedia.

OK. It doesn't change anything. I never suggested to calculate Ksp from the solubility directly either. Ksp is not calculated using total (formal) concentrations, but equilibrium concentrations.

Huh? Well, the Ka values we were working with only had 2 sig figs. So right from the beginning that limited us in how many sig figs we can extract for phosphate ion hydrolysis. Most (make that all) Ka values I've seen only have two sig figs - i.e. #.# * 10#.

pKa1 = 2.148, pKa2 = 7.199, pKa3 = 12.35. 4 sig figs for each. I have a reference for that. But it doesn't matter at all.

3 sig figs for the molar solubility would be a problem if the only way to calculate concentration of PO43- left in the solution was "total minus eaten by hydrolysis", and the amount of PO43- left was below 0.1%. Then yes, we don't have enough accuracy for the calculations. However, it is not the only way. Actually it is the worst way I can think of, there are much better methods of calculating PO43- concentration.

You can start with assumption that hydrolysis was complete, use it to calculate pH and concentration of HPO42-, plug these numbers together with pKa3 into the dissociation constant definition

$$K_{a3} = \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$$

and solve for PO43-. You can then check if the assumption was correct - and as long as [PO43-] << [HPO42-] it holds well enough. That means concentration that you calculated is perfectly valid, and its accuracy is limited by either accuracy of molar solubility or accuracy of pKa3 - 3 sig figs vs 4 sig figs, so we can safely assume we got 3 sig figs from these data.

Note that if the assumption doesn't hold we are in the area where 3 sig figs of the molar solubility are not a problem for the "total minus eaten by hydrolysis" approach.

Borek said:
OK. It doesn't change anything. I never suggested to calculate Ksp from the solubility directly either. Ksp is not calculated using total (formal) concentrations, but equilibrium concentrations.

The values taken from Wikipedia are necessarily not the equilibrium concentrations? But why does my prof ask us to go from the Wiki values to Ksp? I'm guessing he's wrong in his approach?

Also what's the distinction between formal concentration and molar concentration?

 Okay I seem to have arrived at the fact that formal concentration does not take into account the actual chemical species which are solvated but rather only gives you the concentration of the moles of the original chemical formula in solution.

So if indeed Wiki gave us a formal (F) value rather than a molal (M) value ... how does that change anything? From the F value in grams of silver phosphate per liter we can go to moles of silver phosphate per liter. And since silver phosphate is stoichiometric, we can go to moles of silver ion in solution.

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The values taken from Wikipedia are necessarily not the equilibrium concentrations?

No, they are not. I explained it at least once, your prof explained it as well, you are still surprised?

But why does my prof ask us to go from the Wiki values to Ksp? I'm guessing he's wrong in his approach?

You can calculate Ksp from the molar solubility, just not directly. Knowing just molar solubility you can easily calculate formal concentrations, but not equilibrium concentrations - for these you need additional information about dissociation constants. Once you find equilibrium concentrations, you can calculate Ksp value.

"Naive" approach - treating formal concentrations as equilibrium concentrations - is a sure way of getting wrong Ksp value.

We are going in circles, I feel like I am wasting my time repeating the same again and again.

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