If the solubility products of the following ionic compounds all have the same numerical value, and all solubilities have values between zero and one, which one would have the lowest solubility in moles of solute per litre?





Their chemical equations and $\ce{Ksp}$ values would be:

  1. $\ce{PQ <-> P +Q}$ and $\ce{Ksp=[P][Q]=[P][P]=[P]^2}$

  2. $\ce{RS2 <-> R +2S}$ and $\ce{Ksp=[R][2S]^2=[R][2R]^2=4[R]^3}$

  3. $\ce{T2U2 <-> 2T +2U}$ and $\ce{Ksp=[2T]^2[2U]^2=[2T]^2[2T]^2=16[T]^4}$ (although the textbook says $\ce{[T]^4}$ which I don't understand)

  4. $\ce{VW4 <-> V + 4W}$ and $\ce{Ksp=[V]^2[4W]^4=[W]^2[4W]^4=256[W]^5}$ (although the textbook says $\ce{256[V]^5}$ but I am pretty sure its equivalent whether it is written in terms of V or W)

FYI my thought process here is that:

  1. Based on the chemical equation RS2⟷R+2S
  2. The concentration of one product (S) is twice that of (R).
  3. So if there is X moles of RS2, and there is X moles of R and 2X moles of S
  4. Hence in the Ksp=[A]^n⋅[B]^m , A=X and B=2X

Now the textbook says

As $\ce{[P][P]=4[R]^3=[T]^4=256[V]^5}$
[P] must be the lowest because $\ce{[P]^2}$ has the same value as $\ce{4[R]^3 and [T]^4 and 256[V]^5}$

I can't get my head around this. I would have thought this proves $\ce{[P]}$ has the highest value not the lowest. Why is [P] the LOWEST?

E.g. noting that all values are between $0$ and $1$ (as defined in the question)

$\ce{[P]^2=256[V]^5}$ (the other two options are in the middle so don't matter)

If both sides equaled e.g. $0.25$ (which is between $0$ and $1$)

$\ce{[P]^2=0.25 and 256[V]^5=0.25}$

${[P]=0.5}$ and ${[V]=(0.25/256)}^{1/5} = 0.25$

Showing [P]>[V], so [P] (moles/L) is the greatest solubility, not the lowest.

Is anyone able to spot my error?

Question and answer as shown in book below (I have typed up the relevant parts above but sometimes I interpret parts of the textbook question incorrectly (making my stack exchange question invalid or misleading): Question Answer

  • $\begingroup$ You could help yourself out a lot if you wrote out the final concentrations in terms of the original solid, instead of arbitrarily (?) picking one of the ions to express it in. $\ce{PQ}$ in the aqueous phase is essentially the solubility. My guess is that book is missing that factor of 16 for $\ce{T2U2}$. $\endgroup$
    – Zhe
    Nov 14, 2016 at 21:11
  • $\begingroup$ Thanks for the suggestion Zhe. You are right that I did arbitrarily pick one. I am not sure what the textbook's reasoning was in doing that. If I write them in terms of the original solid I get [PQ]^2=4[RS2]^3=16[T2U2]=256[VW4]^3 and so am stuck with the same problem I was having (although it does help give me a more systematic approach). The issue being that [PQ]^2=256[VW4]^3 to me shows that [PQ] > [VW4] whilst the solution says [PQ] is the lowest solubility. $\endgroup$
    – K-Feldspar
    Nov 14, 2016 at 21:18
  • $\begingroup$ Also thank you for confirming what I thought about the factor of 16 for T2U2. Do you have any idea as to the other part of my question regarding why the book says [P] < [VW4] or [V] ? (Because the 16*T2U2 answer is somewhat irrelevant as if falls between the other two ([PQ] and [VW4]) options and can be excluded). $\endgroup$
    – K-Feldspar
    Nov 14, 2016 at 21:24
  • $\begingroup$ Here's the issue. The value $a^{x}$ gets bigger for bigger values of $x$ only when $a > 1$. For small values of $a$, it's the reverse. Notice the key phrase "all solubilities have values between zero and one." This means for a fixed small value of $K_{\mathrm{sp}}$, $a$ will be bigger if $x$ is bigger. This means that your assertion of $\ce{[PQ]} > \ce{[VW4]}$ is incorrect based on the previous equation. $\endgroup$
    – Zhe
    Nov 14, 2016 at 21:29
  • $\begingroup$ I think you have spotted my mistake, but when I test it numerically I can't seem to prove it. E.g. if I assumed Ksp was 0.25 (i.e. a small value i.e. between 0 and 1 as specified). $[PQ]^2=Ksp=0.25=256[VW4]^5$ ... This simplifies to $[PQ]=sqrt(0.25)$ = $0.5 & [VW4]$ = $(0.25/256)^(1/5)=0.25$. Hence $[PQ]$ (which is 0.5) > $[VW4]$ (which is 0.25) (please ignore the box, I can't seem to fix the formatting) $\endgroup$
    – K-Feldspar
    Nov 14, 2016 at 21:41

2 Answers 2


Easier just to migrate my comment into an answer. Your assumption $$\ce{[PQ]} > \ce{[VW4]}$$ is not valid based on $$\ce{[PQ]^{2}} = 256\ce{[VW4]}^{5}$$

For $y = a^{x}$, if $a>1$, larger $x$ leads to larger $y$. But for $a<1$, larger $x$ leads to smaller $y$, and the question indicates that solubilities are between zero and one, i.e., small.

If you plug in $K_{\mathrm{sp}}= 10^{-10}$, you get $\ce{[PQ]}=10^{-5}$ and $\ce{[VW4]} = 3.3\cdot 10^{-3}$.

As I said in my comment, I believe that this is a very poorly designed question: the constraints are too broad and there is a typo in the key. Someone was very sloppy...


Is anyone able to spot my error?

I think so. There is a systematic error in your equations when it comes to stoichiometric factors.

What is the solubility product constant $\mathrm{K_{sp}}$ for the reaction $$\ce{A_nB_m(s) <=> nA(aq) + mB(aq)}$$

$$\mathrm{K_{sp}} = [\ce{A}]^n\cdot [\ce{B}]^m$$

  • $\begingroup$ Thank you for the answer Klaus. I agree with that formula. But based on the chemical equation RS2⟷R+2S ... doesn't this show the concentration of one product (S) is twice that of (R). So if there is X moles of RS2, doesn't that mean there is X moles of R and 2X moles of S? Hence in the Ksp=[A]^n⋅[B]^m , doesn't A=X and B=2X? This seems to be what the solution shown at the bottom of the question has done. $\endgroup$
    – K-Feldspar
    Nov 14, 2016 at 19:03
  • $\begingroup$ As they have done here in step 2 i.imgur.com/uUGtdT8.png (Source mhchem.org/223/223PPT/pdf223/III18b01Chapter18bCG.pdf) $\endgroup$
    – K-Feldspar
    Nov 14, 2016 at 21:28

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