[OmniReader]

Write and solve equilibrium relations to calculate solubility (mol·dm-3) of PbSO₄, K_sp=2.53×10^-8 in a solution already containing 0.2 mol·dm-3 of H₂SO₄ (1st dissociation complete, K_a2=0.0102) and 10^-6 mol·dm-3 of PbCl₂.

-----

PbCl₂ is all dissolved so we have 10^-6 M Pb²⁺ in solution before adding the sulphide.
Once we find [Pb²⁺] at saturation, solubility=[Pb²⁺] alone since this is the only species containing lead. Or, once we find [SO₄^2-] at saturation, solubility=[SO₄^2-]+[HSO₄^-]+[H₂SO₄].

Now for the equations.
Of course there is 2nd dissociation equilibrium of H2SO4 governed by Ka2. There is the Ksp equilibrium since we are at saturation. For this confusing system what are the mass and charge balances, and any equilibrium equation I missed?

[magician4]

depending on the precision desired, there are several ways to solve this problem.

(a) usually, you would work out some reasonable approximations, and end up with and expression that can easily be solved.

(b) you could, however, start with the desire to have higher precision,and start with a more global expression like a charge balance:

[H⁺] + 2 [Pb²⁺] = 2 [SO₄^2-] + [HSO₄^-] + [Cl^-] + [OH^-]

... where almost everything including K_Water is included, leaving only exotic species like for example [O^2-] and thatlike away, and still living with a [itex]\approx[/itex] c . Now, you'll just have to fill in the usual secondary conditions and rearrange everything a little bit ... 
this way, you'll end up with a mathematical equation of higher order  (something like x⁸ + ... = 0 in this case, I estimate) which requires numeric solution, but will give you high precision, indeed.


nevertheless, for more practical purposes people almost always would go for path (a), and that's what I would propose in this case, too.
in my opinion,  the very approximation required here would be, that solving additional PbSO₄ to the matrix already given wouldn't change the equilibrium [SO₄^2-] in any meaningfull way  (pls. ask yourself why I believe we're entitled to do this), and that [SO₄^2-] hence is governed by the K_a2 of sulphuric acid, exclusively.

so, we'll start with calculating this [SO₄^2-] :

[tex] K_{a2} \ =  \frac {[SO_4^{2-}]  \cdot  [H^+]}{[HSO_4^-]} \ =  \frac {[SO_4^{2-}]  \cdot  (c_0(H_2SO_4)  +  [SO_4^{2-}] )}{c_0(H_2SO_4)  -  [SO_4^{2-}] }[/tex]

from this, it follows:

[tex] [SO_4^{2-}] \ = \ - \ \frac {K_{a2} \ +   c_0(H_2SO_4)}{2} \ +   \sqrt { \left( \ - \ \frac {K_{a2} \ +   c_0(H_2SO_4)}{2} \ \right)^2 \ + \ K_{a2} \cdot c_0(H_2SO_4) } [/tex]

and now: just fill in the numbers and calculate...

the rest is easy:

[tex] K_{sp}  = [Pb^{2+}] \cdot [SO_4^{2-}]  \approx  ( [Pb^2+]_{PbCl2}  +  [Pb^{2+}]_{PbSO4} ) \cdot [SO_4^{2-}]  \to  [Pb^{2+}]_{PbSO4}  =  \frac {K_{sp} }{[SO_4^{2-}] }  -  [Pb^2+]_{PbCl2} [/tex]

... and from this , the required additional solubility of PbSO₄ is no problem at all for you, I take it


regards

Ingo

[OmniReader]

Let us start from high precision and then afterwards can think about approximations

[H⁺] + 2 [Pb²⁺] = 2 [SO₄^2-] + [HSO₄^-] + [Cl^-] + [OH^-]

is the charge balance, thanks. Equilibria expressions:

K_sp = [Pb²⁺]  * [SO₄^2-]
K_a2 = [SO₄^2-] * [H⁺] / [HSO₄^-]
K_water = [H⁺] * [OH^-]

We can say that [Cl^-] = 2 * 10^-6 M since it's original concentration is 2 * 10^-6 M and not involved in any equilibria (assume dissociation for HCl is complete).
What other equation(s) can we write for high precision? Looks like we have 5 variables right now so we need 1 more equation. Do we need K_sp of PbCl₂, since the solution is at saturation - will it hold to say that [Pb²⁺]  * [Cl^-] = K_sp of PbCl₂?

Approximations to come next

[magician4]

Looks like we have 5 variables right now so we need 1 more equation.

working with charge balances more often than not requires to work with more equations than variables*⁾ to be calculated: sometimes the additional secondary conditions will force to introduce new "variables" to the game, which is too bad, but can't be avoided, as those equations are the only game in town.
to me, it looks like exactly this might be the case here.

Do we need K_sp of PbCl₂, since the solution is at saturation - will it hold to say that [Pb²⁺]  * [Cl^-] = K_sp of PbCl_2?

the solution we're asked to consider will be at saturation, all right, but only with respect to lead sulfate PbSO₄.
with respect to lead chloride PbCl₂ , the solubility of this stuff is in the ballpark of 10 g/L , and hence lightyears beyond anything we have to deal with here (given the initial conditions reported): no, I don't think that K_sp(PbCl₂) should be allowed to show up here**⁾

btw.: the expression for a K_sp of PbCl₂ would have to be K_sp=[Pb²⁺]*[Cl^-]² , in case it should be required  (which it isn't: au contraire, it's forbidden to use this here)

What other equation(s) can we write for high precision?

from a first glance, my impression is that we'd need:

- a sulfate balance : [SO₄^2-]_{total, eq.} = [SO₄^2-]_{H2SO4, eq.} + [SO₄^2-]_{PbSO4, eq.}

- a lead-ion balance: [Pb²⁺]_{total, eq.} = [Pb²⁺]_PbCl2 + [Pb²⁺]_{PbSO4, eq.}

- the relation of [Pb²⁺]_PbCl2 to [Cl^-] :
[Pb²⁺]_PbCl2 : [Cl^-] = 1 : 2
(maybe even as absolute value(s), as those are obvious from initial conditions)

- a dissociation equation for lead sulfate: [PbSO₄]₀  [Pb²⁺]_{PbSO4, eq.} + [SO₄^2-]_{PbSO4, eq.}
(and subsequent relations resulting thereof)


and now, the nasty part is about to start: fumbling all those equations together, eliminating one variable after the other by expressing it indirectly  (until there are only known values left, plus the one and only variable you're looking for), and solving for [PbSO₄]₀ (or [SO₄^2-]_{PbSO4, eq.} , respectively )

  have fun 


regards

Ingo


*⁾
usually, it takes n equations to solve for n variables, as long as the variables (or parts of them) aren't geometrically coupled, as as in sin x = cos y .
for charge balances, you'll find that this "n" often isn't equal to the variables as suggested per the "naked" basic equation, and for the reasons explained above

**⁾
every secondary condition we're introducing has to be "true" for the final equilibrium  ( as the equilibrium results of all aspects denoted by those equations to be fulfilled at the same time).
using K_sp(PbCl₂) anyway, but without on the other hand the solution being saturated with respect to this substance, most likely will lead to a crash of this mathematical construction, i.e. will give you only negative, complex or otherwise impossible solutions, respectively,  for the final expression

[OmniReader]

Thanks. Looks like we're still under-determined though: from 5 variables and 4 equations, now we've got 8 variables and 6 equations? As your two mass balances also happen to introduce 4 new values (sulphate from the acid, sulphate from the salt, lead from lead chloride and lead from lead sulphate) where 1 only (lead from lead chloride) is a known constant (=10^(-6) M). I ask because I've never seen approach of adding the contributions from different places before (e.g. total sulphate = sulphate from acid + sulphate from salt), bit confusing how to deal with this, and why we need it rather than standard total equilibrium concentrations as before?

[magician4]

(...), and why we need it rather than standard total equilibrium concentrations as before?

what would those be?  everything directly or indirectly is connected to everything - that's the problem here!
  without detailled calculation we simply don't know (if we wish for precision, that is. else: choose method (a) instead, as shown earlier)

(...)bit confusing how to deal with this,

pls. allow for a moment : later tonight I might find time to give you my proposal 'bout how to get a single expression from the values / equations mentioned

see ya !


regards

Ingo

[OmniReader]

If you don't mind small interruption:

if we just wanted [Pb²⁺] total, could we write a straightforward mass balance(s)? then take away 10^-6 M from PbCl2 (fully dissolved), and we would get solubility.

[magician4]

if we just wanted [Pb2+] total, could we write a straightforward mass balance(s)? then take away 10-6 M from PbCl2 (fully dissolved), and we would get solubility.

it depends on what you'd call a "straightforward mass balance"

besides, in my opinion that's exactly what I did in my proposal ref. (a), when I subtracted [Pb²⁺]_PbCl2 from K_sp/[SO₄^2-]

with respect to my proposal ref. charge balance: I did begin, but it seems to take a little longer...

tnx for your patience

regards

Ingo

[OmniReader]

it depends on what you'd call a "straightforward mass balance"

besides, in my opinion that's exactly what I did in my proposal ref. (a), when I subtracted [Pb²⁺]_PbCl2 from K_sp/[SO₄^2-]

I mean, equation which maintains high precision approach, but changes problem to ask "equilibrium concentrations of all species" not "solubility of lead sulfate", and writes mass balance(s) in terms of total eq concentrations. and you do show that in (a), but we still need extra mass balance to help find [Pb²⁺]_eq in the high precision way. Once values of [Pb²⁺]_eq are known, subtracting 10^-6 M is exact, no? since PbCl₂'s K_sp is not approached here

Maybe I''m just asking distracting side questions. please do carry on with the approach from before, if it cannot be done this way. I much appreciate your help

[magician4]

here we go....

... and I hope I don't get into trouble for posting quite lengthly calculations...


basic equation:
[tex][H^+] + 2 [Pb^{2+}] = 2 [SO_4^{2-}] + [HSO_4^-] + [Cl^-] + [OH^-][/tex]

introducing:
[tex](i) \ K_w = [H^+] * [OH^-][/tex]
hence:
[tex](I) \ [H^+] + 2 [Pb^{2+}] = 2 [SO_4^{2-}] + [HSO_4^-] + [Cl^-] + \frac {K_W}{[H^+]}[/tex]

introducing:
[itex][H^+] [/itex]will be composed of a fixed socket from complete first dissociation of sulfuric acid, plus an additional  variable amount I'll call [itex]H'^+[/itex] :
[tex](ii) \ [H^+] = c_0(H_2SO_4) + H'^+[/tex]
(consequence of given initial conditions)
hence:
[tex](II) \ (c_0(H_2SO_4) + H'^+) + 2 [Pb^{2+}] = 2 [SO_4^{2-}] + [HSO_4^-] + [Cl^-] + \frac {K_W}{c_0(H_2SO_4) + H'^+}[/tex]

introducing:
[tex](iii) \ c_0(PbCl_2) = [Pb^{2+}]_{PbCl2} = 0.5 [Cl^-][/tex]
(dissociation of initial [itex]PbCl_2[/itex])
hence:
[tex](III) \ (c_0(H_2SO_4) + H'^+) + 2 [Pb^{2+}] = 2 [SO_4^{2-}] + [HSO_4^-] + 2 c_0(PbCl_2) + \frac {K_W}{c_0(H_2SO_4) + H'^+}[/tex]

introducing:
[tex](iv) \ [Pb^{2+}] = [Pb^{2+}]_{PbSO_4} + [Pb^{2+}]_{PbCl2}[/tex]
(lead ion balance)
hence (including (iii) ) :
[tex](IV) \ (c_0(H_2SO_4) + H'^+) + 2 [Pb^{2+}]_{PbSO4} = 2 [SO_4^{2-}] + [HSO_4^-] + \frac {K_W}{c_0(H_2SO_4) + H'^+}[/tex]

some side calculations

(A) ref. sulfuric acid
[tex](v) \ c_0(H_2SO_4) = c_0(HSO_4^-)_{H2SO4}[/tex]
(given from initial problem)

(B) theorem:
we consider the sulfate, hydrogensulfate species as to be composed as follows:
[tex](vi) \ [HSO_4^-] = c_0(HSO_4^-)_{H2SO4} \ - \ [SO_4^{2-}]_{H2SO4}[/tex]
[tex](vii) \ [SO_4^{2-}] = [SO_4^{2-}]_{H2SO4} + [SO_4^{2-}]_{PbSO4}[/tex]

(C) sulfuric acid, [itex]K_{a2}[/itex]
[tex](viii) \ K_{a2} = \frac {[SO_4^{2-}] \cdot [H^+]_{total}}{[HSO_4^{-}]} = \frac {( [SO_4^{2-}]_{H2SO4} + [SO_4^{2-}]_{PbSO4}) \cdot (c_0(H_2SO_4) + H'^+)}{c_0(HSO_4^-)_{H2SO4} \ - \ [SO_4^{2-}]_{H2SO4}}[/tex]

[tex]\to [SO_4^{2-}]_{H2SO4} = \frac {K_{a2} \cdot c_0(HSO_4^-)_{H2SO4} - [SO_4^{2-}]_{PbSO4} \cdot (c_0(H_2SO_4) + H'^+)}{c_0(H_2SO_4) + H'^ \ + \ K_{a2}}[/tex]

(D) solubility lead sulfate (including (iv), (v) (vii) ):
[tex](ix) K_{sp} = [Pb^{2+}] \cdot [SO_4^{2-}] = ( [Pb^{2+}]_{PbSO_4} + c_0(PbCl_2) ) \cdot ([SO_4^{2-}]_{H2SO4} + [SO_4^{2-}]_{PbSO4} )[/tex]
[tex]\to [SO_4^{2-}]_{H2SO4} = \frac {K_{sp}}{[Pb^{2+}]_{PbSO_4} + c_0(PbCl_2) } - [SO_4^{2-}]_{PbSO4}[/tex]

(E) combining (viii) and (ix)

[tex]\frac {K_{a2} \cdot c_0(HSO_4^-)_{H2SO4} - [SO_4^{2-}]_{PbSO4} \cdot (c_0(H_2SO_4) + H'^+)}{c_0(H_2SO_4) + H'^ \ + \ K_{a2}} = \frac {K_{sp}}{[Pb^{2+}]_{PbSO_4} + c_0(PbCl_2) } - [SO_4^{2-}]_{PbSO4}[/tex]

introducing:
[tex](x) \ [Pb^{2+}]_{PbSO4} = [SO_4^{2-}]_{PbSO4}[/tex]

combining with (E)

[tex]\frac {K_{a2} \cdot c_0(HSO_4^-)_{H2SO4} - [Pb^{2+}]_{PbSO4} \cdot (c_0(H_2SO_4) + H'^+)}{c_0(H_2SO_4) + H'^ \ + \ K_{a2}} = \frac {K_{sp}}{[Pb^{2+}]_{PbSO_4} + c_0(PbCl_2) } - [Pb^{2+}]_{PbSO4}[/tex]

[tex]\to [/tex]

[tex]\frac {K_{a2} \cdot c_0(HSO_4^-)_{H2SO4}  \cdot  ([Pb^{2+}]_{PbSO_4} + c_0(PbCl_2))}{K_{sp} - ([Pb^{2+}]_{PbSO_4} + c_0(PbCl_2) ) \cdot [Pb^{2+}]_{PbSO4}} \ - \ K_{a2} = (c_0(H_2SO_4) + H'^+) \cdot (1 +  \frac {[Pb^{2+}]_{PbSO4}   \cdot  ([Pb^{2+}]_{PbSO_4} + c_0(PbCl_2))}{K_{sp} - ([Pb^{2+}]_{PbSO_4} + c_0(PbCl_2) ) \cdot [Pb^{2+}]_{PbSO4}} )[/tex]

[tex]\to[/tex]

[tex](xi) \ c_0(H_2SO_4) + H'^+ = \frac {K_{a2} \cdot c_0(HSO_4^-)_{H2SO4}  \cdot  ([Pb^{2+}]_{PbSO_4} + c_0(PbCl_2)) \ - \ K_{a2} \cdot (K_{sp} - ([Pb^{2+}]_{PbSO_4} + c_0(PbCl_2) ) \cdot [Pb^{2+}]_{PbSO4})}{K_{sp} - ([Pb^{2+}]_{PbSO_4} + c_0(PbCl_2) ) \cdot [Pb^{2+}]_{PbSO4} +  [Pb^{2+}]_{PbSO4}   \cdot  ([Pb^{2+}]_{PbSO_4} + c_0(PbCl_2))}[/tex]

(F) calculation of equilibrium sulfate
from  (ix)  it follows that

[tex](xii) \ [SO_4^{2-}] = \frac {K_{sp}}{[Pb^{2+}]_{PbSO_4} + c_0(PbCl_2) }[/tex]

(G) calculation of equilibrium hydrogensulfate
from (vi), (viii) and (x) it follows that

[tex](xiii) \ [HSO_4^-] = c_0(HSO_4^-)_{H2SO4} \ - \frac {K_{a2} \cdot c_0(HSO_4^-)_{H2SO4} - [Pb^{2+}]_{PbSO4} \cdot (c_0(H_2SO_4) + H'^+)}{c_0(H_2SO_4) + H'^+ \ + \ K_{a2}}[/tex]


so much, so good.

what we'd do next (but I won't make that explicit no longer: I am sure this "pure mathematics" thing you'd be able to work out for yourself)

- introducing (F) and (G) to (IV), gaining (V)
- multiplying all (V) with [itex](c_0(H_2SO_4) + H'^+)[/itex] and gaining a square equation (VI)
- solving this equation, and gaining [itex]"c_0(H_2SO_4) + H'^+ = ... "[/itex] thereof: (VII)
- combining (VIII) and (xi) , i.e. eliminating [itex]"c_0(H_2SO_4) + H'^+" [/itex](and having both the right sides equal)

... which will lead you to a pretty nasty expression (IX), which, however,  aside from [itex][Pb^{2+}]_{PbSO4} [/itex] will consist exclusively of other known values and constants.

we're pretty much there: just rearrange this equation untill you've nothing left but a summ of powers of [itex][Pb^{2+}]_{PbSO4}[/itex](weighted by combinations of said constants and known values) equaling zero

finally, do a numeric solution of this expression: that's how it's done


... and I pretty much prefer the approximated aproach (a) over this, as I think you'll understand by now


any blunders that might have happened are mine exclusively, and I apologize

and no, I don't know any "way in the middle" between (a) and (b): either highest precision or not

...and (a) is really , really , really good enough for all practical purposes, trust me

regards

Ingo


addendum:
I just see, that I forgot to eliminate "[itex]c_0(H_2SO_4) + H'^+[/itex]" from (part of ) the denominator of (xiii) (right side).
  you'll have to replace it with (xi)

[OmniReader]

Thank you so much! Your help has been incredibly detailed.

Slightly off-topic question: for  definition of solubility PbSO4, does Solubility = [Pb²⁺], or something else, if something else, what is the mathematical definition of molar solubility? I thought it was total concentration of an ion from the salt, in all forms that contain that ion (inc hydrolysed forms or complexed), divided by the ion's coefficient in the salt, is solubility of the same salt.

Maybe, we can change this. total concentration of an ion from the salt, in all forms that contain that ion (inc hydrolysed forms or complexed), minus the concentration of the ion before your salt was added (in this case this is 10^-6 M from PbCl2), divided by ion’s coefficient in the salt. (This step seems high-precision to me) Is this universal?

[magician4]

Slightly off-topic question: for  definition of solubility PbSO4,(...)

"solubility" being a more general expression with many specific meanings, I think your question is a good one in it's own right.

I am aware of at least 6 different more precise subjects summarised by "solubility", amongst them "mass concentration of a saturated solution of named substance" , "molar concentration of a saturated solution of named substance" an a more indirect expression denoted by K_sp or pK_sp , respectively (i.e. referring to the law of mass action expression belonging to the saturated solution)
(the latter is the one I was dealing with in # "D")

Maybe, we can change this. total concentration of an ion from the salt, (...)

I see your point, but this wont work here, because in my reading, here*⁾ we're asked for the (hypothetical) concentration of the undissociated salt c₀(PbSO4)  that still might be added (if it was alone in the solvent), but under the side conditions as expressed by the problem.
here, this is not identical with the concentration of (dissolved) lead sulfate in the system, as the presence of both lead chloride and sulfate (from the sulfuric acid) is tantamount to a partially pre-saturated system, with respect to PbSO₄


regards

Ingo

*⁾ "here" meaning:

"(...) to calculate solubility (mol·dm-3) of PbSO₄,(...)"

(from your original question)

[OmniReader]

I see your point, but this wont work here, because in my reading, here*⁾ we're asked for the (hypothetical) concentration of the undissociated salt c₀(PbSO4)  that still might be added (if it was alone in the solvent), but under the side conditions as expressed by the problem.
here, this is not identical with the concentration of (dissolved) lead sulfate in the system, as the presence of both lead chloride and sulfate (from the sulfuric acid) is tantamount to a partially pre-saturated system, with respect to PbSO₄

thanks. I covered the PbCl2 in "minus the concentration of the ion before your salt was added (in this case this is 10-6 M from PbCl2),", which is same approach as your high-precision solution takes as you let c₀(PbCl₂)+[Pb²⁺]_PbSO4 = [Pb²⁺]_tot. so, [Pb²⁺]_tot+(species containing Pb²⁺ tot conc at equilibrium)-[Pb²⁺]_PbCl2 = solubility. is this ok as general expression for solubility (assuming given problem can be solved at all ...)?

[magician4]

I would go for that, too

let's put it like this: you might define "solubility" with respect to this problem as (1) "referring to PbSO₄ that might still be added" or as (2) "total final concentration of PbSO4 in solution, denoted by equal (!) concentrations of Pb²⁺ and SO₄^2- ions, no matter what the sources, and neglecting any excess of any of those ions whatsoever (in this case: sulfate)" *⁾

by the end of the day, your professor has to make clear what he wishes to be understood by his question

anyway: no matter what your choice might be (and you might give him both solutions, just to be on the safe side, as it's no big deal to calculate them from each other), you'll have to take either path (a) or path (b) for the very calculation : a different choice of what is to be calculated ( i.e. (1) or (2), respectively) is not at all related to the mode of calculation, both (1) and (2) being of the same difficulty and demands  (i.e. you won't gain nothing from going for the one, but not for the other instead)


regards

Ingo

*⁾
interestingly, this definition would not cover the situation you had from preparing a saturated solution of pure PbSO4 in pure water  ( as immediately it would react to a defined solution of lead sulfate, flanked by lead hydrogensulfate and lead hydroxide  from partial reprotonation of sulfate).

[OmniReader]

ok, if we want for some oddball reason to define solubility in terms of sulphate ion, maintaining high precision method - and ok with the amount of work (please DON'T go to the trouble this time round, if I wanted solution for this way I'd use mathematica!) - can we do something like S = [SO₄^2-]_tot,eq + [HSO₄^-]_tot,eq - ([SO₄^2-]_acid,eq + [HSO₄^-]_acid,eq)... where [SO₄^2-]_acid,eq, [HSO₄^-]_acid,eq are equilibrium conc of these two species calculated as the system was before the adding of PbSO₄. will this hold? - would it bring the same answer as our definition of solubility from S = [Pb²⁺] - 10^-6?

looking to see how all these definitions interconnect, and whether they're self-consistent