[UG]

Hey everyone, I'm a bit stuck on a pH/solubility calculating question. It goes:
Carbonic acid has the first and second ionization constants equal to 4.4x10^–7 and 5.6x10^–11, respectively. The solubility product constant for barium carbonate is 5.5x10^–10. Calculate the solubility of BaCO₃ taking into account the hydrolysis of the ions.
So I've worked out that without ionisation, the solubility would be (5.5x10^-10)^0.5 = 2.345x10^-5 mol L^-1. But now I have no idea what to do with the carbonate, I've never done a calculation like this before. From what I can gather, [Ba²⁺] = [H₂CO₃] + [HCO₃^-] + [CO₃^2-] and the charge, 2[Ba²⁺] + [H⁺] = [OH^-] + [HCO₃^-] + 2[CO₃^2-] Can someone please give me a way to get this problem started?

[Borek]

So far so good.

In general - there are four more equations describing the solution, can you write them out?

Can you try to guess what is the solution pH? Not precisely, but at least approximately? That will give you a hint about which ions can be safely ignored.

[UG]

Thanks Borek,
Two of them should be the K_a values

K_a1 = [HCO₃^-][H₃O⁺]/[H₂CO₃]

K_a2 = [CO₃^2-][H₃O⁺]/[HCO₃^-]

Are the other two like K_w=[H₃O⁺][OH^-] and K_w = K_a x K_b?

I think the pH will be somewhere around 9.5 - 10, is that a good enough guess? 

[Borek]

Two of them should be the K_a values

Right.

Are the other two like K_w=[H₃O⁺][OH^-] and K_w = K_a x K_b?

Not exactly. Kw is OK, but Kw=KaKb doesn't give any new, additional information about what is happening in the solution. In other words - Kb can be already calculated using known Ka and Kw.

However, you are still missing very important equilibrium, one that was listed in the original question.

I think the pH will be somewhere around 9.5 - 10, is that a good enough guess?  

I would put it in the same range - slightly alkalic.

This is a very important information. In your mass and charge balances you have sums of concentrations. However, some of the concentrations there are much smaller than others - so you can safely ignore them. Knowing that solution is slightly alkalic, can you tell which ones can be neglected?

Note: at this stage I am not yet sure if we will not finally get polynomial of too high degree. If it happens, we will be forced to look a step back. But so far we are on the right track.

[UG]

Oh right, you mean like

BaCO₃ (s) Ba²⁺ (aq) + CO₃^2- (aq)

And for [Ba²⁺] = [H₂CO₃] + [HCO₃^-] + [CO₃^2-] and  2[Ba²⁺] + [H⁺] = [OH^-] + [HCO₃^-] + 2[CO₃^2-] I think we can ignore [H₂CO₃] and [H⁺], how's that? I'm afraid I still do not quite see where this is leading to.

[Borek]

Oh right, you mean like

BaCO₃ (s) Ba²⁺ (aq) + CO₃^2- (aq)

Yes, Ksp is the last equilibrium that we should take into account.

I think we can ignore [H₂CO₃] and [H⁺], how's that? I'm afraid I still do not quite see where this is leading to.

System is described by system of several equations - if you can solve them, you will know concentrations of all ions. You already know all equations, but solving the system is difficult, as equations are nonlienar and solving them leads to high degree polynomials. What we are trying to do is to find out a way of simplifying the system of equations to make it easier to solve.

You are right about [H₂CO₃] and [H⁺]. Note that if we assume [H₂CO₃] is neglectalbly small, we can also ignore Ka1 - as [H₂CO₃] is present only in this one equation, we can use Ka1 to calculate [H₂CO₃] AFTER we know all other concentrations.

Now, try to write system of all the remaining equations ignoring [H₂CO₃] and [H⁺] in all sums (but not in products and quotients). Does it look solvable? Are there any other approximations you can think off?

[UG]

Now, try to write system of all the remaining equations ignoring [H₂CO₃] and [H⁺] in all sums (but not in products and quotients). Does it look solvable? Are there any other approximations you can think off?

I'm not quite sure what you mean by this, but here goes anyway...

If we have [Ba²⁺] = [HCO₃^-] + [CO₃^2-]           (1)
and 2[Ba²⁺] = [OH^-] + [HCO₃^-] + 2[CO₃^2-]           (2)

Using the K_a expression then [CO₃^2-] = K_a2[HCO₃^-]/[H₃O⁺]                (3)
Substituting 1). into 2). gives
2[HCO₃^-] + 2[CO₃^2-] = [OH^-] + [HCO₃^-] + 2[CO₃^2-]
which simplifies to [HCO₃^-] = [OH^-]       (4)
Substituting 3 and 4 back into 1 gives

[Ba²⁺] = [OH^-] + K_a2[HCO₃^-]/[H₃O⁺]
Which can also be written [Ba²⁺] = K_w/[H₃O⁺] + K_a2[HCO₃^-]/[H₃O⁺]

Is this right?

[Borek]

Unfortunately, seems like we will get a high degree polynomial, just like I was afraid of. Sigh.

Your working is mostly OK, however, final equation that you get still contains several unknowns, so it doesn't move us ahead.

These were the equations that we had just a step back:

Ksp = [Ba²⁺][CO₃^2-]

Kb = [HCO₃^-][OH^-]/[CO₃^2-]

2[Ba²⁺] = [OH^-] + [HCO₃^-] + [CO₃^2-]

[Ba²⁺] = [HCO₃^-] + [CO₃^2-]

(note I have changed Ka to Kb - to eliminate [H⁺] and Kw, we don't need them anymore). Assuming [OH^-] = [HCO₃^-] we get three equations in three unknowns:

Ksp = [Ba²⁺][CO₃^2-]

Kb = [HCO₃^-]²/[CO₃^2-]

[Ba²⁺] = [HCO₃^-] + [CO₃^2-]

(charge balance equation is now identical to mass balance, so I have left it out; your approach yielded the same effect and I can't see anything wrong about it).

Many ways of eliminating variables now. My approach was to solve first equation for [Ba²⁺], solve second equation for [HCO₃^-] and substitute them into third equation, yielding

Ksp = [CO₃^2-]*sqrt(Kb*[CO₃^2-]) + [CO₃^2-]

This euqation is at the same time good and bad. Good - it contains only one variable, so technically when we solve it we have the answer (well, almost, we have to use it to calculate [Ba²⁺], but this is pretty simple). Bad - it is difficult to solve. Substituting x=sqrt([CO₃^2-]) we get fourth degree polynomial

x⁴ + sqrt(Kb)*x³ - Ksp = 0

and fourth degree polynomial is not a thing that I like to solve manually (at this stage mathematicians say something like "the rest is obvious" and they disappear without trace). However, it can be realatively easily solved numerically (even on handheld, or using spreadsheet) so in many cases this approach would be acceptable.

More later.

[Borek]

OK, now for a small trick.

Such equations (like x⁴ + sqrt(Kb)*x³ - Ksp = 0) can be often easily solved iteratively. That is we will express it as x'=f(x), we will strat with some guessed value for x and we will check if iterations converge to some value. Let's see. Our polynomial can be expressed as

x³(x + sqrt(Kb)) - Ksp = 0

so our x'=f(x) can take form

x' = third_degree_root(Ksp/(x+sqrt(Kb)))

This is easily implemented in spreadsheet - see attached image. A1 contains initial value (10^-13, but it works for 1 and 10^-30 as well), then each next cell in a row is a function of the value of the cell above. There is no third root function, so I used exp and ln for calculations.

As you can see convergence is almost instant - after third step following results are identical.

But, what have we calculated? Ah, yes. Our x is a square root of the [CO₃^2-]. That means [CO₃^2-] = (3.21*10^-3)² = 1.03*10^-5, and solubility of the barium carbonate (which equals concentration of dissolved barium) is

[Ba²⁺] = Ksp/[CO₃^2-]

or

[Ba²⁺] = 5.33*10^-5 M

Interestingly, when we compare this result with the result of full calcuation, taking into acount all equilibria - our result is perfectly accurate (see the second image).

But I guess you may not like this approach 

[UG]

Whao, thanks Borek  That was intense but I'm pretty sure I get it now, so thanks again!

[UG]

One more thing, so now how do I find the [H⁺]? Do I find [HCO₃^-] and then substitute values into K_a2 = [CO₃^2-][H₃O⁺]/[HCO₃^-] ?

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Edit, never mind, I noticed that all the numbers are there on your attchments.

[Borek]

One more thing, so now how do I find the [H⁺]?

Any way you like - for example [OH^-] = [HCO₃^-] and [H⁺] = Kw/[OH^-]. Once you have one unknown calculated it is not difficult to calculate all other values.

Note that this approach - while precise - is not always the best and the fastest. Especially during quizzes you may have not enough time to follow whole procedure. But it is really worth to know that it can be done this way.