A problem regarding solubility and concentration

Question

A solution is $\pu{0.10 M}$ $\ce{Ba(NO3)2}$ and $\pu{0.10 M}$ $\ce{Sr(NO3)2}$. If solid $\ce{Na2CrO4}$ is added to the solution, what is the $\ce{[Ba+]}$ when $\ce{SrCrO4}$ begins to precipitate?
$\mathrm{K_{sp}}=\pu{1.2 \times 10^{-10}} \text{ for } \ce{BaCrO4}$
$\mathrm{K_{sp}}=\pu{3.5 \times 10^{-5}} \text{ for } \ce{SrCrO4}$

The solution given in my textbook is:

$\mathrm{K_{sp1}}=\ce{[Sr^2+][CrO4^2-]}$ $$\ce{[CrO4^2-]}=\frac{3.5\times 10^{-5}}{0.1}=3.5\times 10^{-4} \pu{M}\tag{1}$$ $\mathrm{K_{sp2}}=\ce{[Ba^2+][CrO4^2-]}$
$\ce{[CrO4^2-]_{total}}\approx\ce{[CrO4^2-]}\text{ from } \ce{SrCrO4}$ $$\ce{[Ba^2+}]=\frac{1.2\times 10^{-10}}{3.5\times 10^{-4}}=3.4\times 10^{-7}\pu{M}$$

First of all, in equation $(1)$ why did they put $\ce{[Sr^2+]}$ same as the initial concentration (before the addition of $\ce{Na2CrO4}$)?
When $\ce{Na2CrO4}$ is added, $\ce{[Sr^2+]}$ will react with $\ce{CrO4^2-}$ (obtained from $\ce{Na2CrO4}$) to form precipitate. The concentration of $\ce{[Sr^2+]}$ will undoubtedly change until the equilibrium is established.

This given solution is totally confusing, I would appreciate if you could tell me the proper way of solving this problem. I tried this on my own, but after calculating the concentration of $\ce{CrO4^2-}$, I couldn't make sense out of this question.

Answer 1

It is supposed addition of solid $\ce{Na2CrO4}$ would cause negligible volume change, so we can afford to consider the same $[\ce{Sr^2+}]$ as the initial one.

The task is rather theoretical, supposing the time point when $\ce{[Sr^2+][CrO4^2-]}$ just reached $K_\mathrm{sp,\ce{SrCrO4}}$, while $K_\mathrm{sp,\ce{SrCrO4}}$ and $\ce{[Sr^2+]}$ are given.

The proper computation procedure is already presented in you solved task. What else is unclear ?

Answer 2

I am going to assume that you already understand the theory behind this equilibrium problem. I wanted to first say that you are absolutely correct in that the [Sr2+] will "undoubtedly change until the equilibrium is established".

The short answer to your question is that the solution in the textbook used an assumption that we often use in equilibrium calculations for simplification.

Different textbooks have different rules, but the general idea is that, for an equilibrium constant K, such that K = [x][I + x], we can approximate K ~ [x][I] given that K << [I]. I have often seen that the condition for K << [I] holds for when [I] is at least 10^2 times greater than K.

So, basically, your textbook just simplified the work. Instead of writing [Sr2+] = 0.10 + x, they wrote the initial conc (0.10) because Ksp = 3.5*10^-5 << 0.10 to avoid solving an annoying quadratic.

Hope that helps.

Answer 3

Before we begin, there are two implicit assumptions:

• That there is not a significant change in the volume of the solution by either adding $\ce{Na2CrO4}$ nor the precipitation of the $\ce{BaCrO4}$
• That concentrations can be used instead of activities.

First of all, in equation $(1)$ why did they put $\ce{[Sr^2+]}$ same as the initial concentration (before the addition of $\ce{Na2CrO4}$)?

The problem statement asks for the $\ce{[Ba^{2+}]}$ when $\ce{SrCrO4}$ "begins to precipitate." Given that "some" $\ce{SrCrO4}$ must precipitate to reach the point where $\ce{SrCrO4}$ has already begun to precipitate, let's define some variables:

$\ce{[Sr^{2+}]_{init}}$ -- The initial concentration of $\ce{Sr^{2+}}$, 0.10 molar.
$\ce{[Sr^{2+}]_{ppt}}$ -- the concentration of $\ce{Sr^{2+}}$ removed by ppt.
$\ce{[Sr^{2+}]_{final} = [Sr^{2+}]_{init} - [Sr^{2+}]_{ppt}}$ -- the concentration of $\ce{Sr^{2+}}$ after some $\ce{SrCrO4}$ has precipitated.

Now here comes the rub. Chemistry isn't like math. The value of $\pi$ has been calculated to millions of digits. $\ce{[Sr^{2+}]_{init}}$ was given as 0.10 molar. So there are only two significant figures in this value. If 0.01% of the $\ce{Sr^{2+}}$ is used to form the initial precipitate then $\ce{[Sr^{2+}]_{final} \approx ([Sr^{2+}]_{init}} = \pu{0.10 M})$

Note that the problem could have been written differently. The problem could have asked for $\ce{[Ba^{2+}]}$ when $\ce{SrCrO4}$ reaches its $\mathrm{K_{sp}}$ (or some variation thereof). This would be right at the equilibrium point, before any $\ce{SrCrO4}$ has formed.

When $\ce{Na2CrO4}$ is added, $\ce{[Sr^2+]}$ will react with $\ce{CrO4^2-}$ (obtained from $\ce{Na2CrO4}$) to form precipitate. The concentration of $\ce{[Sr^2+]}$ will undoubtedly change until the equilibrium is established.

Obviously when $\ce{SrCrO4}$ nucleates there is some finite size of the $\ce{SrCrO4}$ particle which will be stable in an aqueous solution. So if we use a nanoliter of the cation solution then the given answer will be wrong. Thus there is a implied assumption that a macro volume of the solution is being used. "Macro" in this case meaning that creating a detectable amount of the $\ce{SrCrO4}$ does not change the $\ce{[Sr^2+]}$ from its initial nominal value of 0.10 molar.

I would appreciate if you could tell me the proper way of solving this problem.

The book gives the proper answer albeit without a full explanation of the additional assumptions.

The notion is that $\ce{BaCrO4}$ is more insoluble than $\ce{SrCrO4}$, so most of the $\ce{Ba^{2+}}$ will be removed before any $\ce{Sr^{2+}}$ starts to precipitate.

I'll add that this assumption isn't really true. As the $\ce{BaCrO4}$ precipitates, it will incorporate some of the $\ce{Sr^{2+}}$ into the $\ce{BaCrO4}$ precipitate. Again, not an appreciable amount compare to what is left in solution since initially $\ce{[Sr^{2+}] = [Ba^{2+}]}$. However if initially $\ce{[Sr^{2+}] = 0.01\cdot [Ba^{2+}]}$ then there would be a significant loss of $\ce{Sr^{2+}}$ in the solution.