Calculating Precipitate and Moles: Na2S + Co(NO3)2

[STEMucator]

Homework Statement

$300.0 mL$ of $1.50 \times 10^{-10} M$ $Na_2S_{(aq)}$ is combined with $200.0 mL$ of $1.50 \times 10^{-10} M$ ${Co(NO_3)_2}_{(aq)}$.

Determine what precipitates.
Determine how many moles precipitate.

Homework Equations

The Attempt at a Solution

The first question was easy. I simply found the moles of each substance before mixing, which are equal to the moles of the ions $Co^{+2}, S^{-2}$. Then I divided the moles of ions by the new total volume of $500.0 \times 10^{-3} L$ to get the new ion concentrations after mixing.

I found $K_{sp} = 4.0 \times 10^{-21}$ in my reference material and I calculated the ion product to be $Q = 5.40 \times 10^{-21}$. Since $Q > K_{sp}$, the solution is supersaturated and so $CoS_{(s)}$ precipitate will form until $Q = K_{sp}$.

When the question asks how many moles precipitate, is it simply asking me to find the limiting reagant that produces fewer moles of $CoS$?

 

[Borek]

No, you need to check how much can precipitate before Q=Ksp.

 

[STEMucator]

No, you need to check how much can precipitate before Q=Ksp.

EDIT: Thinking...

$Q > K_{sp}$ is telling me the ion concentration is too high and the reaction will shift to produce more $CoS$... So that means:

$CoS ⇌ Co^{+2}+S^{-2}$

Will shift to the left.

Okay I think I got the answer now. I'm getting $n_{CoS} = 3.00 \times 10^{-11} mol$. Does that sound reasonable?

 

[Borek]

That's not what I got.

After precipitating 3e-11 moles of CoS there would be no Co left in the solution, and solution would be not saturated.

Assume x moles precipitated. What is concentration of Co²⁺ now? What is concentration of S^2- now? What should their product be equal to?

 

[STEMucator]

That's not what I got.

After precipitating 3e-11 moles of CoS there would be no Co left in the solution, and solution would be not saturated.

Assume x moles precipitated. What is concentration of Co²⁺ now? What is concentration of S^2- now? What should their product be equal to?

Bah! My original idea about using the ICE strategy with concentrations was the way to go then.

So at the moment the two solutions are mixed, $Q > K_{sp}$ and some of the ions are going to be used up to form a precipitate.

Considering the equilibrium: $Co^{+2}+S^{-2} ⇌ CoS$

Initial: $9.00 \times 10^{-11}, 6.00 \times 10^{-11}, 0$
Change: $-x, -x, +x$

So at equilibrium:

$K_{sp} = \frac{1}{[Co^{+2}][S^{-2}]}$
$4.0 \times 10^{-21} = \frac{1}{(9.00 \times 10^{-11} - x)(6.00 \times 10^{-11} - x)}$

This does not make sense though, if I use the quadratic formula from here I get an absurd value. What am I ignoring?

EDIT: If the reactant terms could be moved to the numerator... i get a reasonable answer, but it's always supposed to be [products]/[reactants].

If I solve it this way (which obviously makes more sense logically):

$4.0 \times 10^{-21} = (9.00 \times 10^{-11} - x)(6.00 \times 10^{-11} - x)$

I get $x = 1.0 \times 10^{-11}$ and $x = 1.4 \times 10^{-10}$

Though I'm not certain which value I would pick.

Any advice on this one?

 

[Saitama]

Initial: $9.00 \times 10^{-11}, 6.00 \times 10^{-11}, 0$
Change: $-x, -x, +x$

Can you show how did you got those values for initial moles? I don't think that's right.

$4.0 \times 10^{-21} = (9.00 \times 10^{-11} - x)(6.00 \times 10^{-11} - x)$

It looks as if you plugged in the moles instead of the concentrations of ions.

 

[Borek]

Bah! My original idea about using the ICE strategy with concentrations was the way to go then.

That's the correct approach.

Initial: $9.00 \times 10^{-11}, 6.00 \times 10^{-11}, 0$

As you were already told, these are wrong.

$K_{sp} = \frac{1}{[Co^{+2}][S^{-2}]}$

This is wrong too. What is the definition of K_sp?

 

[STEMucator]

Here is the working of the first question:

I know now that I want the ion product concentration minus 'x' from each ion. That's like asking 'how much precipitates before $K_{sp}$ is reached'.

EDIT: I got the answer now thanks guys.