# Homework Help: Moles of ions

1. Apr 19, 2014

### Zondrina

1. The problem statement, all variables and given/known data

$300.0 mL$ 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.

2. Relevant equations

3. 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 $CoS$?

2. Apr 19, 2014

### Staff: Mentor

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

3. Apr 19, 2014

### Zondrina

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?

Last edited: Apr 19, 2014
4. Apr 20, 2014

### Staff: Mentor

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 S2- now? What should their product be equal to?

5. Apr 20, 2014

### Zondrina

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.

Last edited: Apr 20, 2014
6. Apr 20, 2014

### Saitama

Can you show how did you got those values for initial moles? I don't think that's right.
It looks as if you plugged in the moles instead of the concentrations of ions.

7. Apr 20, 2014

### Staff: Mentor

That's the correct approach.

As you were already told, these are wrong.

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

8. Apr 20, 2014

### Zondrina

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.

Last edited: Apr 20, 2014