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

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In summary: What is concentration of Co2+ now? What is concentration of S2- now? What should their product be equal to?
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## 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.

## 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##?

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

Borek said:
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?

Last edited:
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?

Borek said:
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?

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:
Zondrina said:
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.

Zondrina said:
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 Ksp?

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:

## 1. How do you calculate the moles of a compound?

To calculate the moles of a compound, you need to know the mass of the compound and its molar mass. The formula is moles = mass / molar mass. For example, if you have 2 grams of Na2S and its molar mass is 78.04 g/mol, then the moles of Na2S would be 2 g / 78.04 g/mol = 0.0256 moles.

## 2. What is the balanced chemical equation for the reaction between Na2S and Co(NO3)2?

The balanced chemical equation for the reaction between Na2S and Co(NO3)2 is 2Na2S + 3Co(NO3)2 → 2NaNO3 + 3CoS.

## 3. How do you determine the limiting reactant in a precipitation reaction?

To determine the limiting reactant in a precipitation reaction, you need to calculate the moles of each reactant and compare them to the mole ratio in the balanced chemical equation. The reactant with the smaller number of moles is the limiting reactant.

## 4. How do you calculate the mass of the precipitate formed in a reaction?

To calculate the mass of the precipitate formed in a reaction, you need to know the moles of the precipitate and its molar mass. The formula is mass = moles * molar mass. For example, if the reaction between Na2S and Co(NO3)2 produces 0.0256 moles of CoS and its molar mass is 118.73 g/mol, then the mass of CoS formed would be 0.0256 moles * 118.73 g/mol = 3.04 grams.

## 5. What is the purpose of calculating the moles in a precipitation reaction?

Calculating the moles in a precipitation reaction allows you to determine the limiting reactant, which is important for understanding the amount of product that can be formed. It also helps in performing stoichiometric calculations to determine the mass of the product formed.

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