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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.

## 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##?