Solubility of substances

[rasen58]

 

So I did #4 correctly because I predicted that they would mix and produce PbCrO4 and NaNO3. So that means that an insoluble yellow solid and a soluble white solid would be produced. So I picked A - Yellow solid and colorless solution, which is correct.

 

But for #5 and 6, I don't understand how to use the amount of moles to determine which is produced.

I made the balanced equation for 5, but the coefficients for each product and reactant are 1, so I don't see how the moles change.

[studiot]

 

So I did #4 correctly because I predicted that they would mix and produce PbCrO4 and NaNO3. So that means that an insoluble yellow solid and a soluble white solid would be produced. So I picked A - Yellow solid and colorless solution, which is correct.

 

There is more to it than this.

 

When the lead nitrate and sodium chromate dissociate in solution, there is enough lead to totally precipitate the yellow chromate ion as lead chromate, leaving only white ions in solution making it colourless. Note this still leaves lead ions in solution as nitrate.

 

So look at the relative amounts of each ion and see what is left in solution after precipitation (if any).

Also ask if there is enough of both reactants to fully precipitate the insoluble product.

Edited May 2, 2014 by studiot

[rasen58]

Oh, that makes sense now.

But do you only look at the relative amounts of the anion? Because for #5, there are 3 mol of the Na ion and 3 mol of the CrO4 ion right? So do you only consider the anion because there is only 1 mol of the Pb to make PbCrO4?

[Hailstorm]

For #5 there are 3 moles of Na₂CrO₄: so you would have 3 x 2 = 6 moles of Na+ and 3 x 1 = 3 moles of CrO_4, be careful with your stoichiometry. I'm not entirely clear what you mean when you're referring to only considering the anion, in this particular case you have an excess of anion so all (you will learn about partial soluble salts later), of the cation is capable of precipitating. As Studiot pointed out, in #4 you have an excess of cations so once you've used all the anions you have remaining cations in solution. It is necessary to look at the relative concentrations of each and find the limiting reagent, you can "ignore" the excess reagent, but only after you've correctly identified it as such.