Lead(II) ions can be removed from solution by precipitation
with ...

Question

Lead(II) ions can be removed from solution by precipitation
with sulfate ions. Suppose that a solution contains lead(II) nitrate.
Write complete ionic and net ionic equations for the reaction
of aqueous lead(II) nitrate with aqueous potassium sulfate
to form solid lead(II) sulfate and aqueous potassium nitrate.

Accepted Answer

hey everyone in this example we have a quick solutions of lithium sulfide and copper to nitrate being mixed and we need to give the complete ionic equation for the reaction if there is one. So writing out the first compound lithium sulfide within that compound we have the L. I. Plus one carry on. And then the S two minus sulfide and ion so that will form L. I. Two S. When we combine them into a compound and this based on our cell viability rules for sulfides is going to be soluble so it gets the Acu Aquarius label. For our second compound we have copper to nitrate. So we have C. U. Two plus and then recall that nitrate is N. 03 minus one the an ion. And combining those two, we have our second reactant which is going to be see you And '03. And then we have parentheses around the nitrate with the subscript of two when we crossed the charges. So based on our sustainability rules for nitrates, all nitrates are soluble. So this would also get the labels. So we will have a reaction because we have to Aquarius react mints. And so when we combine the ions we combine the opposite charges. So we combine these two ions together and then we combine these two ions together to form our products. So what we're going to form is lithium nitrate. So L. I N. 03 and we're going to recall again the nitrates eligibility rule means that this compound is going to be a quiz because all nitrates are soluble and then for our second product we have copper sulfide which is going to be see you us and this is going to be based on our cell viability rules for sulfides insoluble because sulfides are generally insoluble. So this is a solid precipitate product. And now the next step is to make sure that this is balanced. And in order to balance this out we're going to place a coefficient of two in front of our lithium nitrate and that's going to leave everything as balanced now. And this is going to be our molecular equation. So what they want us to give is the complete ionic equation and that's going to be our next step. So within the complete ionic equation we want to recall that we're going to include Spectator ions. So we're listing out all of the ions as well as our Spectator ions. But what we should have is for our first compound lithium lithium sulfide we should have L. I. Plus catalon and specifically we have two moles of this Catalan because we have that subscript of two there and then this gets an accu label because all ions are soluble then for a second adam in that compound we have one mole of the S two minus an ion for a second reactant we have one mole of the copper, two plus cat ion And then two moles of our nitrate and ion. So this completes our ions on our reactant side. And now on our product side we have two moles of our lithium plus one cat ion And then two moles of our nitrate and ion. And so for our second product we have copper sulfide, but we will write that out as a compound because it's insoluble. And so this remains as a compound, we can't break it up into ions. So our next step is to make sure that we identify our spectator ions and those are going to be the ions that match on both sides of our reactant and product side. So here we have two modes of nitrate on our product side. We also have that on the reactant side. So this is going to be a spectator ion as well as our two moles of lithium kati on on the product and reactant side. So these here are going to be all of our spectator ions that we've identified. And so this is actually going to complete this example as our final answer here. This entire equation is going to be our complete ionic equation where we marked our spectator ions with the asterix boxed in yellow and so we would technically be canceling out these in order to get to our net ionic equation if the question did ask for that in the next step, but in this example we're only asked for the complete ionic equation. So this completes this question. If you have any other questions, please leave them down below. And I will see everyone in the next practice video.

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