How could you use a precipitation reaction to separate each of the following pairs of cations? Write the formula for each reactant you would add, and write a balanced net ionic equation for each reaction. (a)

Video transcript

Hello everyone today. We are the following question identify the reactant that you can use to separate the Catalans strontium and lead provide the chemical formula of the reaction and give the balanced net ionic equation. So two separate carry ons. You need insoluble reactant. So when you react a reactant with these two they are to dissociate and so off the bat we can get rid of the answer choices E and F. Because when it comes to nitrates all of them are soluble and therefore are not effective reactant. Now if we start with our hydrogen sulfide and we dissociated You would get two hydrogen ions and two sulfide ions. And so then we would react our lead and we would react our strontium with both of these. So first we would have our lead two plus as an Aquarius form Reacting with our two hydrogen As well as our one strong team or our sulfide From our H two s. And that would be the quickest form and that would go ahead and form our lead sulfide which is actually a solid as well as two of our H plus ions. And so if we look at what we can cancel out, we noticed that we have two hydrogen ions on the left and two on the right and those can cancel. And so the net equation of this would be lead two plus in the acquis form plus our sulfide in the acquis form and then on the product side we would just have led sulfide and this is actually going to be insoluble which makes it a great reactant a very effective reactant. So now we know that we are most likely going to be working with H2 S. or hydrogen sulfide. Let's however react our strontium with our to react ints to see what we would produce. So we have strong Tm plus our two reactant which is our hydrogen and our sulfides. And so this would give us strontium sulfide which is an Aquarius form and two hydrogen ions. And it's important to note that this strontium sulfide is soluble or it's in the acquis form because it is soluble. It's important to note That when it comes to H. two s. specifically or hydrogen sulfide these this is going to be insoluble on less it is paired with calcium berry, um or strontium. And in this case we have our sulfur founded with our strontium making it soluble. So so far we have an insoluble reacted in H. two s. And we have a soluble and age to us. And so the key is to find one that would be soluble and insoluble. And so far we have that with our H. To us. And we already stated that our net ionic equation would be the highlighted equation or reaction here. And so the only one that actually matches up with that is anti choice A. And that is actually our answer. So we're going to use the reactant hydrogen sulfide to produce the following net ionic equation. And thus it will be able to separate the Catalans strontium and lead overall. I hope this helped, and until next time.