In quantitative analysis, Ag+, Hg2+, and Pb2+ are seperated from other cations by the addition of HCl. Calculate the concentration of Cl-ions required to just begin the precipitation of (a) AgCl, (b) Hg2Cl2, (c) PbCl2 in a solution hav-ing metal-ion concentrations of 0.030 M. What fraction of the Pb2+ remains in solution when the Ag+ just begins to precipitate?

Video transcript

Hello everyone today. We have the following question based on the quantitative analysis, the addition of hydrogen sulfide sulfide can separate lead, copper and cadmium from other cantons present in a solution. What is the sulfide ion concentration needed to start? The precipitation of one led to copper and three cadmium ions. If each of the metal ions has a concentration of 30.20 moller. When copper starts to precipitate what fraction of cadmium is still left in the solution. And then we have the following K. S. P. Values for each of these. So let's start with our lead. So if we have lead sulfide it is going to dissociate an irreversible reaction into lead two plus ions as well as to as well as one sulfide to minus ion. We have the K. S. P. Value as follows. Now we're going to solve for the concentration of our sulfide in this material here. So the concentration of our sulfide is going to be equal to the K. S. P. Value divided by the concentration of our lead two plus R. K. S. P. is seven times 10 to the -29. And our concentration for lead is 0. zero Mueller has given in the question stem. This is gonna give us 3.5 times 10 to the negative 27th moller. Now if we look at our second option here which is copper, we're gonna have copper sulfide in the solid form. In a reversible reaction is gonna disassociate into copper Two plus ions as well as one sulfide to minus ion as before. We're going to solve for the concentration of sulfide, that's going to be our K. S. P. Divided by the concentration of our copper two plus That K. S. P. is eight times 10 to the -37. And that copper concentration was also 0.020 moller. This calculation is going to yield us with four times 10 to the negative 35 molar. And then we have our cadmium, we will have our cadmium sulfide in the solid form to associate into Academy. Um two plus ions as well as sulfide to minus. Our concentration of sulfide is going to be the K. S. P. Divided by the concentration of our cadmium two plus the K. S. P. Was one times 10 to the negative 28. And once again our concentration was 0.020 moller. This is going to yield us five times 10 to the - moller. So we've answered parts 12 and three. Now we need to find the fraction of academy and that's still left over. So how are we going to do that? Well we need to calculate what's known as Q. Or the Q. Quotient for cadmium sulfide. So our cue is going to be equal to our concentration of sulfide. Our concentration of sulfide in the cadmium sulfide reaction which we found was four times 10 to the negative 35 molar divided by the concentration overall. In the solution. When we do that, we get a quotient of eight 0.0 Times 10 to the -37. Then we must compare that with our K. S. P. For that. In this event, our q quotient is less than the k S P. For this reaction here, the 8 to 8 times 10 to the negative 37 is less or smaller than one times 10 to 28. And so therefore we will have no precipitation, so no precipitation. So for this last portion we can say that all of the academy um is going to remain and so we have our three answers here overall, I do hope that this helped, and until next time.