For each pair of compounds, use Ksp values to determine which has the greater molar solubility: (b) PbCO3 or BaCrO4.

Video transcript

Hello everyone. So in this video we're going to go ahead and see which compound has a greater molars eligibility and were given its respective Ks. P values. So first molecule that we're dealing with is MGF two. That is magnesium fluoride. So M G F two and we also have PB I two which has led to iodide. Alright, so let's first go ahead and write out the equation. So for the magnesium fluoride we have the starting material being MGF two. It'll go ahead and associate into MG two plus ion and two moles of f minus an eye on next one for P. B. I two or lead to iodide, we have the, sorry, materials being P B. I too will go ahead and associate into PB two plus and two moles of I minus. So we can see here that both salts have an equal number of islands in the solution. So we can go ahead and actually compare our KsB values and see which one has a greater molar molars eligibility. So it is given to us in the problem. The K. S p of magnesium fluoride is 5.16 times 10 to the negative 11 and the K. S. P. Value of lead to iodide is 9.8 times 10 to the negative nine. So if we're comparing these two Ks P values, we can see that the K. S. P of lead to iodide is much greater. So in conclusion then We can say that PB. i. two has a greater Moeller Sawyer ability and the reason being that the K. S. P value is greater. So this right here is going to be my final answer for this problem. Thank you all so much for watching.