The following pictures represent solutions of Ag2CrO4, which also may contain ions other than Ag+ and CrO42- that are not shown. Solution 1 is in equilibrium with solid Ag2CrO4. Will a precipitate of solid Ag2CrO4 form in solutions 2-4? Explain.

(1) (2) (3) (4)

Video transcript

Hello. We are told the images shown below show solutions of copper one carbonate. Where the orange spheres represent the carbonate ions and the violet spheres represent the copper one. Irons. We are told to note the other islands may be present the solution but are not shown if image one shows the solution in equilibrium with the solid copper carbonate. Which of the solutions shown in images two through four will form a solid copper carbonate precipitate and were asked to explain. Let's begin by writing the equation then that describes are solid copper carbonate, an equilibrium with its ions. So we have two moles of copper one And one mole of carbon eight. So our scalability product constant then will be given by concentration of copper one Squared times the concentration of carbonate. Looking at image one. So we're told that the violet spheres are copper and the orange spheres are carbonate. And so we will have Three coppers. We'll square that and we have four carbonates. So this works out to then 36. We then look at our other figures. So, image to then We have five copper ions and three carbonate annan's image. Three We have two copper ions and six carbonate anons and an image four we have four copper ions and two carbon eight times recall that for a precipitate to form our reaction quotient Q is greater than our scalability product constant. That means then that the reaction will move in the reverse direction to reach equilibrium. And that is towards the formation of our precipitate. So for number one then number two, then our Q. Is equal to our copper iron concentration. So that's five squared times that carbonate, which is three. So that works out to 75, then is greater than 36, which is our scalability product constant. So that means that we will form a precipitate in and that's three. Our reaction Potion Q. Then we have the number of copper ions which is to that squared times the number of carbonates, which is six. That works out to 24, which is less than 36. So no precipitate will form and an image for a reaction potion. We have the number of copper ions which is four squared times the number of carbonate. This works out to 32. is less than 36. And so again we will not form precipitated. So the only solution that will form a precipitate Is the one shown in Diagram two. And that's because we have a reaction quotient greater than our cell ability product constant. And this corresponds to answer a thanks for watching. Please help