The following three solutions are mixed: 100.0 mL of 0.100 M Na2S...

Question

The following three solutions are mixed: 100.0 mL of 0.100 M Na2SO4, 50.0 mL of 0.300 M ZnCl2, and 100.0 mL of 0.200 M Ba(CN)2. 
(b) What is the molarity of each ion remaining in the solution assuming complete precipitation of all insoluble compounds?

Accepted Answer

Hello everyone today. We have the following problem. Consider the following solutions. 100 mL of 1000.1 50 moller of cadmium chloride, 1 50 mL of 500.2 50 moller of calcium hydroxide and 150 mL of 1500. potassium sulfate. If the three solutions were mixed, calculate the molar itty of the ions remaining in the solution. After all of the insoluble compounds precipitate completely. So first we have to write the dissociation reactions. So we have our cadmium chloride here which would dissociate into cadmium, two plus ions and two chloride ions. We within have our calcium hydroxide which would associate into our calcium two plus and our hydroxide are two hydroxide ions. And then lastly we have our potassium sulfate which would disassociate into two potassium ions as well as just one sulfate ion. And so next we have to find our moles of our gratifying moles of our cadmium. And for chlorine we already have the equation here that we saw that we found the dissociation for academy. Um So what we're gonna do is we're gonna take our volume that we were given, which was 100 mL. We're going to convert that to leaders by using the conversion factor that one. Middle leader is equal to 10 to the negative third Leaders. We are then going to multiply by the polarity of our calcium chloride which is in units of moles per liters. So you can say that this is equal to . malls of cadmium chloride Is equal to one leader and then we can use our multiple ratio that in one mole of our a cadmium chloride. We have one cadmium or one mole of cadmium. When our units canceled out, we were left with 0.0150 moles of cadmium. And then we are going to do the same process with our chlorine. We were given 100 ml. We're going to convert that to leaders using a similar conversion factor and then multiplying by the polarity of our academy of chloride once again which was 0. moles of can't be in chlorine over one leader. And this time we're going to say that in one mole of our cadmium chloride, we have two chlorine or two moles of chlorine. And what our units cancel. We're left with 0.0300 moles of chloride. So now we have the molds for that And so now we must find our moles of calcium two plus And hydroxide. And so we're essentially going to apply the same concept as well. We're going to take the volume that we were given of that which was ml convert that to leaders by using the same conversion factor. We're gonna multiply by the polarity of our calcium hydroxide which is .250 moles of our calcium hydroxide over one leader And then in our one mole of our calcium hydroxide. We have one calcium or one mole of calcium. And this is going to leave us with 0. moles of calcium. We're gonna play the same thing with our hydroxide or 1 mL. Multiplying by that conversion factor then by the polarity and then we're going to see that in one more of our calcium hydroxide, We have two moles of hydroxide And this is going to leave us with 0.0750 moles of hydroxide. And so we have those values there. Next we're going to find the moles. We're going to find the moles of our potassium ions as well as our sulfate ions. We're gonna take the volume that you were given, which was one of 50 ml convert that to leaders And then multiply the polarity which was 0.250 moles Of potassium sulfate over one leader. They were going to say that in one mole of our potassium sulfate We have two moles of potassium. This is gonna give us 0.0750 moles of potassium. And then lastly we're going to do the same thing but with sulfate supply by the polarity here and then in one mole of our potassium sulfate, We have one mole of sulfate Giving us 0.0375 moles of sulfate. And so we have these answers here. The next thing that we're going to do is we are going to draw our calcium and our sulfate, combining with one another to form calcium sulfate in the solid form, We see that this is a 1-1 ratio. And therefore, If we have 0.0375 moles of calcium, We also have 0. moles of sulfate. And so the concentrations are going to be essentially equal to zero. If we were to do the same with our cadmium and our hydroxide you would get cadmium hydroxide in the solid form of course. And we see here that this is a actual 1-2 ratio. So for every one mole of cameo we have two moles of hydroxide. So what does this mean? This 1-2 ratio? So we have our 0.01 fishery moles of cadmium. And when that is used up we have none left. However, we would still Have R 0. moles of hydroxide left over when it has been used up. And so that's going to cause us to need to find our moles of hydroxide. And so we're simply going to take our 0.150 malls of cadmium two plus And multiply that by our two moles of hydroxide ratio over one mole of cadmium two plus. And that's gonna give us 0.03 moles of hydroxide next to find. However leftover molds of hydroxide that we have, we're simply going to take that number that point zero 750 moles of hydraulic side that we had before and subtract that by the amount that we just found. And that is going to yield us 0. moles of hydroxide left over after the reaction has occurred. And so one of the final steps that we have to do is find out which ion forms of precipitate. So we have our potassium ion which does not form a precipitate. This essentially says after the reaction has occurred, there is no solid left over. Then we can use our chloride and we can say that with chloride, chloride also does not form a precipitate. And then we did acknowledge that we had this .045 moles of hydroxide left. And so to find the clarity, we have to find out our total volume. And so our total volume is going to be the 100 middle leaders plus the 1 50 mL plus the other 1 50 mL. Which will give us 400 middle leaders. However, we have to convert this to leaders. And so we use that conversion factor to give us 0.4 liters. And so our last and final step we can find the Moleketi of the concentration of ions left, the concentration of ions. So for potassium we're going to use the moles that we have which is 0.750 moles over the volume we found which is 0.4 liters giving us 0.188 molar for our chloride. We're going to use our 0.3 moles over our 0.4 liters, Giving us 0.075 Moller. And then for our hydraulic sign, we're going to take our 0.0450 moles over our 0.4 leaders, giving us 0.113 molar. And with that we have found the leftover concentrations of our ions after their respective reactions overall, I hope this helped, and until next time.

Watch next