Suppose that a 10-mL sample of a solution is to be tested for I- ion by addition of 1 drop (0.2 mL) of 0.10 M Pb1NO322. What is the minimum number of grams of I- that must be present for PbI21s2 to form?

Video transcript

Hello everyone today. We are being given the following problem. This is a 15 millimeter sample of solution is tested for the presence of chloride ions to drops with one drop equaling about 2.2 mL of 0.15 silver nitrite nitrate were added to the solution, calculate the minimum mass in grams of chloride ions that should be present in order for silver chloride, which has a K. Sp of 1.8 times 10 to the negative eight to form. So the first thing you wanna do because you want to take into account what the dissolution of this silver chloride would look like. So you have solid silver chloride which associate into silver Plus science as well as chloride minus science. And this would have a subsequent K sp of 1.8 times 10 to the negative 10th. We didn't want to take into account what the dissolution of this silver nitrate would look like. And so in the acquis form here it would readily disassociate into regular silver plus Aquarius violins as well as nitrate equally. Silence 30. Want to use our minority and volume equation, which is that the majority of our first solution times the volume of our first solution to the polarity of our second solution, times the volume of our second solution. And so if we want to find the volume of our the polarity of our silver plus ions or the concentration of it. We're gonna go ahead and do 0.15 polarity which would account for the silver nitrate Times our 0. mil leaders times are 15.2 mL and this is all given in the question stem and this is going to yield us with 3.9474 times 10 to the negative third polarity. We then must find the K. S. P. For this. So we have our K. Sp set up and we have our silver ions and brackets and then multiply that by the chloride ions which is also in brackets. And so you know the K. S. P. Is 1.8 times 10 to the negative 10. And we know that we have our concentration of silver ions that we just found 3.9474 times 10 to the negative third times are unknown concentration of chloride ions. And so through simple math, we can say that our concentration of chloride ions is going to equal 4. times 10 to the negative eight polarity. Using this, we can finally calculate the mass the minimum mass of chloride ions. And so by doing that we can do 15.2 mL convert that to leaders by using the conversion factor that one. Middle leader is equal to 10 to the negative third. Leaders multiplying that by our polarity so 4.56 times 10 to the negative eight moles over. Leaders that we just saw it for times our molar mass Of Chloride which is 35.45 per the periodic table per one mole. And once our unit's canceled out, we'll be left with an answer of two times 10 to the negative eighth grams of chloride ions as our final answer. Overall, I hope this helped, and until next time.