Calculate the molar solubility of AgI in: (a) Pure Water (b) 0.10 M NaCN: Kf for [Ag(CN)2]- is 3.0 x 10^20

Video transcript

Hello everyone today. We have the following question what is the eligibility of silver chloride in pure water? And in .20 molar of sodium sulfate solution. Use the following values provided. So First we need to find what the moral responsibility is in pure water or H20. And so how we can do this is we can show the dissolution of our silver chloride. So we have our silver chloride here as a solid and a reversible reaction dissociating into silver plus acquis ions as well as chloride ions. And then we're gonna draw an ice table here to show the concentrations our solid is not incorporated in this equilibrium table so we don't have to include it. Our initial concentrations of our ions here are going to be both zero. Any change is going to be X. And we're going to be adding something. So it's going to be plus X. And then equilibrium is just adding the first two columns up. So it's just going to be X. And X. Now we have to construct what R. K. S. P. Formula would look like. And so that's usually going to be the products over the reactant. However, we only have two products here that can be incorporated into this equation. The reactant cannot because it is a solid. So therefore we have our silver and we have our chloride. We were given the K. s. p. value of 1.8 times 10 to the -10. So what we're gonna do is we're gonna say that we're gonna substitute that in for R. K. S. P. And then we said that our silver and our chloride are both going to be presented by X. So we can shorten this two X squared, Solving for X. We get that X. is equal to 1.3 times 10 to the negative 5th moller. So this is the motor ability in pure water. So let's find it in the other solution. So in this other solution we show the dissolution once again and this time it's going to be a quiz because it's already in a solution and it's gonna associate into two sodium ions As well as just one soul fight ion. So sodium is actually a neutral ion. So we're not going to include that in the reaction. It doesn't affect the scalability. What we can do now is we can show our initial equation or reaction with our silver chloride here and we can try to find a spectator ion between these two. So what does that mean? So we have our silver chloride here and irreversible reaction giving us our silver ion and our chloride ion. And then what we're gonna do is we're going to actually add this silver, this positive ion to this cell fight here when we do that, we're gonna get silver plus reacting with those ions there in a reversible reaction to form this silver cell fight complex here. And so we have our K. F. Value and we have our K. S. P. Value. Once again for this reaction, we're going to need to solve for our equilibrium constant. So we're gonna have our equilibrium constant being equal to R K. S. P. Times our K. F value. RKSP was 1.8 times 10 to the -10, and RKF value was 4.7 times 10 to the 13th. When we multiply those two together, we're going to get 8.46 times 10 to the third, this is going to be our equilibrium value. So we're gonna do is we're gonna need to construct an ice table from this reaction that we just did. So we're gonna have our silver and our soul fight giving us our silver so fight complex. We're gonna have our ice chart here. Our initial concentrations for the solid are not going to be accounted for, which is the silver. However, our initial concentrations for the sole fight is going to be 0. for our product is going to be zero as well and we also had chloride as a product That is going to be zero. Our change is going to be -2X. Because we are taking away from our reactant and because of this, we're adding an X. To each of our products. The next thing that we're gonna do is we're gonna have our equilibrium table which is gonna be .2 - x. And that's simply adding the first two rows. Now we need to solve for our equilibrium expression for a second time to get the final answer. That's gonna be the equilibrium constant is eager to once again our products overreacting and we said that our products were both represented by X. So it's going to be X squared Divided by 0.2 minus two X. And then we're gonna square that as well. We already have a K, E. Q. Or the equilibrium constant, which was 8.46 times 10 to the third. And we're gonna equal it to this expression here, Solving for X. We get .1 molar as our final answer. And so these are two answers overall, I hope this helped and until next time.