Here it says the formation of the complex ion created by the combining of silver ion and cyanide is given below. Here we have one mole of silver ion reacting with two moles of cyanide ion to give us our complex ion. Here it has a formation constant of 1.0×1021.

Now if 100 M LS of 0.010 molar of silver per chlorine is mixed with 220 M LS of 0.25 molar cyanide, what is the concentration of silver ion once equilibrium has been reached? Alright, so the setup is we set up our ice chart with the given formation equation. Here we have to remember that there's two moles of cyanide ion. Here we're dealing with a nice chart which stands for initial change equilibrium.

We're going to say here Step 2. We're going to determine the moles of the metal cation and the ligand anion and divide them by the total volume using the chemical reaction to determine their initial concentrations. Remember ice charts use molarity as one of its units. Moles equals liters times molarity. If I divide this by 1000 and divide this by 1000 and multiply by their molarities will have the moles of our reactants.

So here I'd have 0.100 liters times 0.010 molar. Here silver perchlorate is in the same ratio as silver ion, so the moles of silver perchlorate equal the moles of silver ion. So liters times molarity gives me an initial amount equal to 0.0012 moles of silver ions, so 0.0012 moles of silver ion here. We need the molarity of it so we divide it by the total volume used. We have 100 MLS here and 220ML here, so that's a total of 320 M LS. Dividing that by 1000 gives me my liters. This comes out to be 0.00375 molar for my silver ion, so that's your initial amount, the cyanide ion.

Divide the MLS by 1000 to get liters and multiply by the molarity. That'll give me initially 0.055 moles of cyanide ion. Again, divide by total volume used, which is 0.320 liters. So my molarity here is 0.171875 molar. Take those molarities and plug them in. Now here using the initial rows, place the initial concentrations of the metal cation and the ligand anion. And here we don't have any information on our product. So initially it's zero.

Now using the change row, looking at the reactants subtract from their initial amounts by the smaller mole amount. So what do I mean by that? Well, out of the two, well, two molarity amounts that we've got, we're going to subtract by the smaller ones, we're going to do subtract 0.00375 and then here there's a 2 here, so it's 2X or 2 * 0.00375 molarity moles. Remember we're using molarity here, so we can say the same thing will be applied to molarity as they would have been with molts.

Now using the law of conservation of mass, whatever you lose as a reactant, you gain or you add that amount to products. So here we add 0.00375 molar here. Now this is important because this is quite different from your customers seeing with ICE charts. Using the equilibrium row, set up the equilibrium expression and solve for X. Here we're going to say the amount of the metal cation in reality will never reach 0, so set it to the X variable. So here, although it looks like these are subtracting each other to give me a 0 total at the end, we never quite get there. So this is X at equilibrium.

Here when we subtract points 171875 by two times this 0.00375, I have a remainder of 0.164375 molar. Here we just bring this down 0.00375 molar so we have everyone at equilibrium except for the ion we need to find. Here They gave us a formation constant for this equation. So we're going to say KF, which is an equilibrium constant, is equal to products overreactants, just like all other equilibrium constants. We know what KF is, we know what the product is at equilibrium, and we know what this ion is at equilibrium. The only variable missing is the silver island, so just isolate it.

Rearranging my equation, I'm going to see that silver ion concentration equals the product concentration divided by KF times the cyanide ion concentration squared. Here, plug in the values that we know. So silver ion concentration equals 0.00375 divided by the KF, which in this question was 10×1021 and then CN minus was 0.164375. And don't forget that it's squared punch. Punch that all into your calculator, you'll see the silver ion concentration at the end is 1.39×10−22 molar. So this would represent the molarity of my silver ion.