Complex Ions: Formation Constant - Video Tutorials & Practice Problems

Complex Ion and Ligand

Now, a complex ion is a structure containing a metal cion that acts as a luis acid and poly bonded to a ligand. Now, remember luis acid is an electron pair acceptor. Now, a ligand is a molecule ion that acts as a Lewis base and donates a long pair to the metal cat I. Here in this example, we have the ammonia molecule with its lone pair and we have the cadmium ammonia with its lone pair can donate this lone pair to the cmi ammonia is acting as our Lewis face cadmium ion is accepting the electron uh pair. So it's acting as our Leis acid. Now, in reality, more than one of these ammonia can connect with the cadmium. Here, we have four of them connecting to the academy. Now, you don't need to be able to figure out how many would attach. You'll learn that much later on in a in a later chapter or right now, in this example, we have four ammonia molecules connecting to the cadmium, ammonia itself is neutral, it has no charge but cadmium here has a two plus chart. So overall, the structure has a two plus chart. Now, in the past, we've talked about Lewis acid base chemistry. In the past, we would have called this an ad duct because we added them together. But in this case, we can also call it a complex ion. This would represent our complex ion structure. Now connected to the idea of complex ions is the um equilibrium constant of the formation constant. Here, we're gonna say the formation constant is capital K sub F, it is another equilibrium constant like other equilibrium constants. It is a ratio of product to reactive concentrations. But now they deal with complex ions. Here we're gonna say like other equilibrium constants. It can be calculated by setting up an expression and ignoring the phases of solids and liquids. So just remember we have a Lewis acid Lewis space reacting together adding together to create a complex ion associated with this complex ion is a KF value known as our formation constant.

Complex Ions: Formation Constant Example

Here, it says the formation of a 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 times 10 to the 21. Now, if 100 mls of 0.010 molar of silver per chlorate is mixed with 220 mls of 0.25 molar cyanide. What is the concentration of silver ion once equilibrium has been reached? All right. 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 iron. Here we're dealing with a nice chart which stands for initial change equilibrium. We're going to say here, step two, we're gonna determine the moles of the metal cion and the ligand anna 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. Remember moles equals liters times molarity. If I divide this by 1000 and divide this by 1000 and multiply by their molarity. We'll have the moles of our reacts. So here, I'd have 0.100 L times 0.010 molar. Here, silver per chlorate is in the same ratio as as silver ion. So the moles of silver per chlorate equal the moles of silver ion. So liters times molarity gives me an initial amount equal to uh point zero oops 0.010012 moles of silver ion. So 0.0012 moles of silver ion here, we need the molarity of it. So we divided by the total volume used. We have 100 MLS here and 220 MLS here. So that's a total of 320 MLS dividing that by 1000 gives me my leaders. 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 L. So my molarity here is 0.171875 mode, take those molarity and plug them in. Now here using the initial rows, place the initial concentrations of the metal C ion and the lid an ion. And here we don't have any information on our products. 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 gonna subtract by the smaller ones. We're gonna do subtract 0.00375. And then here, there's a two here. So it's two X or two times 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 moles. 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 M here. Now this is important because it is quite different from what the custom is 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 cion in reality will never reach zero. So set it to the X variable. So here, although it looks like these are subtracting each other to give me a zero 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 gonna say KF which is an equilibrium constant is equal to products over reactants, 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 ion. So just isolated, rearranging my equation, I'm gonna 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 1.0 times 10 to the 21. 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 times 10 to the negative 22 more. So this would represent the molarity of my silver eye.

If your equilibrium constant K is equal to the product of K_sp and K_f, find the solubility of AgCl in 2.0 M NH₃. K_sp of AgCl = 1.77 × 10⁻¹⁰; K_f of Ag(NH₃)₂⁺ = 1.7 × 10⁷.

• A. 1.04 × 10⁻¹⁷ M
• B. 9.78 × 10⁻³ M
• C. 9.89 × 10⁻² M
• D. 3.01 × 10⁻³ M

A solution is composed of 3.20 × 10⁻⁴ M Co(NO₃)₃ mixed with 0.200 M NH₃. Determine the [Co³⁺] that remains once the solution reaches equilibrium in the formation of Co(NH₃)₆³⁺.

• A. 2.30 × 10⁻³³ M
• B. 7.03 × 10⁻³⁷ M
• C. 2.76 × 10⁻³⁸ M
• D. 8.40 × 10⁻⁴² M