Calculate the percent ionization when 73.2g of sodium hypoiodite are dissolved with 500 MLS of solution. We're told that the K value of hypoiotis acid, which is Hio, is 2.3 * 10^-11. All right, so we're going to say that hypoiotis acid is a weak acid. Since its KA is less than one. Naio represents its conjugate base. It has one less hydrogen because it's the conjugate base. It represents the weak base O.

We're going to have to use the steps from 1:00 to 3:00 to set up our ice chart for this weak base. Remember, for ionic bases we ignore the neutral metal cation, in this case the sodium ion. This leaves us with the hypoiodite ion alone. Since it's a base, it will react with water. Water here will be in its liquid phase. It is the base, so water acts as the acid, meaning that water will donate an H plus to it. This creates Hio plus OH minus as products. We're dealing with a weak base, so this is our ice chart setup.

So we have initial change equilibrium. Now remember, the units of an ice chart with a weak base have to be in molarity. So we're going to take the 73.2g of sodium iodide and we're going to convert that into molds. So we're going to stay here. For every one mole of sodium iodide, we have a mass of 165.89 grams of sodium hypoiodite. That's the mass of the sodium, the iodine and oxygen together. So here grants cancel out and we're going to have here for everyone mole. We have a sodium iodide. We have one mole of iodide ion which equals .441 moles.

We need molarity for the ice chart, so molarity here will be moles over liters should be 0.441 moles divided by .500 liters, which comes out to .882 molar. Now with the ice chart, we ignore solids and liquids, so the water would be ignored. We don't know anything about the products initially, so there's zero. We lose reactants in order to make product, bring down everything. Now here, using the equilibrium row, set up the equilibrium constant expression and solve for X. Here we'll check if a shortcut can be utilized, and if so, we can avoid the quadratic formula.

Since this is a weak base, we're going to have to utilize KB. KB would equal products over reactants. However, we're not given KB initial, we're giving Ka. So we have to remember that KA times KB equals kW. So we're gonna say KB here equals kW divided by ka. So here this would be 1.0 * 10^-14 divided by RKA, which we're told is 2.3 * 10^-11. Doing this gives our KB value. Our KB value here would equal 4.3 * 10^-4.

Now that we know the KB value, we can do the 500 approximation method when the ratio of the initial concentration divided by, in this case, KB. If the ratio happens to be greater than 500, then we'll be able to ignore the minus X within our equilibrium expression. So our initial concentration is .882 molar, and again our KB is 4.3 * 10^-4. When we punch that in, that gives us 2051.2, so we got a number greater than 500. So we can ignore the minus X within our equilibrium expression.

So here's our equilibrium expression when we plug in the values and. Again, because the ratio is greater than 500, I can ignore this minus X. So coming up here we can say this is X² divided by 0.882. By ignoring the minus X we can avoid the quadratic formula. We're going to cross multiply these two, so when we do that we're going to X² = 3.7 times 10^-4 and then take the square root of both sides. X will give me approximately 0.01947 when we find X.

Here X gives me OH concentration, so that takes us to step five. We would use the X variable which we just found to help us calculate the percent ionization or dissociation. So remember here we're going to say percent ionization or percent association for a week base equals the the concentration of hydroxide ion at equilibrium, which we find is .01947 molar UMM divided by the initial concentration of my base which is 0.882 molar times 100. So when we do that we get 2.21%. So this will represent the percent ionization of my hypoiodite ion.