Here it says to calculate the hydroxide ion concentration of a 0.55 molar potassium fluoride solution at 25¬∞C. Here we're told the acid association constant of HF hydrofluoric acid is 3.5×10-4. All right, so step one is we're going to set up an ice chart for the weak base that has it reacting with water. Now we know that this is a weak base because this is an ionic compound, so we break it up into its ions. So we have K positive and F minus. Based on our understanding of ionic salts, we know that potassium is a main group. Metal main group metals need to be +3 or higher in terms of charge in order to be acidic. Since it doesn't meet that minimum requirement, it's neutral F minus.
On the other hand, adding H2 it creates HF, which is a weak acid, which means that this F minus is basic. But since it comes from a weak acid, it is stronger. But it's still relatively weak, right? So F minus here is our weak base. Now for ionic bases, which this one is, we ignore the neutral metal cation. Here we ignore the potassium ion. Now use the Bronsted-lowry definition to predict the product's form. Make sure that KB is used in the presence of the weak base. So right now we are only given the Ka of its acid form, but we're going to need to utilize its KB later on to find our final answer.
So remember we're going to set up our equation so F minus here will react with water, which is a liquid. Since it is a base, water is going to act as the acid. Water would therefore give an H plus over to F minus. This would create HF aqueous plus OH minus aqueous as products. Step 2. Use the initial row. So remember for this I stands for Initial change equilibrium. Using the initial row, place the amount given for the weak base. So the weak base we're told it was .55. Molar places 0 for any substance not given an initial amount in an ice chart. We ignore solids and liquids, so the water is ignored. We're not told anything about our products initially, so they're both 0.
Step three, we lose reactants to make products, so using the change row, place A -, X for the reactants and A + X for the products, so minus X since we're losing it, and plus X for the product since we're making them. We bring down everything, so .55 -, X + X + X. Now, using the equilibrium row, set up the equilibrium constant expression with AB again 'cause it's a weak base and solve for X. We checked if a shocker can be utilized to avoid the quadratic formula. Alright, so the shortcut that we're going to utilize is called the 500 approximation method. Basically, we're going to take the ratio of initial concentration of our weak base and divide it by our KB value. If we get a number greater than 500, then we can ignore the minus X within our equilibrium expression, right?
So here we're going to say kW equals KA times KB. We need to isolate KB, so KB equals kW divided by ka. So this would be 1.0×10-14 divided by our ka, which is 3.5×10-4. Doing that gives us a KB value of 2.857×10-11. Alright, so our initial concentration is 0.55 molar. We found our KB as 2.857×10-11. When we punch that in, we get 1.925×1010. So we get a number much greater than 500, which means we'll be able to ignore the minus X within our equilibrium expression.
Remember your equilibrium expression for this weak basis KB equals products over reactants, so we plug in the products over reactant. Water is a liquid so we ignore it. So bringing down this expression and inserting values for what we have AB we said is 2.857×10-11. Both the products are X at equilibrium, so multiply together as X ^2 and this will be oh .55 -, X. Well remember we just did the ratio of initial concentration divided by KB and we saw a number much greater than 500, which means we can ignore this minus X and avoid the quadratic formula that we have here.
So then all it becomes is cross multiplying my KB with .55. So I'm going to do that up here. So when I do that, I'm going to hit X ^2 = 1.57135×10-11. I don't want X ^2, I just want X. We'll take the square root of both sides X = 3.96×10-6 molar. Now this is our answer because we're asked to find the hydroxide ion concentration, and at equilibrium that's equal to X. So we just found out what my X value is, which is equivalent to my hydroxide ion concentration. So my final answer will be 3.96×10-6 molar.
