Determine the percent ionization of a 0.225 M solution of benzoic acid.

Video transcript

Hey everyone we're asked what is the percent ionization of? 0.150 molar solution of Ben's OIC acid. And they gave us our K. A. First, we need to recognize that Ben's OIC acid is a weak acid. So when Ben's OIC acid dissociates, we're going to end up with the following reaction. We will get our Ben Zoe ion plus our protons. Now let's go ahead and create our ice chart. Initially we had 0.150 moller of our Ben's OIC acid and we had zero of our products formed. Our change is going to be a minus X on our reactant side and a plus X on our product side. Since we're losing reactant and gaining products at equilibrium, we have 0.150 minus X. For our react inside and an X. And an X. For our product side. Now, to solve for X, we can go ahead and use our K. A. Of 6.5 times 10 to the negative five. And this will be equal to our products over our reactant. In this case it will be X times X. All over 0.150 minus X. Now we can go ahead and check if we can disregard our X. In the denominator by taking 0.150 and dividing it by R. K. A. Of 6.5 times 10 to the negative five. Now, if we get a value greater than 500 then we can safely disregard our X. Since it is negligible. In this case it is greater than 500. So we can disregard the X. In our denominator. Now let's go ahead and solve for X. So we have 6.5 times 10 to negative five equals X squared over 0.150. We're going to multiply both sides by 0. and then take the square root of both sides. This will get us to an X Of 3. times 10 to the negative 3rd molar. And this is also going to be the concentration of our protons. Now to find our percent ionization, we've learned that percent ionization is equivalent to the concentration of our protons divided by the initial concentration of our acid. And since we want this in percentages we will multiply it by 100, plugging in those values, we get 3.1225 times 10 to the negative three. And we're going to divide that by our Initial concentration of our acid which was 0.150. Then multiplying that by 100 We end up with a percent ionization of 2.08%. And this is going to be our final answer. Now, I hope that made sense. And let us know if you have any questions