(b) Calculate the percent ionization of 0.0075 M butanoic acid in a solution containing 0.085 M sodium butanoate.

Video transcript

Hey everyone we're asked what is the percent ionization of? 0.550 moller peru vic acid with a K. A. Of 4.1 times 10 to the negative three When it is in an aqueous solution with 0.0 800 moller potassium piru 1st. Let's go ahead and write out our reaction. We can have terrific acid be represented by H. A. And potassium piru be represented by a minus. Writing out our reaction. We have our peru vic acid and this is going to react with water. When these two react, we're going to get our hydro ni um ions plus our potassium piru creating our ice chart. Initially we had 0.00550 moller of our Pyro vic acid. Since water is a liquid, we do not include this. And for our products we had zero of our hydro ni um ions initially but we had 0.800 molar of our potassium piru. Our change is going to be a minus X. In our reactant side and a plus X on our product side. Since we're losing reactant and gaining products at equilibrium for our react inside we will have 0.550 minus X. And for our hydro ni um we will have X. And for our potassium piru bait we will have 0.0 minus X. Now we can go ahead and solve for R. X. Using R. K. A. Which was provided to us. We had 4.1 times 10 to the negative three. We know that this will be equivalent to our products over our reactant plugging in these values we have X times 0.0 minus X. All over 0.550 minus X, Solving for X. We end up with a value of 2.67 times 10 to the -4. Now to solve our percent ionization, we know our percent ionization is equivalent to the concentration of our hydro ni um ions divided by the concentration of our acid initially. And this will be multiplied by Plugging in our values we get 2.67 times 10 to the negative four and this will be divided by 0. And we will multiply this by 100%. This gets us to a percent ionization of 4.86%. Which is going to be our final answer. Now, I hope that made sense. And let us know if you have any questions