Butyric acid is responsible for the foul smell of rancid butter. The pKa of butyric acid is 4.84. (c) Calculate the pH of a 0.050 M solution of sodium butyrate.

Video transcript

Hey everyone, we're told that appropriate nick acid has a P. K. A. Of 4.886 were asked what is the ph of a 0.65 molar sodium appropriate solution. So we were told that we had sodium appropriate and this is going to disassociate into our sodium ions and our appropriate ion and are appropriate. It ion is the conjugate base of our weak acid, which is pro pre onek acid. As we can see from our reaction, we have a 1 to 1 ratio between our sodium appropriate it and our appropriate ion. So this means that the concentration of our appropriate ion Is going to be 0.065 molar. So let's go ahead and create our ice chart. So we have our conjugate base which is appropriate ion. And this is going to react with water. And our products are going to be our appropriate onek acid plus our hydroxide ions creating our ice chart. We know that we had 0.065 molar of our appropriate ion. We can completely disregard our water since it is a liquid and we initially had zero of our appropriate nick acid and our hydroxide ion. Now 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 0.65 minus X. In our react inside and an X. And an X. In our product side. Now to find R. K. A. We can find our K. A. By taking 10 to the negative of R P. K. So plugging in those values, we get 10 to the negative 4.886 which was provided to us in our questions. Then Solving for RK. A. We end up with 1.3 times 10 to the -5. Now to find our KB we can go ahead and use our K. A. And R K W. We know that our K W is equivalent to our K A. Times R K be solving for K. B. We get K B. Is equivalent to K. W over K. A. Plugging in those values. We get one point oh times 10 to the negative 14 which is R K W. And we're going to divide that by R K A. Which is 1.3 times 10 to the negative five, Solving for RKB. We end up with a value of 7.69-3 times 10 to the -10. Now we can go ahead and use our KB to solve for our X. As we've learned, we know that our KB is equivalent to our products over our reactant. In this case it will be our appropriate onek acid times our hydroxide ions, all divided by our appropriate ions. And we can further simplify this by simply writing an x times and x all over 0.065 -1. And again this is going to be equal to R K. B. Which is 7.69 to 3 times 10 to the negative 10. Now we can go ahead and check if our X. Is negligible in order to do so all we need to do is take our 0.65 and divide it by 7.69 to Times 10 to the -10. And if we get a value greater than 500 this means our X. is negligible and in this case it is so solving for X, we get X squared equals 7.6923 times to the negative 10. And we're going to multiply that by 0.065, Taking the square root of both sides, we end up with an ex of 7.0711 times 10 to the -6 Molar. And this is also going to be the concentration of our hydroxide ions. We can go ahead and use the concentration of our hydroxide ions to find our P. O. H. As we've learned we know that our P. O. H. Is equivalent to the negative log of our concentration of hydroxide ions Plugging in these values, we get the negative log of 7.0711 times 10 to the -6. This gets us to a p. h. of 5.15. Now, since our original question asked us for the ph We simply need to take 14 and subtract 5.15, and this will get us to a ph of 8.85 and this is going to be our final answer. Now, I hope that made sense and let us know if you have any questions.