Use the Henderson–Hasselbalch equation to calculate the pH of each solution. c. a solution that contains 10.0 g of HC2H3O2 and 10.0 g of NaC2H3O2 in 150.0 mL of solution

Video transcript

Welcome back everyone to another video. They were going to solve another problem. He used the Henderson Hasselback equation to calculate the ph of each solution. Part C A solution that contains 10 g of acetic acid and 10 g of sodium acid date in 150 mL of solution. We're given for answer choices. A 7.02 B 4.61 C 3.56 and D 8.23. So let's recall the equation according to the Henderson Hasselback equation, we can calculate the ph of a buffer solution by using our PK value of the weak acid. And we're going to add the logarithm of base, some of the ratio of moles of our conjugate base. So we can say base and then we're going to divide that by the number of moles of the acid. Now, what is our base? Well, essentially that's sodium acetate, right? It has that iron spectator. So if we eliminate our iron spectator, we will get our acetate anion C two H 302 negative. And now we have acetic acid, they differ by one hydrogen. So we can just add HC two H 302. And we noticed that this is a pair of a weak acid and a base. So now let's calculate their number of moles starting with the acid. We can essentially say that would be the number of moles of acid. And we can calculate the number of moles of acid using the mass. And the molar mass, we have 10 g of each. So we're going to take 10.0 g of acetic acid. And to calculate the number of moles, we need to divide that by the molar mass of acetic acid. Now the molar mass of acetic acid would be XT, let's write it down, that would be 60.05 g per mole. And similarly, we can use the same formula to calculate the number of moles of days. We still have 10 g for sodium acetate. We're going to use the molar mass of sodium acetate. In this case, that would be 82.03 graham from all. Now let's get our results for the number of most of each. For the first one, we get 0.1665 moles. We're going to carry that extra significant figure. And now for the number of moles of the base, we get a result of 0.1219 moles. Now we need the K A value of the assets. That's acidic assets. We can say that according to the tables, K A value is equal to 1.8 multiplied by 10 to the power of negative fifth. And now we are ready to plug in everything into our equation. We sold, the PK value would be equal to the negative log of KA. We are seeking negative log of 1.8 multiplied by chance the power of negative fifth. And we're going to add the logarithm of the ratio. So first of all, we have our base on top that would be 0.1219 moles. Notice that we are not using volume. Of course, we can calculate the molarity dividing our moles by the volume in liters. But that's not necessary. We can just use the number of moles. So the volume doesn't matter because both of our components are in the same solution with the same volume and the ratio will be the same even though we get different number for moles and molarity. So if you want to try, you can calculate the concentrations and plug in those concentrations into the equation. But you will end up with the same answer. Instead, we are making it shorter and using the number of moles for our numerator and for our denominator. So on top, we use the number of moles of base and on the bottom, we used the number of moles of the acid. And what we notice is that the answer that we get is 4.61. So now that we have four answer choices, let's identify the correct answer. And that would be option B 4.61 is the correct answer for the ph of the solution. Thank you for watching.