Give a recipe for preparing a CH3CO2H-CH3CO2Na buffer solution that has pH = 4.44.

Video transcript

Welcome back, everyone. How would you prepare a buffer solution made up of Ben's OIC acid and sodium benzoate with a ph of 4.10 were given the acid dissociation constant K A of Ben's OIC acid equal to 6.5 times 10 to the negative fifth. Power recall that Ben's OIC acid is not on our list of strong acids and so therefore, is a weak acid. Next, we want to figure out what its conjugate base is and we want to figure out the conjugate base because we want to recall that a buffer solution consists of our concentration of conjugate base equal to our concentration of weak acid. And so for the conjugate base of Ben's OIC acid, we would form a compound with one less hydrogen atom than our weak acid has. Originally. Meaning that if Ben's OIC acid donates a proton HC 7 H502, if it donates a proton to water, it would be acting as an acid being a proton donor. And so it would be an equilibrium because it's a weak acid with its eye on products where we form the contract base being our Benzo IQ an ion. So C seven H 502 minus as well as hydro knee um as our second product. And so we would label our benzoate an ion as the conjugate base of our weak acid benzoate acid because it has that additional proton. We're sorry because it has the loss of a proton since Arbenz OIC acid donated a proton to form our product. Hydro ni. Um and so hydro ni um would be our conjugate acid of water. Now, let's focus on the fact that we know our conjugate base. And next, we want to recall our Henderson Hasselbach equation in which we would calculate the PH set equal to R P K A added to the log of which is multiplied by our quotient where we take the concentration of our conjugate base divided by our concentration of our weak acid. Now, in order to figure out how we would prepare our buffer solution, we need to solve for what this ratio between our conjugate base and weak acid would be. And so let's plug in all the variables that we know Were given the ph of the solution being 4.10. So we can say 4.10 is equal to RPKA. Now note that we haven't figured out our peak A. So let's do the work for that to the side. Here. We're given our K A for Ben's OIC acid as 6.5 times 10 to the negative fifth power. And so recall that we can calculate R P K A as set equal to the negative log of R K. And so we would have the negative log of our given K A for Ben's OIC acid, our weak acid as 6.5 times 10 to the negative fifth power. This is going to result in a PKA equal to a value of 4.18709. So going back to our henderson household back equation, let's plug in that PKA which is added to the log of the concentration between our conjugate base divided by our concentration of our weak acid, which we're trying to figure out what this ratio is. So we need to simplify to solve for that ratio. And so we're going to go ahead and subtract 4.18709 RPKA from both sides of our equation So that it cancels out on the right hand side. And the difference on the left hand side would result in a value of negative 0.08709 equal to the log of our ratio between our concentration of conjugate base and are divided by our concentration of weak acid. Now, we need to get rid of that log term recall that we can cancel out log by making both sides exponents to the base of 10. And so for the left hand side when we take 10 to the negative 0.08709, we have a value equal to 0.8183. And so that would equal our ratio between our concentration of conjugate base and our concentration of divided by our concentration of weak acid. This is our concentration ratio. And so we would say that therefore, the buffer must have a ratio of conjugate base too weak acid As 0.818321. So now let's get even more specific and figure out the mass of our salt, sodium benzoate that we would use to create this buffer. So because we know that Our concentration of our conjugate bases in the numerator and .8183 is our concentration ratio. That means that we would need this amount of our conjugate base. And so if we understand that our sodium benzoate NAC seven H 502 is the salt of our weak acid benzoate acid. That means that we can assume that we have the same amount or that we would need the same amount of our salt to our concentration of conjugate base. And so we can say that we have, And I'll move this over, we can say that we have 0.8183 moles of our salt, which we want to convert to a mass. And so we're going to utilize our molar mass where on our periodic table for our for every atom in our compound of sodium benzoate, we have a total molar mass of 144. g per mole of sodium benzoate. And so, canceling out moles of sodium benzoate were left with g. And this will result in our mass of our salt equal to 117.9, g of sodium benzoate. And so we can round this to about 118 g as 366. Now, our concentration of weak acid is in the denominator of our ratio. And so we can say that we can use one leader and we'll say one point leader. So it's clear We can use 1.0 L of our one Moeller Ben's OIC acid To get one mole of men's OIC acid. And so overall, for the conditions of creating a buffer made of Ben's OIC acid and sodium benzoate, we would say overall Dissolve 118 g of sodium benzoate in one liter of one Moeller Ben's OIC acid. And so this entire statement That we've outlined is going to represent our final answer to complete this example corresponding to choice B on how we would prepare a buffer solution made up of Ben's OIC acid and sodium benzoate with a ph of 4.10. I hope everything I reviewed was clear. If you have any questions, please leave them down below. And I'll see everyone in the next practice video.