You are asked to prepare a pH = 4.00 buffer starting
from 1.50 L ...

Question

You are asked to prepare a pH = 4.00 buffer starting
from 1.50 L of 0.0200 M solution of benzoic acid
1C6H5COOH2 and any amount you need of sodium benzoate
1C6H5COONa2. (a) What is the pH of the benzoic acid
solution prior to adding sodium benzoate?

Accepted Answer

Hi everyone for this problem it reads boric acid has a P. K. A. Of 9.27. What is the P. H. Of a 0.45 molar solution of boric acid. So for this problem we're looking for P. H. And the first thing we're going to need to do is identify whether or not this is a strong acid or weak acid. Because that's to determine how we solve this problem, boric acid does not fall on our list of strong acid. So that makes this a weak acid. Okay, so when we have a weak acid, so here we have our boric acid, it equilibrium, it dissociates into, its ions are hydro ni um ion and its conjugate base. Okay, so for our ice table we're going to have our boric acid which is represented by H A R boric acid and this a minus is our conjugate base. For our ice table we're going to have our weak acid react with water. So our reaction is going to be boric acid plus water yields hydro ni um plus our conjugate base. Okay, so we'll go ahead and write ice on the side. I like to draw this line here to divide our reactant from our products and we're going to ignore water because we don't include liquids or solids in our equilibrium expression. Okay, so the iro of our ice table represents our initial concentration in the problem. We're told the initial concentration of boric acid is 0.45. So we'll put that there and we have zero product. Okay, so that means our reaction is going to be moving to the right which means our concentration of reactant is going to decrease and our concentration of products is going to increase and we have one mole of everything. So everything is just going to be minus X. And then plus X. Our equilibrium row is going to be the sum of R I n E row, so 0.45 minus X. And then this is just X. Now the equilibrium row of our ice table is going to tell us what our equilibrium concentrations are. And that means we need to solve for X. X. Is the variable that we want to solve for in our ice table. So let's go ahead and write out our equilibrium expression R K A. Okay, so K A. Is equal to the sum of our products over reactant. Okay, so we have our concentration of hydro knee um times our conjugate base over our concentration of are conjugate acid. Okay, so this is going to equal a value and we can solve for this value because we were given P K A. In the problem we're told peak A is 9.27. So we need to go from P K A two K A. And for us to do that we need to remember that A K A is equal to 10 to the negative P K A. So here we are going to be able to solve for R K A. Because we know what the P. K. Is. So R K A. Is 10 to the negative 9.27. Okay so that means R. K. A. When we solve is equal to 5.3703 times 10 To the negative 10. Alright so we're going ahead. We're going to go ahead and write that here. RK. is equal to 5.37 times 10 to the negative 10. And I'll go ahead and move this up. Alright so now that we know what our equilibrium expression is, we're going to plug in what's on our equilibrium equilibrium row for our concentrations. Okay so for our concentration of hydro ni um that's represented by X. For our concentration of bass, that's represented by X. And for our concentration of our boric acid H. A. That's represented by 0.45 minus X. So we just took everything that's in our equilibrium row and plugged it in. Okay so this is equal to R. K. A. So 5.3703 times 10 to the negative 10. So now we're going to solve for X. But one thing we can do, so we see this X here and the denominator we need to check whether or not this X. Is negligible. If this X. Is negligible, then we can go ahead and ignore it and the bottom will just be 10.45 but we need to first check if it's negligible. And the way we check it is by dividing Okay, so we're right here is X negligible. Okay, and the way we do that is by dividing our initial concentration of our weak acid. So 0.45 by R k a value. Okay, so by 5.3703 times 10 to the negative 10. If this value is greater than 500 then X will be negligible. However, if it is not greater than 500 then X is not negligible. So here This X is going to be greater than 500, so that means X is negligible. Okay, so we just simplified this a little bit. So we'll have X squared over 0.45 is equal to 5.3703 times 10 to the negative 10. Alright. And remember, our goal here is to solve for X. So let's go ahead and simplify. So when we simplify we get X squared, we get X squared is equal to 5.3703 times 10 to the negative 10 times 0.45. Let's simplify the right side. We get 2. times 10 to the negative 11. We need to get rid of this square on the left side. So we'll take the square root of both sides. So X is going to equal the square root of 2.416635 times to the negative 11. Alright so we get a final answer of X. Is equal to 4. times 10 to the negative six. Now remember in our ice table we're solving for X. And X tells us what is our concentration of hydro ni um ions. And remember we can find P. H. Because we can find ph knowing our concentration of hydro knee um ions because P. H. Is equal to the negative log of our concentration of hydro knee um ions. Okay now that we know what X. Is. Will plug in our value of X. For a concentration of hydro knee um ions. Because in the table we see that our equilibrium row or our concentration of hydrogen ions is equal to X. So let's finish this off. So P. H. Is equal to the negative log of 4.9159 times 10 to the negative six. Okay so we get a final answer of P. H. Is equal to 5.31. And this is our final answer. This is the ph of a 0. molar solution of boric acid with a P. K. A. Of 9.27. That's the end of this problem. I hope this was helpful

You are asked to prepare a pH = 4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid 1C6H5COOH2 and any amount you need of sodium benzoate 1C6H5COONa2. (a) What is the pH of the benzoic acid solution prior to adding sodium benzoate?

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