Okay, This is a really long problem. Um, first thing is, when you solve for the ph of the buffer solution to do this, pH will be equal to the negative log of the base that is President. I'm sorry. The acid that is president, which is Benzo Kassid. We can look up the K value of Bensel kassid 6.3 times 10 to the negative five and then plus the log of the moles of benzoate over the moles of Ben's OIC acid. The moles of benzoate can be calculated by taking the massive of sodium benzoate divided by the molar mass of sodium benzoate. And then divide that by the moles of Ben's like acid, which will be the massive benzo kassid divided by the molar mass of Ben's OIC acid. And we get an initial ph of 4.13 now for the second part, which buffer component must be added in order to and in what quantity to change the pH to four with the pH is 4.13 and we want to lower the pH. We need to add an acid, and the acid component is Ben's OIC acid, so we'll start over with our Henderson hassle bolt equation. Now, the target pH is going to be equal to the negative log of K A for benzo kassid plus the log of the moles of benzoate that we start with divided by the moles of the moles of the, um Benda Look, acid we started with plus the amount that we're adding. So what I did is I recognize that we started with 1.5 grams of benzo kassid. I'm gonna add on an unknown amount of Ben's Ilic acid that we need to add in order to achieve our ph of four and then divide that whole thing by the molar mass of Ben's OIC acid. Now I've set up my equation to solve for X. I'll take the negative log of this value subtracted from both sides and then take the anti log of both and I get 10 to the negative. 100.20 is equal to miss ratio here after taking the anti log of both sides. Then what I'm going to dio is I'm going to combine these, uh denominators into a single value and then multiply that by both sides. Well, it'll divide that by both sides. So 10 to the negative point to is 100.0.63 And then when I take my, uh 1 44 and my 1 22 I get 0.847 and then I still have. After I pulled those out, what would be left? 1.50 divided by 1.50 plus X. Now I'm gonna divide both sides by 0.847 and multiply both sides by this denominator 1.5 plus x, I get this 1.12 plus 0.743 X equals 1.50 and my ex is 0.52 grams. If we need to add 0.52 grams of benzo CASS it in order to achieve a Ph of four instead of a pH of 4.13 So we're still not done as I mentioned, this is a very lengthy problem, Phil. Part C. Part C asks us to calculate the amount of hydrochloric acid or sodium hydroxide we need to add to achieve a ph of four. We just established that we need to add an asset Yes, before we added the Benz OIC acid, but Now we're going to add a different asset HCL in orderto lower the pH from the 4.13 to the fore, so we'll still have our target Ph four, which will be equal to the negative log of Benzo Kassid, plus the log of the moles of benzoate. Now the moles of Ben's await we start with is 1.50 divided by the molar mass of sodium benzoate. The moles of Ben's OIC acid we start with is 1.5 divided by the molar Mass have been so look acid. But if we add HCL to drop the pH, we're going to decrease the moles of benzoate when we add the acid by the amount of HCL we add, which will be an unknown volume multiplied by its concentration. Then we will add to Ben's OIC acid the amount of HCL that we add because every mole of HCL we add will convert a mole of benzoate into a mole of Enzo Kassid. So that's why we're subtracting the moles of HCL. We adhere and adding on the moles of HCL we had here. So then what I'll do is I'll take the negative log of this value subtracted from both sides and then take the anti log. So when I take the negative log of this and I subtracted from both sides, I get negative 0.20 I then take the anti log and it's 10 to the negative 100.20 And then the anti log of what's left over here is just going to be what's in the parentheses. So I'll rewrite again. What's in the parentheses? Okay, so then if I go 10 to the negative 100.20 I get 0.63 and then this ratio right here is 1.4 times tend to negative two minus two X. This ratio right here is 1.23 10 to the negative two and then plus two X. Now I'm gonna multiply both sides by this denominator down here and I'll get this. Then I'll collect my like terms and divide both sides by negative 3.26 and I get 8.2 times 10 the negative four leaders or 40.82 milliliters.