Citric acid, which is present in citrus fruits, is a triprotic acid (Table 16.3). (a) Calculate the pH of a 0.040 M solution of citric acid. (b) Did you have to make any approximations or assumptions in completing your calculations? (c) Is the concentration of citrate ion 1C6H5O7 3-2 equal to, less than, or greater than the H+ ion concentration?

Video transcript

Hi everyone, this problem reads arsenic acid is a tri protic acid. What is the ph of a 0.35 molar solution of arsenic acid. So two things we need to pay attention to here is our acid and that its try product. Okay, so arsenic acid is a weak acid. It does not fall onto our list of strong acids. So that means it's ions are going to disassociate and it's not going to dissociate completely into its ions. Okay. And the tri product means that it's going to have a first ionization and second ionization in which we're going to need to calculate the concentration of hydrogen ions. Alright, so let's start off by writing out our reaction. Okay, So we have our reaction. And what we're going to do here is calculate our concentration of hydrogen ions. And this is going to be from the first ionization. So we're going to create an ice table in order for us to find out the ionization or the excuse me, the concentration of hydro ni. Um Okay, and so we're told that we have a initial concentration of 0.35. So this is zero here and this is zero here. So that means our reaction is going to move in the right direction. So our reactant are going to be consumed and we're going to have products produced here. So when we combine our two rows, we end up with the following and this X. Is going to represent our concentration of hydrogen ions. So we're going to need to solve for X. And we're going to write out our K. A. Expression for our first ionization. Okay and so that's why we have a one here. So R K expression is our concentration of products over our concentration of reactant. Since and this is going to be equal to when we look up the acid dissociation constant or the K. A. It is equal to 5.5 times 10 to the negative three. So we're going to plug in what we have in our equilibrium row of our ice table into our K. A. Expression. So we have X times X over 0.35 minus X. Is equal to 5.5 times 10 to the negative three. So remember our goal here is to solve for X. Because that's going to give us our hydro ni um ion concentration. So when we simplify we get x squared over 0.35 minus X. Is equal to 5.5 times 10 to the neck three. Now this minus X. In the denominator. We can check to see if it's negligible. If this X. Is negligible then we can ignore it. And the way we check to see if it's negligible is we take our initial concentration so 0.35 and divide it by R. K. A value. So 5.5 times 10 to the negative three. If this value is greater than 500 that means X is negligible. If it's less than 500 then X. is not negligible and we can't ignore it. So this value is less than 500 so X is not negligible. So we have to leave it here. Alright, So that means now we're going to cross multiply to get rid of our fraction. Alright, so we get X squared is equal to 5.5 times 10 to the - time 0.35 minus X. All right, so we're going to foil out the right side. Okay, So we get X squared is equal to 1.925 times 10 to the negative four minus 5.5 times to the negative three X. So we're going to want to bring everything over to one side. So we're gonna bring everything on the right side of our equal sign to the left side. So when we do that we get X squared plus 5.5 times 10 to the negative three X minus 1.925 times 10 to the negative four is equal to zero. So we have what's called a quadratic equation here. And so we're going to need to use the quadratic formula to solve for X. And when we use the quadratic equation we get X is equal to 0.1139. And based off of our ice table, we know X is equal to our concentration of hydro ni um ions and our concentration of our conjugate base. Okay, so with that information we can now calculate our concentration of hydro knee um ion from our second ionization because this X. Represents our concentration of hydro ni um from the first ionization. So we're going to repeat the same steps but now we're going to calculate it for the second ionization. So we're going to rewrite out our equation. Okay? And I'll write that note here. This is first ionization because we're dealing with a tri product. Yes, we're dealing with a tri protic acid. This is why we're doing this. All right, so this is equal to r hydro ni um ion concentration plus our conjugate. So we're going to do the same thing, creates a ice table. Alright. So now we know that the the concentration we just solved four is what we're going to put here. Okay so that 0.01139, that's our that's our starting concentration now, up above it was 0.035 but now it is 0.01139. And it's going to be the same for our hydro ni um concentration and this is zero. So our change now we're going to use Y. So we have minus Y plus Y. And plus y. So when we combine we have 0.1139 minus Y. We have 0.01139 plus Y. And then we just have Y. So we're gonna go ahead and write R. K. Expression for our second ionization and that is K A two. And it's going to be written in the same way our products over our our concentration of products over our concentration of reactant. And this is going to be equal to 1.7 times 10 to the negative seven. Okay, So we're going to plug in our equilibrium row into our K expression. So we have 0. plus Y times why? Over 0.1139 minus Y. Is equal to 1.7 times 10 to the negative seven. Okay, so now we're going to cross multiply to get rid of our fraction. And when we cross multiply we're going to get 1.7 times 10 to the - time 0. minus Y is equal to 0.1139 plus Y times Y. Alright, so we're going to foil here and when we foil this out we get 1.9363 times 10 to the negative nine minus 1. times 10 to the negative seven. Why is equal to And we're going to also foil this out. Okay and this is equal to 0.1139 Y plus Y squared. Alright, so our goal here is to solve for y. So let's go ahead and continue. So we're going to add this 1.7 times 10 to the negative seven Y. To both sides. And when we do that we get 1.9363 times 10 to the negative nine is equal to 0.1139 plus Y squared. So we want to bring everything over to one side. Okay? And set it equal to zero. So when we do that we get y squared minus 1.9363 times 10 to the negative nine plus 0. is equal to zero. So this is a quadratic formula. So we're going to use the quadratic equation to solve for Y. And when we use the quadratic equation we get Y is equal to 1.6999 times 10 to the negative seven. Okay, so the value for Y is very small compared to X. So that means the concentration of hydro ni um ions from the second ionization energy is not significant. So we're going to use the value that we got for X. Alright and that is what we're going to you to solve for P H. P H is equal to the negative log of our hydro ni um ion concentration and we're going to use the value from our first ionization. Okay, so that means we have P H. Is equal to the negative log of 0.1139. And when we solve this out we get a p h is equal to 1.94 as our final answer. And that is it for this problem. I hope this was helpful.