The acidity of lemon juice is derived primarily from citric acid (H3Cit), a triprotic acid. What are the concentrations of H3Cit, H2Cit-, HCit2-, and Cit3- in a sample of lemon juice that has a pH of 2.37 and a total concentration of the four citrate-containing species of 0.350 M?

Video transcript

Welcome back. Everyone commonly found in Cola drinks. Phosphoric acid is atriotic acid. If a sample of Cola has a Ph of 2.80 and the total concentration of phosphoric acid species is 0.500 molar. What are the concentrations of H three po 4h 2 po four negative HPO 42 negative and po 43 negative. The K A values for phosphoric acid are given to us, we have a total of three K A values those essentially represent the asset dissociation constant. So we are going to set a lot of equations for this problem to identify each of our unknowns. And first of all, we can start with the given Ph right, we know that Ph is equal to the negative log of the concentration of hydro. And therefore, we can easily determine the concentration of hydro in the solution, taking the an clog. So that we sense the power of negative Ph. In this case, Ph is 2.80. So we're taking 10 to the power of negative 2.80 which essentially gives us a one point 58 multiplied by it. Sense the power of negative third. And that will be the molarity of hydro num. Now that we have molarity of hydro num, let's express the K A one valley. This is the first acid dissociation constant. And we know that in the process H three po four donates hydrogen to water to produce H two po four negative. So the expression of K A one is simply the product between hydro, the conjugate base of H three po four, which is H two po four negative. And then we are dividing by the reactant side. That would be the concentration of phosphoric acid itself. Now, from here, we have a total of two unknowns, right? So we can essentially label them as our unknowns. That would be the concentration of H two po four negative and the concentration of H three po four. So let's derive the relationship between them because we know the concentration of hydro and the K A one valley. So if we continue our equation, we can substitute K A one that would be 7.5 multiplied by sense, the power of negative third, the concentration of hydro is 1.58 multiplied by sense, the power of negative third, we're multiplying by the concentration of H two po four negative. We can essentially label it as our first unknown. Let's begin with H three po four. Let's say that this is our X and let's say that H two po four negative is our Y right. So it would make things easier for us. Therefore, we have 7.5 multiplied by sense, the power of negative third equals 1.58 multiplied by it. Sense the power of negative third Y divided by X. That's our first equation. Now, let's continue, if we express the concentration of phosphoric acid in terms of H two po four negative, we're going to obtain X equals 0.21 Y. So it would be better to use this equation. So we essentially obtain it after manipulation. Typically, what we are doing here is taking the top side and dividing by the left hand side of the equation to get the value of X in terms of Y. And that would be our first equation that would be more accurate to use it later. Now, let's continue with K A two. That be our third step. We know that K A two is the second acid association constant. And once again, that would be the concentration of hydro num multiplied by the concentration of the conjugate base produced. So in this case, that would be HPO 42 negative divided by the concentration of H two P four negative. Let's label our variables H two po four negative is our Y, we have a new variable which is Z right. So if we look at this specific equation, we can write down the K A two value that would be 6.2 multiplied by 10. The power of negative A is equal to 1.58. That's the concentration of hydro multiplied by it. Sense the power of negative third is C divided by Y. And from here, we can express the value of Z in terms of Y. So if we multiply both sides by Y and divide by 1.58 multiplied by 10 to the power of negative third, we get that Z equals 3.92 multiplied by sensitive power of negative fifth. Why let's label it as equation number two, we currently have three nos and two equations. So we need to use an additional equation of course or an additional equation. We are going to assume that the concentration of phosphate is approximately equal to zero. Now, essentially the reason why we're doing this is because this is the third the association step, right? And it has a very low value. So it will be negligible. What we're going to say here is that if we add the concentrations of all species acid starting with H three po four, then it's conjugated base H two po four negative. And then if we add H concentration of HPO four too negative and the concentration of phosphate that would be equal to 0.500 molar just as given in the problem. And we can say that the concentration of phosphate is approximately zero based on the fact that this would be the final ionization step, right. And it has a negligibly small K value. And of course, due to the common ion effect, so that's the total concentration that we have. And essentially this equation tells us that if we take X and Y and add C, we should end up with 0.500 which essentially gives us equation number three. So we have three nouns and three equations. Let's go ahead and introduce our system of linear equations. We can essentially see that X is equal to 0.21 Y. We can also see that C is equal to 3.92 multiplied by 10, the power of negative fifth Y. And we can also see that X plus Y plus C is 0.500. So we want to solve this equation using the subs substitution method. We can express everything in terms of Y and substitute in equation number three. So we get 0.21 Y plus Y plus 3.92 multiplied by sense. The power of negative fifth Y must be equal to 0.500. Now, from here we end up with a 1.21 Y notice how the lost value is really negligible equals 0.500. And if we divide both sides by the leading coefficient, we get Y of 0.413 right? So we have one of our unknowns. The next one is X. That would be 0.21 Y. So 0.21 multiplied by zero point, what should we get? We obtain 8.67 multiplied by 10 to the power of negative second, if we express this to two significant figures, that will be 8.7 multiplied by 10, the power of negative second. And of course, our C value is 3.92 multiplied by sense, the power of negative fifth multiplied by Y which is 0.413. And from here we get 1.6 multiplied by sense, the power of negative fifth. So from here we are ready to express our answers. Starting with phosphoric acid. The concentration of H three po four, that's our ax. We're going to state that X is equal to 8.7 multiplied by 10. The power of negative second molar followed by the concentration of H two po four negative. That's our Y which is 0.413 molar followed by HPO or too negative. That's our word C 1.6 multiplied by 10. The power of negative fifth molar. And of course, the concentration of phosphate was negligible po 43 negative was approximately zero. However, we can actually identify the value, what we're going to do is use the K A three value, we were given 4.8 multiplied by, since the power of negative 13th and we're going to equate it to the product between the concentration of hydro multiplied by the concentration of phosphate bo 43 negative. And what is our acid in that step? Well, it's essentially HPO 42 negative. What we're going to do here is essentially assume a neg negligible ionization of HPO 42 negative, right? And we're going to assume that this is equal to 1.6 multiplied by 10, the power of negative fifth. So if we solve for the concentration of phosphate po 43 negative, we are essentially going to take 4.8 multiplied by 10. The power of negative 13 multiplied by the concentration of HPO 42 negative which is 1.6 multiplied by sense the power of negative fifth. And we're going to divide it by the which is 1.58 multiplied by it sense the power of negative third. And here we are getting 4.9 multiplied by 10 to the power of negative 15th molar. And those are the final answers to this problem. Thank you for watching.