Citric acid (H3Cit) can be used as a household cleaning agent to dissolve rust stains. The rust, represented as Fe(OH)3, dissolves because the citrate ion forms a soluble complex with Fe3+ (a) Using the equilibrium constants in Appendix C and Kf = 6.3 x 10^11 for Fe(Cit), calculate the equilibrium constant K for the reaction. (b) Calculate the molar solubility of Fe(OH)3 in 0.500 M solution of H3Cit.

Video transcript

Well every once in this video are being asked to consider a scenario where any two H two P 207 is added to solution and we're trying to solve the salt here because C two plus cat ions or ions form a soluble complex with our pyro phosphates. So this ion here. So for the first part of this problem here being given the K. S. P. For the salt which is at 4.7 times 10 to the negative six, were also given the KF value. We're being asked to determine the equilibrium constants. Okay, cue of the reaction, we're sort of doing hess's law in a sense here, starting off with the K. S. P. Value, what that is going to be the K. Or the C. I A oh H two which is a solid state equilibrium with its eye on so that see a two plus which is since it's a ion and two moles of R O H minus. So this for our K. S. P value of 4.7 times 10 to the negative six. Then we're gonna go ahead and add some K values in here. Starting off with K A two. So we have the H two P 2072 minus the ion. So that's we're gonna go ahead and hydrate this with one mole of H 20. This is an equilibrium with hydro ni um so H 30 plus which is obvious as well as each P 2073 minus which against ions. And that is also Aquarius, we're probably gonna go ahead and put these constants for them to their rights. So for here we have a K. A two. This K two value is equal to 2.51 times 10 to the negative seven. We're gonna go ahead and move forward with K. A. Three. So this chemical reaction here is H P 2073 minus. Of course that is acquis. We're gonna go ahead and hydrate this with one mole of H 20. This is an equilibrium with the products of again hydro knee. Um So H 30 plus As well as our eye on P Again, we're doing this for K. A. three. This has a value of 3.89 times 10 to the -10. We're gonna go ahead and combine RCA two plus With our an eye on that we just produced and that is P- four minus. This will go ahead and be in equilibrium with our complex. That's C. A. P 207 two minus that stone ion. So that is a curious, this has a K information equal to 4.0 times 10 to the four is something that were given to us in the problem. And lastly we have two moles of our hydro ni um ions reacting With two moles of Rh and ions. And this is equilibrium with four moles of our H 20. Alright and this has a constant equal to one over K. W. praise the power of two. We actually know that information. So go ahead, fill that in. Let's just one over K. W. Is 1.0 times 10 to the negative 14. That's raised to a power of two. That gives me a value equal to 1.0 times 10 to the 28. All right, so let's go ahead and scroll down here. So we see here that if we cancel some stuff out, we can cancel the two moles of R. O. H minus with this right here. We can also cancel out this one mole of H 20. As well as this one right here With two of the waters. We also cut out the H. 30. Plus. So we'll have one of those here. We have another H. 30. Plus. So completely be erased. We also cancel out this here as well as this ion here with here. So we see here what we have remaining then one more thing is our study on soc two plus here and here. So what we should have remaining is that R. C. A. Oh each. We used to hear that. We forgot one eye on here. That needs to be canceled with. Alright, so again the final equation is that the C. E. O. H two in its solid state will go ahead and react. Actually we don't need to cancel this Let's say 2 -. This does not repeat itself on the product side. Alright so this reacts with our an ion H two P 2072 minus which is of course a chris. And that's an equilibrium with our complex and that is C. A. P 207 The charge of two And our remaining two moles of H. 20. In its liquid state. And to solve for our constant of the equation. So they were just doing green here in the bottom the K. E. Q. Equals to the K. S. P. Value multiplied by R. K. A. To value multiplied by R. K. A. Three value multiplied by the K. So then applying all the information, we also need to multiply this value by two which we did. Okay so let's see here. Okay S. P. Value This 4.7 times 10 to the -6. My K. Two value is 2.51 times 10 to the negative seven. My K. Three value is 3.89 times 10 to the negative 10. Continuing on we have 4.0 times 10 to the four. Lastly for our K. Is just 1.0 times 10 to the 28. So putting everything into my calculator, I see that my K. E. Q. Is equal to 1.8 times 10 to the 11 power. So this is my answer for part eight. And moving on to part B. Which is being asked what is the Mueller Celje bility of the salt and 2.50 molars of N. A to H two P 207 solution were given the K. Two and K three value. So we're gonna go ahead and recognize That the concentration of my n. A two again, this is for B. So my concentration of N. A. To H. Two P 207 is also equal to the concentration of my H two, P 207 to minus this has a concentration of 0.200 molars. Now going ahead to write out my final net equation and that was C. A oh H two and a solid state that reacts with our H two P 2072 minus which is of course this is an equilibrium with our complex. So that's C. A. P 207, The charge of two reacting with the remaining H 20 molecules. So we're gonna go ahead and form a nice table. So I for initial concentration C for a change in concentration and E. Is the concentration during equilibrium. So we can see here for our first material, it's a solid so we're gonna go ahead and ignore this. And then for my product side we have HBO its liquid state. So we'll go ahead, ignore this for the ice table. So for my first eye on here, the initial concentration is 2.50 molars. And then for my complex that's concentration of zero. For the change here, we don't know this. So it's X. For my sur materials it's always minus and for my products always plus. So then for my change for my an ion here that's minus X. For my complex, that's a plus X. So at equilibrium can go ahead and combine the two. So for my first one here that's 2.50 minus X. For products, that's just simply X. I'm gonna go and scroll down for more space here. So for my K. E. Q. Is equal to the concentration of my C. A. P two minus. So that's my complex over the concentration of my certain materials which is H. Two P. 2072 minus. So we know the cake. You already. That's 1. times 10 to the 11. We don't know the concentration of the complex. That's ex that's over 2.5 zero minus X. Here are basically solving for X. And we do. So we see that we eventually get the answer for X equaling 22.50 molars. So this is going to be the more reliability and the solution. So this here is my final and last answer for this problem