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Ch.17 - Acids and Bases

Chapter 17, Problem 113

Write chemical equations and corresponding equilibrium expressions for each of the three ionization steps of phosphoric acid.

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Hey everyone. We're asked to provide the K expressions for the ionization of citric acid. And as we can see right here, citric acid is a tri protic acid which means that we have three ionized herbal protons. So first let's go ahead and start with our citric acid. We have H three C six H 507. And this is going to react with water. Now when this reacts with water we get the conjugate base of our citric acid with the formula of H two C six H minus. And we also have our hydro ni um ion. Now for our K. A. One this is going to be the concentration of our products over our reactant. In this case it will be the conjugate base of our citric acid which is H two C six H 507 minus times the concentration of our hydro knee um ion all divided by the concentration of our citric acid. And we do not include water in our expression since it is a liquid. So this will be our first K. A expression. Now let's go ahead and look at our 2nd 1. So we're going to start off with the conjugate base of citric acid. So the formula of H two C six H minus. And this is going to react with water. Now when this reacts with water We're going to get HC 6 H507 with a -2 charge. And we're also going to get our hydro ni um ion again, we're going to take the same steps and we know that our K. Expression for K. Two is going to be our products over our reactant. So we have the concentration of H C. Six H 507 to minus times the concentration of our hydro knee um ion. And this is going to be divided by the concentration of H two C six H 507 minus which is the conjugate base of our citric acid. Again water is not included since it is a liquid. Now let's go ahead and look at our last K. Expression. We're going to take the formula H C six H 5072 minus and we're going to react it with water when this reacts with water. We get our citrate ion which has a formula of C six H 507 with a minus three charge and our hydro ni um ion. Now to get our last K expression, this is going to be the concentration of our citrate ion with the formula of C six H five times the concentration of our hydro ni um ion. And this will all be divided by the concentration of HC six H 507 with a - charge. And this will be our los que expression for our try protic acid citric acid. Now I hope this made sense. And let us know if you have any questions