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Ch.17 - Additional Aspects of Aqueous Equilibria

Chapter 17, Problem 100a

Tooth enamel is composed of hydroxyapatite, whose simplest formula is Ca51PO423OH, and whose corresponding Ksp = 6.8 * 10-27. As discussed in the Chemistry and Life box on page 746, fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, Ca51PO423F, whose Ksp = 1.0 * 10-60. (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite.

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Hello everyone today. We have the following question. Carbonate minerals are minerals that contain carbonate as one of their major components. Two examples of these carbonate minerals are malachite with the K. S. P. Of 6.9 times 10 to the 34th and as a right with a K. S. P. Of 1.3 times 10 to the negative provide the expression for the salt ability, product constant for maliki. And as right. So first we will work with malachite and so malachite has the following chemical formula that's copper too, Hydroxide, two of those and one carbonate per the question. We essentially have to find the K. S.P. expression for this product. And so if we do that, we must know the charges for each of these. And so per the question, we know that carbonate has a charge of -2. Using our oxidation rules, we know hydroxide has A charge of -1. And so what we can do to find the charge of copper is begin to note the charge with our variable X. And so if we have two coppers we have two X minus the two hydroxide times it's negative one charge. So we have It would be -1 times the fact that we have two of them minus the carbonate charges, negative two. And this would equal zero. We get two, X -2 -2 equals zero two, X -4 equals zero two, x equals four and x equals two. So we know that the charge for this copper must be plus two. And so our K. Sp expression is as follows, we have our copper and we have our two plus there inside of our brackets We're going to use an exponent of two since we have two coppers right? For hydroxide, that will be O h minus the minus for the negative one sign. And then we have two hydroxide. So we have an exponent of two and last but not least for carbonate, we have C. 032 minus. There's no exponents. Carbonate is its own ion as pie atomic ion. And so we have our first K sp expression for our second one. As you're right, it's going to be a very similar process as right is C. U. Three or copper three With two carbonate ions and two hydroxide ions. I said it before we know the charge of carbonate is negative two and the charge of hydroxide is negative one. To find the charge of copper, we denote it with an X. And so we have three X. Since we have three coppers We then have -2 for the charge of carbon eight times the amount of carbon that we have which is to minus one which is the charge of hydroxide times two hydroxide ions. And this will equal zero Algebra will lead us with three X -4 -2 equals zero. We will get three x -6 equals zero three. X is equal to zero and X will equal to so once again our X will equal positive two and r k s p expression will be as follows, we will have our copper Two plus since there are three coppers, that will be our exponent for copper outside of the brackets for carbonate, we have CO and then we will have our hydroxide, our o H minus. And since we have two hydroxide that will go on the outside as well. And now we have solved for both malachite and as a right overall, I hope this helped until next time.