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University of Maine

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Problem 99

Hydroxyapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH}),$ is the main mineral component of dental enamel, dentin, and bone. Coating the compound on metallic implants (such as titanium alloys and stainless steels) helps the body accept the implant. When placed in bone voids, the

powder encourages natural bone to grow into the void. Hydroxyapatite is prepared by adding aqueous phosphoric acid to a dilute slurry of calcium hydroxide. (a) Write a balanced equation for this

preparation. (b) What mass $(\mathrm{g})$ of hydroxyapatite could form from $100 . \mathrm{g}$ of 85$\%$ phosphoric acid and $100 . \mathrm{g}$ of calcium hydroxide?

Answer

a)

$3 \mathrm{H}_{3} \mathrm{PO}_{4}+5 \mathrm{Ca}(\mathrm{OH})_{2} \longrightarrow \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})+9 \mathrm{H}_{2} \mathrm{O}$

b)

$\mathrm{m}\left(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})\right)=\mathrm{Mr}^{*} \mathrm{n}=135,62 \mathrm{g}$

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## Discussion

## Video Transcript

given equation in which phosphoric acid reacts with calcium hydroxide to form a new compound hydroxy appetite. We first need to write the balanced equation, and it's important to note that this is an acid reacting with the base. So in addition to the compound formed, we also form water. Once we know that we can write what's called the skeletal equation or formulas instead of words. Phosphoric acid is age three p 04 Calcium hydroxide is made up of the calcium ion, which is a plus to charge in the hydroxide, and I and which is a minus one charge. To make a neutral compound, we need to hide rocks, eyes for every calcium. So the formula is C A. Oh, h two. You're given formula for the new compound. See a five p o for three. Ohh. And of course, water is H 20 Once you have your skeletal equation, weaken balance by changing the coefficients. When the product side there are five calcium. So we need to multiply on this side by five. There are also three phosphate ions and only one here. So we multiply by three. This gives us a total of 10 Oxygen's plus three more or 12 So 22 whereas on this side we have 12 13 and one more. So we need to multiply this by nine to get the correct number of oxygen's. You can then see that you also have the correct number of hydrogen because here you have nine plus 10 and here we have 18 plus one. Once we have the balance equations, we can find the amount of product formed. Since both amounts of reactions are given, we need to identify the limiting re agent. The limiting re agent is the reactant that produces this smaller amount of product. So in this case, we have ah 100 grams of 85% phosphoric acid. This is equivalent to 85 grams, and we have 100 grams of calcium hydroxide to find the amount of product formed. Our first step is to change two moles, and we do this using the molar mass of phosphoric acid, which will come from the periodic table once we have moles, plus for a gas that we can change from one substance to another and find moles, uh, our compound and we do that using the mole ratio from the balanced equation. And finally, once we have moles, we confined grams or mass using the molar mass of our compound. You then go through the same steps with calcium hydroxide, converting two moles and then converting two moles of product and then finally grams a product. And again, whichever produces the smaller quantity is called the limiting re agent. It's helpful to find the molar mass of each substance before we start. So if we look at our molar mass of phosphoric acid, we know that it's equal to three times the molar mass of hydrogen, which we find on the periodic table, plus the molar mass of phosphorus, plus four times the molar mass of oxygen or one mole of phosphoric acid equals 97 0.994 grams. Similarly, we find the molar mass of calcium hydroxide to be 74 went 092 grams, and finally, our product has a molar mass of 502 0.307 grams was starting off with 85 grams phosphoric acid. We changed moles by dividing by the molar mass and then multiply by the molar ratio, putting our number of moles of product on top. So one mole of see a three see a five p o for three. Well, h for every three moles, phosphoric acid and again that comes from this took coefficients of the balanced equation. And then finally we multiply by the molar mass of our product so the phosphoric acid will produce 145 Grand's If we look at our calcium hydroxide Ah 100 grams following the same steps. First changing two moles by dividing by the molar math 74.92 and then using our balanced equation to convert to product one more. Oh, see a five p o for three. Ohh! For every five moles, calcium hydroxide and then finally multiplying by the molar Mass of the product equals 136 grams. Smaller amount is produced, so the final answer is 100 and 36 grams

## Recommended Questions

Artificial Bones for Medical Implants The material often used to make artificial bones is the same material that gives natural bones their strength. Its common name is hydroxyapatite, and its formula is $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}$

a. Propose a systematic name for this compound.

b. What is the mass percentage of calcium in it?

c. When treated with hydrogen fluoride, hydroxyapatite becomes fluorapatite $\left[\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\right]$, an even stronger substance. Does the percent mass of Ca increase or decrease as a result of this substitution?

Tooth enamel is actually a composite material containing both hydroxyapatite and a calcium phosphate, $\mathrm{Ca}_{8}\left(\mathrm{HPO}_{4}\right)_{2}\left(\mathrm{PO}_{4}\right)_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\left(K_{1 \mathrm{p}}=1.1 \times 10^{-47}\right)$.

a. Is this calcium mineral more or less soluble than hydroxyapatite $\left(K_{\mathrm{sp}}=2.3 \times 10^{-59}\right) ?$

b. Calculate the solubility in moles per liter of hydroxyapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH}), K_{4 p}=2.3 \times 10^{-59}$ in water at $25^{\circ} \mathrm{C}$ and $\mathrm{pH}=7.00$

c. What is the solubility of hydroxyapatite at $\mathrm{pH}=5.00 ?$

Tooth enamel is actually a composite material containing both hydroxyapatite and a calcium phosphate, $\mathrm{Ca}_{8}\left(\mathrm{HPO}_{4}\right)_{2}\left(\mathrm{PO}_{4}\right)_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\left(K_{\mathrm{sp}}=1.1 \times 10^{-47}\right)$

a. Is this calcium mineral more or less soluble than hydroxyapatite $\left(K_{\mathrm{sp}}=2.3 \times 10^{-59}\right) ?$

b. Calculate the solubility in moles per liter of hydroxyapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH}), K_{\mathrm{sp}}=2.3 \times 10^{-59} \mathrm{in}$ water at $25^{\circ} \mathrm{C}$

"c. Explain why the production of weak acids by bacteria on teeth and gums increases the solubility of hydroxyapatite.

Tooth enamel is composed of a mineral known as hydroxyapatite, which has the formula $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH}) .$ Explain why tooth enamel can be eroded by acidic substances released by bacteria growing in the mouth.

Tooth enamel is composed of hydroxyapatite, whose simplest formula is $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}$ , and whose corresponding $K_{\mathrm{sp}}=6.8 \times 10^{-27}$ . As discussed in the Chemistry and Life box on page $746,$ fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F},$ whose $K_{s p}=1.0 \times 10^{-60}$ (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

Teeth enamel is hydroxyapatite $\left[\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\right]$ When it dissolves in water (a process called demineralization), it dissociates as follows: $$\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH} \longrightarrow 5 \mathrm{Ca}^{2+}+3 \mathrm{PO}_{4}^{3-}+\mathrm{OH}^{-}$$ The reverse process, called remineralization, is the body's natural defense against tooth decay. Acids produced from food remove the $\mathrm{OH}^{-}$ ions and thereby weaken the enamel layer. Most toothpastes contain a fluoride compound such as $\mathrm{NaF}$ or $\mathrm{SnF}_{2}$ What is the function of these compounds in preventing tooth decay?

Tooth enamel is composed of the mineral hydroxyapatite. The $K_{\mathrm{sp}}$ of hydroxyapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},$ is $6.8 \times 10^{-37}$ . Calculate the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}$ , forms. The $K_{\mathrm{sp}}$ of this substance is $1 \times 10^{-60}$ . Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?

Tooth enamel is composed of the mineral hydroxyapatite. The $K_{\mathrm{sp}}$ of hydroxyapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},$ is $6.8 \times 10^{-37} .$ Calculate

the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F},$ forms. The $K_{\mathrm{sp}}$ of this substance is $1 \times 10^{-60} .$ Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?

Composition of Tooth Enamel Tooth enamel contains the mineral hydroxyapatite. Hydroxyapatite reacts with fluoride ion in toothpaste to form fluorapatite. The equilibrium constant for the reaction between hydroxyapatite and fluoride ion is $K=8.48 .$ Write the equilibrium constant expression for the following reaction. In which direction does the equilibrium lie? $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})(s)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{F})(s)+\mathrm{OH}^{-}(a q)$

Fluoride ions in drinking water and toothpaste convert hydroxyapatite in tooth enamel into fluorapatite:

$$\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})(s)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}(s)+\mathrm{OH}^{-}(a q)$$

Why is fluorapatite less susceptible than hydroxyapatite to erosion by acids?