Section 12: Mineralized Tissues 2. Ionic solids; hydroxyapatite 2/28/06

Solubility of ionic solids the solubility of a salt or ionic solid is a resultant of 2 tendencies:  aggregation or formation of the solid state (crystallization, precipitation) favored by the strength of attraction between ions (ion-ion interaction)  dispersion or formation of the dissolved state (dissolution) favored by attraction between ions and solvent (ion-dipole interaction) 1

Types of ionic solids  slightly soluble salts  stable in the solid state  ions: –have high charge density (e.g., Ca 2+, Al 3+, PO 4 3 – ) –pack so as to: maximize separation of same-charge ions minimize separation of opposite-charge ions  highly soluble salts  stable in the dissolved state  ions: –have low charge density (Na +, NH 4 + ) –pack relatively poorly –are attracted to H 2 O 2

 solid formation & dissolving described as a chemical equation: dissolution [M(H 2 O) 6 ] + + [X(H 2 O) 6 ] -  M + X H 2 O precipitation [M(H 2 O) 6 ] + + [X(H 2 O) 6 ] -  M + X H 2 O precipitation  the ions on the left side of the above equation are shown hydrated (aquo ions) with the common coordination # of 6 [M(H 2 O) 6 ] + : ball & stick Solid–solution transition 3 space-filling

 solid formation & dissolving described as a chemical equation: dissolution [M(H 2 O) 6 ] + + [X(H 2 O) 6 ] -  M + X H 2 O precipitation [M(H 2 O) 6 ] + + [X(H 2 O) 6 ] -  M + X H 2 O precipitation  the ions on the left side of the above equation are shown hydrated (aquo ions) with the common coordination # of 6  usually ion's hydration state not shown: M + + X -  M + X – MX [M(H 2 O) 6 ] + : ball & stick space-filling Solid–solution transition 3

Solubility & rates of precipitation & dissolution  the rates of these processes depend on the surface area of the solid, A solid : rate dsln = k dsln ( A solid ) rate pptn = k pptn ( A solid )[M + ][X – ]  [M + ][X – ] is often called the ion product k dsln & k pptn are rate constants  at equilibrium, rate dsln = rate pptn k dsln ( A solid )= k pptn ( A solid ) [M + ] eq [X – ] eq 4

Solubility: K sp dividing by k pptn (A solid ), k dsln /k pptn = [M + ] eq [X – ] eq k dsln /k pptn = [M + ] eq [X – ] eq this ratio of rate constants is an equilibrium constant K sp = [M + ] eq [X – ] eq K sp = [M + ] eq [X – ] eq  this last equation states that at equilibrium the product of the component ions is equal to a constant, termed the solubility product constant, K sp  the solution is said to be saturated  there is no net dissolution or precipitation, but in general the equilibrium is dynamic  K sp : a measure of solubility larger values indicate greater solubility 5

Deviation from equilibrium often for salts in vivo, applicable ion product ≠ K sp, so under- standing these nonequilibrium conditions is also important: solutiontendency  G' dsln  G' pptn in vivo process condition favored saturated:no net change00 ion product(equilibrium) = K sp supersaturated:precipitation+–mineralization: ion product M + + X –  MX formation of > K sp bone, enamel & calculus undersaturated:dissolution–+demineralization: ion product MX  M + + X – caries, bone < K sp resorption 6

 calcium & phosphate form a variety of salts  at pH  7, the calcium phosphate solid with the lowest solubility is hydroxyapatite  empirical formula: Ca 10 (PO 4 ) 6 (OH) 2  despite its evident complexity, at constant pH a reasonable approximation of its solubility is obtained by the simple equation: K' sp(HA) = [Ca 2+ ][P i ] where [Pi] = [H 2 PO 4 – ] + [HPO 4 =  ] + [PO 4 3 –  ]  in vitro, with only Ca 2+, P i in H 2 O, pH = 7, K' sp  0.01 mM 2  in solutions containing physiological concentrations of other ions, its solubility is higher: K' sp = 0.7 mM 2 Hydroxyapatite (HA): composition, K sp 7

 by comparison, for a salt like NaCl, K sp = 3 × 10 7 mM 2  knowing the K' sp and [ions], one can calculate the tendency of a fluid to dissolve mineral or form it  note that it is the product of the ions' concentrations that matters for example:[Ca 2+ ] × [P i ] = ion product for example:[Ca 2+ ] × [P i ] = ion product mM mM mM 2 mM mM mM  despite differing ion concentrations, these 3 cases are at equilibrium under conditions where K' sp = 0.7 mM 2 8 K' sp of HA compared to [Ca 2+ ] [P i ] ion product

Calcium & P i in vivo Concentrations & ion products of calcium & P i in some biological fluids [Ca 2+ ]×[P i ] = ion product [Ca 2+ ]×[P i ] = ion product mM mM mM 2 mM mM mM 2 ECF: adult child child Saliva: low flow rate155 high "224 high "224  in all cases, ion product > K' sp, so the fluids are supersaturated with respect to HA (as long as the pH is close to 7) 9

Conditions where ion product < K' sp  as the pH of saliva drops toward 5, however, the K' sp of HA increases & becomes larger than the ion product, i.e., saliva becomes undersaturated with respect to HA  HA thus has a tendency or potential to dissolve  this is a necessary but not sufficient condition for caries formation; a number of additional factors determine the actual rate and extent of dissolution K' sp [ H+][ H+][ H+][ H+] Variation of K' sp with [H + ] Ca PO 4 3 – + 2 OH – ↔ Ca 10 (PO 4 ) 6 (OH) 2

Effect of pH on solubility  for most simple salts (e.g., NaCl), solubility does not vary with pH solubility does not vary with pH  because the salt's ions don't react with H + or OH – in the physiological pH range  because OH – is not part of crystal structure  for many other salts (e.g., phosphate salts), solubility is pH-dependent solubility is pH-dependent  because Pi forms are acids/bases with each [form] being pH-dependent due to these pK a values: H 2 PO 4 –  H + + HPO 4 = pK a = 7 HPO 4 =  H + + PO 4 3 –  pK a = 12  because OH – is a component ion of the crystal Which ions account for the pH dependence of HA solubility? 11

Crystal structure: general  unit cell: repeating unit of a crystal smallest sample of a crystal that includes an example of all of the interionic distances & angles which occur in the entire crystal smallest sample of a crystal that includes an example of all of the interionic distances & angles which occur in the entire crystal  relatively simple example: NaCl  unit cell has 4 Na + & 4 Cl – ions (lighter spheres) naclb.jpg 12

HA crystals in enamel rods 44 300x packing of rods in enamel hydroxyapatite crystallite 200 x 500 x 1000 Å  hexagonal prisms that are small & variable in size  small size means large surface area/weight rod fig. is enamrdyl.gif 13

HA crystal structure  composed of unit cells ~10 5 unit cells/crystallite  composed of ions  polar surface  ion exchange with fluids at interface  adhesion of substances due to polar interactions 50× hydroxyapatite crystallite 200 × 500 × 1000 Å hydroxyapatite unit cell 9.4 × 9.4 × 6.9 Å 60º14

 complex  18 ions/unit cell: Ca 10 (PO 4 ) 6 (OH) 2 Ca 10 (PO 4 ) 6 (OH) 2  formation difficult  supersaturation likely  unit cells pack to maximize attractive interactions  ions shared between unit cells  interlocking cells HA unit cell structure OH PO 4 Ca HAuncel.gif HA.gif 15

 2 unit cells stacked together  ions at unit cell surface interdigitate with complementary spaces on adjacent unit cells (knobs into grooves) [molecular LEGOs]  ion sharing Ca 2+ & P i ions on faces are shared with adjacent unit cell HA unit cell structure 16

this view shows how 2 OH – ions (arrows) are shared between 2 adjacent unit cells this view shows how 2 OH – ions (arrows) are shared between 2 adjacent unit cells HA unit cell structure: ion sharing 17

OH – ions shared with 3 adjacent unit cells (total=4) 8 OH – shown ÷ 4 = 2/unit cell HA unit cell structure: ion sharing 18

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