# Calcium Phosphate Dissolution

For what reason does Calcium Phosphate dissolute at low pH? What makes it pH dependent? This question arised after reading about the calcium phosphate bridges in casein micelles and how they 'break apart' at a pH below 6, such that the micelle destabilises and essentially falls apart.

• Would you please provide the said reference? – Mathew Mahindaratne Apr 15 at 15:15
• @MathewMahindaratne : A practical question and application to understand it (authentic source): Milk and Casein – glucose Apr 15 at 15:25
• Calcium (and other divalent cations) are extremely good at bridging phosphate moieties together. I suspect (but it's only my opinion) the following: when lowering the pH, you will partially protonate phosphates and that lets less chances for them to be linked together by calcium. – SteffX Apr 15 at 15:31
• @MathewMahindaratne I read it here: 'uoguelph.ca/foodscience/book-page/casein-micelle-stability'. The source where it says below a pH of 6 I unfortunately cannot find. – Liam Apr 15 at 19:35

The solubility of calcium salts is highly dependent on pH. For example, let’s look at tricalcium phosphate, which presents the solubility equilibrium: $$\ce{Ca3(PO4)2(s) + H2O <=> 3Ca^2+(aq) + 2PO4^3-(aq)}$$ The solubility product constant for this equilibrium ($$K_\mathrm{sp}$$) is vastly varied from source to source. Some listed it as $$\pu{2.0 × 10^–29 mol^5L{–5}}$$, and another listed it as $$\pu{2.07 × 10^−33 mol^5L^{–5}}$$ at $$\pu{25 ^{\circ}C}$$ and $$\mathrm{pH}~7.00$$. However, I found a reliable $$K_\mathrm{sp}$$ value in literature, which reported that $$K_\mathrm{sp}$$ of $$\beta$$-$$\ce{Ca3(PO4)2}$$ in the $$\mathrm{pH}$$ range $$6.0-7.5$$ is $$\pu{1.2 × 10^–29 mol^5L^{–5}}$$ at $$\pu{25 ^{\circ}C}$$ and $$\pu{2.83 × 10^–30 mol^5L^{–5}}$$ at $$\pu{37 ^{\circ}C}$$ (a negative thermal coefficient of solubility; Ref.1). Therefore, the quantitative analysis of $$\ce{Ca3(PO4)2}$$ is near impossible.
Still, we can qualitatively look at the problem: If you decrease the $$\mathrm{pH}$$ of the solution in above equilibrium by adding acid to the solution, some of the phosphate ions would protonate and transform into $$\ce{HPO4^2–}$$ ions, as what happens in phosphate buffer solutions: $$\ce{PO4^3-(aq) + H3O+ (aq) <=> HPO4^2-(aq) + H2O}$$ As a result, the concentration of $$\ce{PO4^3-}$$ ion would be reduced. The system would respond to this reduction by producing more $$\ce{PO4^3-}$$ ions, according to the Le Chatelier principle. To do so, some solid $$\ce{Ca3(PO4)2}$$ would dissolve in first equilibrium, and the equilibrium will be shifted to the right.
Keep in mind that in first equilibrium, $$\ce{Ca/P}$$ ratio is 1.5. The solubility of calcium phosphates is also depend on $$\ce{Ca/P}$$ (Ref.2). For example, both monocalcium phosphate ($$\ce{Ca(H2PO4)2}$$) and monocalcium phosphate monohydrate ($$\ce{Ca(H2PO4)2.H2O}$$) are highly soluble in water at $$\pu{25 ^{\circ}C}$$ ($$\ce{Ca/P}$$ is 0.5). On the other hand, $$K_\mathrm{sp}$$s of dicalcium phosphate ($$\ce{CaHPO4}$$) and dicalcium phosphate dihydrate ($$\ce{CaHPO4.2H2O}$$) are $$1.26 × 10^{–7}$$ and $$\pu{2.57 × 10^{–7} mol^2L^{–2}}$$, respectively at $$\pu{25 ^{\circ}C}$$ (Ref.2). Based on these values, when $$\mathrm{pH}$$ is gradually reduced to a point such that all calcium phosphates would dissolve.
1. T. M. Gregory, E. C. Moreno, J. M. Patel, W. E. Brown, “Solubility of $$\beta$$-$$\ce{Ca3(PO4)2}$$ in the system $$\ce{Ca(OH)2-H3PO4-H2O}$$ at $$5, 15, 25,$$ and $$\pu{37 ^{\circ}C}$$,” Journal of Research of the National Bureau of Standards–A. Physics and Chemistry 1974, 78A(6), 667–674 (http://dx.doi.org/10.6028/jres.078A.042).
It is because the phosphate ion $$PO_4^{-3}$$ is a weak base. In general, ionic substance with anions that are weak bases dissolve better in acidic solution.