Question

Write the principal equilibrium reaction and calculate the equilibrium constant for the base hydrolysis reaction of solid calcium phosphate to form monohydrogenphosphate anion as the principal phosphate containing species. Determine the pH of the solution. Ksp = 1.2 x 10-26. For H3PO4, Ka1, Ka2, and Ka3 = 7.5 x10-3 , 6.2 x 10-8 , and 4.8 x 10-13)

Answer 1

Write the principal equilibrium reaction and calculate the equilibrium constant for the base hydrolysis reaction of
solid calcium phosphate to form monohydrogenphosphate anion as the principal phosphate containing species.
Determine the pH of the solution.
Ksp = 1.2 x 10-26. For H3PO4, Ka1, Ka2, and Ka3 = 7.5 x10-3 , 6.2 x 10-8 , and 4.8 x 10-13)
Answer:
principal equilibrium reaction:
Ca3(PO4)2(s) + 2H2O = 3Ca+2(aq) +2HPO4-2(aq) + 2OH-(aq); Kb
Equilibrium constant: Kb = [OH-]^2*[Ca(+2)]^3*[HPO4(-2)]^2

(i) Value of Kb

H3PO4 = H2PO4- + H+ ; Ka1 = 7.5 x10-3
H2PO4- = HPO4(-2) + H+ ; Ka2 = 6.2 x 10-8
HPO4(-2) = PO4(-3) + H+ ; Ka3 = 4.8 x 10-13

PO4(-3) + H2O = HPO4(-2) + OH-; Kb3 = Kw/Ka3 = 1E-14/(4.8E-13) = 0.021

2PO4(-3) + 2H2O = 2HPO4(-2) + 2OH-; Kb3' = Kb3^2 = 4.34E-4
Ca3(PO4)2(s) = 3Ca+2 + 2PO4(3-); Ksp = [Ca+2]^3*[PO4(3-)]^2 = 1.2E-26
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Adding above two reactions

Ca3(PO4)2(s) + 2H2O = 2HPO4(-2) + 2OH- + 3Ca+2 ;
Kb = Ksp*Kb3' = 1.2E-26*4.34E-4 = 5.21E-30 = 5.21 x 10-30
Kb = 5.21 x 10-30

(ii)pH
Kb = [OH-]^2*[Ca(+2)]^3*[HPO4(-2)]^2 = 5.21 x 10-30
pKb = 2*pOH + 3*pCa(+2) + 2*p[HPO4(-2)] = 29.28

Ca3(PO4)2(s) + 2H2O = 2HPO4(-2) + 2OH- + 3Ca+2 ;
Initial:     --, --, 0, 0, 0
Change:         --, --, 2*x, 2*x,3*x
Equilibrium: --, --, 2*x, 2*x,3*x

Kb = [OH-]^2*[Ca(+2)]^3*[HPO4(-2)]^2   = [2*x]^2*[3*x]^3*[2*x]^2 = 108*x^5 = 5.21 x 10-30
Kb = 108*x^5 = 5.21 x 10-30
x = 5.45E-7
[OH-] = 2*x = 1.1E-6
pOH = 5.96
pH = 14 - pOH = 8.04