Question

1-4 please help

7 points possible Last Name First Name 1 x 1031. It dissociates in water The slightly soluble salt fluorapatite, Cas(POa)3F, has a Ksp according to the following equation: Cas(PO4)3Fi 5Ca2 3PO F 1. Write the Ksp equation for this process: Ksp Calculate the equilibrium concentrations of Ca, PO, and F. 2. a. [Ca2]= b. [PO]= c. [F]= How many grams of fluorapatite will dissolve in 1L of water? 3. 4. The solubility of the related salt hydroxyapatite, Cas(PO4)3OH (Ksp 1 x 1029), is affected by the pH of the water to which it is added. Calculate the number of grams of hydroxyapatite that will dissolve in 0.250L of a. neutral water. b. Calculate the number of grams of hydroxyapatite that will dissolve in 0.250L of water having a pH = 5.5

Answer 1

1. Ca5(PO4)3F(s) ----> 5 Ca+2 + 3 PO4-3 + F-

   Ksp = [Ca+2]^5[PO4-3]^3[F-]

2. Ksp = (5s)^5(3s)^3s = 1*10^-31

         s = 1.02*10^-4 M

   at equilibrium,

a. [Ca+2] = 5s = 5*1.02*10^-4 = 5.1*10^-4 M

b. [PO4-3] = 3S = 3*1.02*10^-4 = 3.06*10^-4 M

C. [F-] = s = 1.02*10^-4 M

3. molarmass of fluorapatite (M.wt)= 504.3 g/mol

   mass of fluorapatite dissolved in 1 L = s*M.wt

                   = (1.02*10^-4)*504.3

                                         = 5.14*10^2 g

4. Ksp = [Ca+2]^5[PO4-3]^3[OH-]

    Ksp = (5s)^5(3s)^3s = 1*10^-29

         s = solubility = 1.7*10^-4 M

molarmass of salt of hdroxyapatite (M.wt)= 502.3 g/mol

mass of fluorapatite dissolved in 1 L = s*M.wt

                   = (1.7*10^-4)*502.3

                                         = 0.085 g/l

solubilit in 0.25 L = 0.085*0.25 = 0.021 g

b) pH = -log[H+]

   5.5 = -log[H+]

   [H+] = 3.16*10^-6 M

same concentration of OH- will be neutralised.so that, solubilit of salt of hdoxapatite increases.

Ksp = (5s)^5(3s)^3(s-3.16*10^-6) = 1*10^-29

         s = solubility = 1.7034*10^-4 M

molarmass of salt of hdroxyapatite (M.wt)= 502.3 g/mol

mass of fluorapatite dissolved in 1 L = s*M.wt

                   = (1.7034*10^-4)*502.3

                                         = 0.0856 g/l