Barium hydroxide powder is stirred in water. Then the excess barium hydroxide is

Question

Barium hydroxide powder is stirred in water. Then the excess barium hydroxide is removed by centrifugation.   The Ksp of barium hydroxide is 5x103

A. If 1 liter of the solution of the above solution is treated with Sodium phosphate to precipitate all of the barium (Ksp of Ba3(PO4)2 is 3x1023) how many grams of barium phosphate would be recovered?

B. Draw an I.C.E. chart and equilibrium expression for the addition of barium hydroxide to a 0.0100 M solution of sodium hydroxide. If the experiment described in question 6 were repeated with this sodium hydroxide solution instead of deionized water, would the quantity of barium phosphate recovered be greater than, less than, or equal to the quantity predicted in part B.

Explanation / Answer

A.)   solubility of Ba(OH)2 = 0.108 M

     Ksp = 4s^3 = 5*10^-3

     s = 0.108 M

3Ba^2+ + 2PO4^3- -----> Ba3(PO4)2(s)

solubility of Ba3(PO4)2 = 1.23*10^-5 M

   ksp = (3s)^3(2s)^2

3*10^-23 = (3s)^3(2s)^2

s = 1.23*10^-5 M

concentration of Ba^2+ precipitated = 0.108 - (1.23*10^-5)

   = 0.10799 M

No of mol of Ba3(PO4)2 produced = 0.10799/3 = 0.036 mol

mass of Ba3(PO4)2 produced = 0.036*601.93 = 21.7 grams

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