For an aqueous solution of Ag_3PO_4 at a fixed pH of 6.00, answer the following

Question

For an aqueous solution of Ag_3PO_4 at a fixed pH of 6.00, answer the following questions: (K_sp = 2.8 times 10^18, K_bl = 2.4 times 10^-2, K_b2 = 1.6 times 10^-7, K_b3 = 1.4 times 10^-12) a) Calculate the molar solubility in pure H_2O. (Ignore activity) b) Calculate the molar solubility in 0.10 M Na_3PO_4. (Ignore activity) c) Using activity, calculate the molar solubility in 0.10 M Na_3PO_4. (Ion size for Ag^+ is 250 pm and for PO_4^3- is 400 pm). d) Calculate the molar solubility using systematic treatment of equilibrium (ignore activity). Show all pertinent chemical reactions and equations.

Explanation / Answer

Ksp (Ag3PO4) = [Ag+]3 [PO43-]

Let the solublility be x

Ksp (Ag3PO4) = (3x)3 (x) = 27x4

Value of Ksp is given, 2.8 x 10-18

Therefore x = (Ksp/27)1/4

                       = (2.8 x 10-18 / 27)1/4

                  = 1.79 x 10-5 M (This is the solubility)

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