1a.) For Ag3PO4, the measured \"s\" cited at 25 degrees C is 6.7 x10^-3 g/L and

Question

1a.) For Ag3PO4, the measured "s" cited at 25 degrees C is 6.7 x10^-3 g/L and one chemical literature table gives Ksp (Ag3PO4) =2.8 x 10^-18. If s is correct (likely since various literaturesources cite the same s), do the calculations which demonstratethat this Ksp value cannot be correct.
b.) A different table gives Ksp (Ag3PO4) = 1.3 x 10^-20(apparently correct). Use s and this Ksp to calculate [PO4^-3] forsaturated aqueous Ag3PO4 solution at 25 degrees C. Then calculate %hydrolysis for PO4^-3.
c.) Now comment on your calculated % hydrolysis. Is it toohigh? Too low? Reasonable? Defend your answer.
b.) A different table gives Ksp (Ag3PO4) = 1.3 x 10^-20(apparently correct). Use s and this Ksp to calculate [PO4^-3] forsaturated aqueous Ag3PO4 solution at 25 degrees C. Then calculate %hydrolysis for PO4^-3.
c.) Now comment on your calculated % hydrolysis. Is it toohigh? Too low? Reasonable? Defend your answer.
b.) A different table gives Ksp (Ag3PO4) = 1.3 x 10^-20(apparently correct). Use s and this Ksp to calculate [PO4^-3] forsaturated aqueous Ag3PO4 solution at 25 degrees C. Then calculate %hydrolysis for PO4^-3.
c.) Now comment on your calculated % hydrolysis. Is it toohigh? Too low? Reasonable? Defend your answer.

Explanation / Answer

part a is this ksp is the constant solubility product forAg3PO4. This means that Ksp=[Ag+]^3[PO4-3] You can use this to calculate the molarity of the twoions. 2.8*10^-18=x^3*x x=4.09*10^-5 mol/L this is the concentration of the PO4 x^3=6.85*10^-14 mol/L Ag+ Now you can convert that into g/L for each one and add ittogether to find s. which will be different from the reportedvalue. 6.85*10^-14 mol/L * 107.87 g/mol = 7.39*10^-12g/L Ag+ 4.09*10^-5 mol/L * 94.974 g/mol= 3.88*10^-3 g/L PO4 add them together, since the silver adds very littleweight. The number for the PO4 is the final answer. andabout half that of the s.

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