If the molar solubility of Silver Phosphate, Ag3PO4, is 1.76×10 - 5 M, what is the solubility product constant, K sp, for this ionic compound?
A) K sp = 2.9×10 - 19
B) K sp = 2.6×10 - 18
C) K sp = 1.5×10 - 13
D) K sp = 3.1×10 - 10
E) K sp = 9.3×10 - 10
Correct Answer:
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