A saturated solution of Ca 3( PO 4)2 contains 2 × 10-8 M Ca 2+ ions and 1.6 × 10-5 M PO 43- ions at a particular temperature. The solubility product of Ca 3( PO 4)2 at that temperature is 2.048 × 10-x . What is the value of ' x '?

Question

A saturated solution of Ca 3( PO 4)2 contains 2 × 10-8 M Ca 2+ ions and 1.6 × 10-5 M PO 43- ions at a particular temperature. The solubility product of Ca 3( PO 4)2 at that temperature is 2.048 × 10-x . What is the value of ' x '? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Ca 3( PO 4)2 leftharpoons 3 Ca 2++2 PO 43- K sp =[ Ca 2+]3 ×[ PO 43-]2 =(2 × 10-8)3 ×(1.6 × 10-5)2 =8 × 10-24 × 2.56 × 10-10 =20.48 × 10-34 =2.048 × 10-33 ∴ Value of x is 33 .

Q. A saturated solution of Ca3​(PO4​)2​ contains 2×10−8MCa2+ ions and 1.6×10−5MPO43−​ ions at a particular temperature. The solubility product of Ca3​(PO4​)2​ at that temperature is 2.048×10−x. What is the value of ' x '?

Ca3​(PO4​)2​⇌3Ca2++2PO43−​ Ksp​=[Ca2+]3×[PO43−​]2 =(2×10−8)3×(1.6×10−5)2 =8×10−24×2.56×10−10 =20.48×10−34 =2.048×10−33 ∴ Value of x is 33.