Question

Solubility of calcium phosphate (molecular mass, M) in water is W g per 100 mL at 25ºC. Its solubility product at 25ºC will be approximately –

• A. 10⁹
• B. 10⁷ $\left( \frac { \mathrm { W } } { \mathrm { M } } \right) ^ { 5 }$
• C. 10⁵
• D. 10³

Answer 1

Answer: (B)

10⁷ $\left( \frac { \mathrm { W } } { \mathrm { M } } \right) ^ { 5 }$

Answer 2

S = $\frac { 10 \mathrm {~W} } { \mathrm { M } }$ mol litre

K_sp of Ca₃ (PO₄)₂ = 108 S⁵

= 108

= 10⁷ $\left( \frac { \mathrm { W } } { \mathrm { M } } \right) ^ { 5 }$ (approximately)