For the following equilibrium, Li3PO4(s)↽−−⇀3Li+(aq)+PO3−4(aq) If Ksp=2.4×10−11, what is the molar solubility of Li3PO4?

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Answer:             1.0  × 10⁻⁵ msolution:                     as we know ph is given as,                                    ph  = -log [h⁺]and,                                    [h⁺]  =  10^-ph    (1)putting value of ph in eq. 1,                                    [h⁺]  =  10⁻⁵                                    [h⁺]  =  1.0  × 10⁻⁵ m