Question

Please calculate the solubility of magnesium phosphate (Ksp = 1x10^-24) in: a) pure water; b) 0.010 M Mg(NO3)2; c) 0.020 M Na3PO4

Answer 1

Step-by-step explanation:

Mg₃(PO₄)₂ -------------------→ 3Mg⁺² + 2PO₄⁻³

x 3x 2x

given (3x)³ x ( 2 x )² = 10⁻²⁴

= 108 x⁵ = 10⁻²⁴

x⁵ = 9259 x 10⁻³⁰

x = 6.21 x 10⁻⁶ mol / l

= 263 x 6.21 x 10⁻⁶ ( mol weight of magnesium phosphate = 263 )

= 1633 x 10⁻⁶ gm .

b )

Mg(NO3)2 = Mg⁺² + 2 NO₃⁻¹

.01 M .01 M .02 M

(.01)³ x [ PO₄⁻³ ] = 10⁻²⁴

10⁻⁶ x (2x)² = 10⁻²⁴

x² = .25 x 10⁻¹⁸

x = .5 x 10⁻⁹ mol /l

= 263 x .5 x 10⁻⁹ gm / l

= 131.5 x 10⁻⁹ gm /l

c )

Na₃PO₄ = 3 Na ⁺¹ + PO₄⁻³

.02 .02

(.02)² x [ Mg ⁺² ] = 10⁻²⁴

.0004 x (3x) ³ = 10⁻²⁴

x³ = 92.6 x 10⁻²⁴

x = 4.52 x 10⁻⁸ mol/l

= 263 x 4.52 x 10⁻⁸ mol/l

= 1188.76 x 10⁻⁸ gm / l