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

Respuesta :

Answer:

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