Respuesta :
The solubility product, Ksp of barium phosphate is helpful to calculate the molar solubility and concentration of ions.
The equilibrium constant for a solid material dissolved in an aqueous solution is the solubility product constant, Ksp. It denotes the concentration at which a substance dissolves in solution. A material's Ksp value indicates how soluble it is - larger the Ksp value, the more soluble it is. Solubility equilibrium is a sort of dynamic equilibrium that occurs when a chemical substance in solid form is in chemical equilibrium with that compound's solution. The solid may dissolve unmodified, by dissociation, or through chemical reactivity with another solution element, such as acid or alkali. The solubility product is determined by the molar concentrations of the ions in saturated solutions, whereas the ionic product is determined by any solution.
Reaction:
[tex]Ba_3(PO_4)_2 \rightleftharpoons 3Ba^{2+} + 2PO_4^{3-}[/tex]
The solubility product of barium phosphate, Ksp = 1.3 x [tex]10^{-29}[/tex]
The expression of solubility product for barium phosphate is given as:
Ksp = [tex](3S)^3[/tex] x [tex](2S)^2[/tex] = 108 [tex]S^5[/tex]
1.3 x [tex]10^{-29}[/tex] x [tex](2S)^2[/tex] = 108 [tex]S^5[/tex]
[tex]S^5[/tex] = 1.3 x [tex]10^{-29}[/tex] / 108
S = 6.5479 x [tex]10^{-7}[/tex]M
Concentration of barium ions = [tex][Ba^{2+}][/tex] = 3 x 6.5479 x [tex]10^{-7}[/tex] = 1.9643 x [tex]10^{-6}[/tex]M
Concentration of phosphate ions = [tex][PO_4^{3-}][/tex] = 2 x 6.5479 x [tex]10^{-7}[/tex] = 1.3096 x [tex]10^{-6}[/tex]M
Molar solubility of [tex][Ba_3(PO_4)_2] \frac{[Ba^{2+}]}{3}[/tex] = 1/3 x 1.9643 x [tex]10^{-6}[/tex] = 6.5479 x [tex]10^{-7}[/tex]M
Solubility product of [tex]Ba_3(PO_4)_2[/tex] = 6.5479 x [tex]10^{-7}[/tex] x 601 = 0.0003935g/L
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