Respuesta :
Answer: a) The solubility product for [tex]SrF_2[/tex] is [tex]7.80\times 10^{-10}[/tex]
b) The solubility product for [tex]Ag_3PO_4[/tex] is [tex]1.80\times 10^{-18}[/tex]
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as [tex]K_{sp}[/tex]
a) We are given:
Solubility of [tex]SrF_2=7.3\times 10^{-2}g/L[/tex]
Solubility of [tex]SrF_2=\frac{7.3\times 10^{-2}g/L}{125.62g/mol}=5.8\times 10^{-4}mol/L[/tex]
The equation for the ionization of the [tex]SrF_2[/tex] is given as:
[tex]SrF_2\leftrightharpoons Sr^{2+}+2F^{-}[/tex]
1 mole of [tex]SrF_2[/tex] gives 1 mole of [tex]Sr^{2+}[/tex] and 2 moles of [tex]F^{-}[/tex]
[tex]K_{sp}=[Sr^{2+}][F^{-}]^2[/tex]
[tex]K_sp=[5.8\times 10^{-4}][(2\times 5.8\times 10^{-4})^2][/tex]
[tex]K_sp=7.80\times 10^{-10}[/tex]
b) We are given:
Solubility of [tex]Ag_3PO_4=6.7\times 10^{-3}g/L[/tex]
Solubility of [tex]Ag_3PO_4=\frac{6.7\times 10^{-3}g/L}{418.58g/mol}=1.60\times 10^{-5}mol/L[/tex]
The equation for the ionization of the [tex]Ag_3PO_4[/tex] is given as:
[tex]Ag_3PO_4\leftrightharpoons 3Ag^{+}+PO_4^{3-}[/tex]
1 mole of [tex]Ag_3PO_4[/tex] gives 3 moles of [tex]Ag^+[/tex] and 1 mole of [tex]Ag^{+}[/tex]
[tex]K_{sp}=[Ag^+]^3[PO_4^{3-}][/tex]
[tex]K_sp=[(3\times 1.60\times 10^{-5})^3][(1.60\times 10^{-5})][/tex]
[tex]K_sp=1.80\times 10^{-18}[/tex]
