Respuesta :
Answer: The solubility product of silver (I) phosphate is [tex]9.57\times 10^{-10}[/tex]
Explanation:
We are given:
Solubility of silver (I) phosphate = 1.02 g/L
To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:
Molar mass of silver (I) phosphate = 418.6 g/mol
[tex]\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L[/tex]
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The chemical equation for the ionization of silver (I) phosphate follows:
[tex]Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)[/tex]
3s s
The expression of [tex]K_{sp}[/tex] for above equation follows:
[tex]K_{sp}=(3s)^3\times s[/tex]
We are given:
[tex]s=2.44\times 10^{-3}M[/tex]
Putting values in above expression, we get:
[tex]K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}[/tex]
Hence, the solubility product of silver (I) phosphate is [tex]9.57\times 10^{-10}[/tex]
