Respuesta :
Answer: The [tex]K_{sp}[/tex] of gold (III) chloride is [tex]3.2\times 10^{-25}[/tex]
Explanation:
We are given:
Solubility of gold (III) chloride = [tex]1.00\times 10^{-4}g/L[/tex]
Molar mass of gold (III) chloride = 303.3 g/mol
To calculate the solubility in mol/L, we divide the given solubility (in g/L) with molar mass, we get:
[tex]\text{Solubility (in mol/L)}=\frac{\text{Solubilty (in g/L)}}{\text{Molar mass}}[/tex]
Putting values in above equation, we get:
[tex]\text{Solubility of gold (III) chloride (in mol/L)}=\frac{1.00\times 10^{-4}g/L}{303.3g/mol}\\\\\text{Solubility of gold (III) chloride (in mol/L)}=3.29\times 10^{-7}mol/L[/tex]
The balanced equilibrium reaction for the ionization of gold (III) chloride follows:
[tex]AuCl_3\rightleftharpoons Au^{3+}+3Cl^-[/tex]
s 3s
The expression for solubility constant for this reaction will be:
[tex]K_{sp}=[Au^{3+}][Cl^-]^3[/tex]
We are given:
[tex]s=3.29\times 10^{-7}mol/L[/tex]
Putting values in above equation, we get:
[tex]K_{sp}=(s)\times (3s)^3\\\\K_{sp}=9s^4\\\\K_{sp}=9\times (3.29\times 10^{-7})^4[/tex][tex]K_{sp}=(s)\times (3s)^3\\\\K_{sp}=27s^4\\\\K_{sp}=27\times (3.29\times 10^{-7})^4=3.2\times 10^{-25}[/tex]
Hence, the [tex]K_{sp}[/tex] of gold (III) chloride is [tex]3.2\times 10^{-25}[/tex]
