Respuesta :
Answer: The decreasing order of [tex]K_{sp}[/tex] is [tex]AgSCN>AgBr>AgCN[/tex]
Explanation:
- For AgBr:
The balanced equilibrium reaction for the ionization of silver bromide follows:
[tex]AgBr\rightleftharpoons Ag^{+}+Br^-[/tex]
s s
The expression for solubility constant for this reaction will be:
[tex]K_{sp}=[Ag^{+}][Br^-][/tex]
We are given:
[tex]s=7.3\times 10^{-7}mol/L[/tex]
Putting values in above equation, we get:
[tex]K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.3\times 10{-7})^2=5.33\times 10^{-13}[/tex]
Solubility product of AgBr = [tex]5.33\times 10^{-13}[/tex]
- For AgCN:
The balanced equilibrium reaction for the ionization of silver cyanide follows:
[tex]AgCN\rightleftharpoons Ag^{+}+CN^-[/tex]
s s
The expression for solubility constant for this reaction will be:
[tex]K_{sp}=[Ag^{+}][CN^-][/tex]
We are given:
[tex]s=7.7\times 10^{-9}mol/L[/tex]
Putting values in above equation, we get:
[tex]K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.7\times 10{-9})^2=5.93\times 10^{-17}[/tex]
Solubility product of AgCN = [tex]5.33\times 10^{-17}[/tex]
- For AgSCN:
The balanced equilibrium reaction for the ionization of silver thiocyanate follows:
[tex]AgSCN\rightleftharpoons Ag^{+}+SCN^-[/tex]
s s
The expression for solubility constant for this reaction will be:
[tex]K_{sp}=[Ag^{+}][SCN^-][/tex]
We are given:
[tex]s=1.0\times 10^{-6}mol/L[/tex]
Putting values in above equation, we get:
[tex]K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(1.0\times 10{-6})^2=1.0\times 10^{-12}[/tex]
Solubility product of AgSCN = [tex]1.0\times 10^{-12}[/tex]
The decreasing order of [tex]K_{sp}[/tex] follows:
[tex]AgSCN>AgBr>AgCN[/tex]
