Question

The molar solubilities of the following compounds (in mol/L) are:

Answer 1

Answer: The decreasing order of $K_{sp}$ is $AgSCN>AgBr>AgCN$

Explanation:

• For AgBr:

The balanced equilibrium reaction for the ionization of silver bromide follows:

$AgBr\rightleftharpoons Ag^{+}+Br^-$

                 s      s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][Br^-]$

We are given:

$s=7.3\times 10^{-7}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.3\times 10{-7})^2=5.33\times 10^{-13}$

Solubility product of AgBr = $5.33\times 10^{-13}$

• For AgCN:

The balanced equilibrium reaction for the ionization of silver cyanide follows:

$AgCN\rightleftharpoons Ag^{+}+CN^-$

                    s      s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][CN^-]$

We are given:

$s=7.7\times 10^{-9}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.7\times 10{-9})^2=5.93\times 10^{-17}$

Solubility product of AgCN = $5.33\times 10^{-17}$

• For AgSCN:

The balanced equilibrium reaction for the ionization of silver thiocyanate follows:

$AgSCN\rightleftharpoons Ag^{+}+SCN^-$

                     s      s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][SCN^-]$

We are given:

$s=1.0\times 10^{-6}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(1.0\times 10{-6})^2=1.0\times 10^{-12}$

Solubility product of AgSCN = $1.0\times 10^{-12}$

The decreasing order of $K_{sp}$ follows:

$AgSCN>AgBr>AgCN$