Question

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

AgBr = 7.3 x 10-7

AgCN = 7.7 x 10-9

AgSCN = 1.0 x 10-6

When these compounds are arranged in order of decreasing Ksp values, what is the correct order?

AgCN > AgSCN > AgBr
AgBr > AgCN > AgSCN
AgSCN > AgBr > AgCN
AgCN > AgBr > AgSCN

Answer 1

Answer: The decreasing order of
`K_(sp)` is
`AgSCN>AgBr>AgCN`

Step-by-step 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* 10^(-7)mol/L`

Putting values in above equation, we get:

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

Solubility product of AgBr =
`5.33* 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* 10^(-9)mol/L`

Putting values in above equation, we get:

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

Solubility product of AgCN =
`5.33* 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* 10^(-6)mol/L`

Putting values in above equation, we get:

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

Solubility product of AgSCN =
`1.0* 10^(-12)`

The decreasing order of
`K_(sp)` follows:

`AgSCN>AgBr>AgCN`