Question

The value of ksp for srso4 is 2.8x10-7. what is the solubility of srso4 in moles per liter? 7.6 x 10-7 1.4 x 10-7 5.3 x 10-4 2.8 x 10-7

Answer 1

Answer : The correct answer is
`5.3* 10^(-4)moles/L`.

Solution : Given,

`K_(sp)=2.8* 10^(-7)`

The balanced equilibrium reaction is,

`SrSO_4\rightleftharpoons Sr^(2+)+SO^(2-)_4`

At equilibrium s s

The expression for solubility constant is,

`K_(sp)=[Sr^(2+)][SO^(2-)_4]`

Now put the given values in this expression, we get

`2.8* 10^(-7)=(s)(s)\\2.8* 10^(-7)=s^2\\s=5.29* 10^(-4)=5.3* 10^(-4)moles/L`

Therefore, the solubility of
`SrSO_4` in moles/L is
`5.3* 10^(-4)`.

Answer 2

Answer: The solubility of
`SrSO_4` is
`5.3* 10^(-4)mol/L`

Step-by-step explanation:

It is given that
`K_(sp)` of strontium sulfate is
`2.8* 10^(-7)`

The balanced equilibrium reaction for ionization of
`SrSO_4` is given by:

`SrSO_4\rightleftharpoons Sr^(2+)+SO^(2-)_4`

At equilibrium: s s

The equation to calculate solubility constant is given as:

`K_(sp)=[Sr^(2+)][SO^(2-)_4]`

Now put the given values in above equation, we get:

`2.8* 10^(-7)=(s)(s)\\2.8* 10^(-7)=s^2\\s=5.29* 10^(-4)\approx 5.3* 10^(-4)moles/L`

Therefore, the solubility of
`SrSO_4` is
`5.3* 10^(-4)mol/L`