Question

Excess Ca(OH)2 is shaken with water to produce a saturated solution. The solution is filtered, and a 50.00 mL sample titrated with HCl requires 11.15 mL of 0.0973 M HCl to reach the end point.calculate ksp for Ca(OH)2?

Answer 1

Answer : The value of
`K_(sp)` is
`5.04* 10^(-6)`

Explanation :

To calculate the concentration of
`Ca(OH)_2`, we use the equation given by neutralization reaction:

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1,M_1\text{ and }V_1` are the n-factor, molarity and volume of
`Ca(OH)_2`

`n_2,M_2\text{ and }V_2` are the n-factor, molarity and volume of HCl.

We are given:

`n_1=2\\M_1=?\\V_1=50.00mL\\n_2=1\\M_2=0.0973M\\V_2=11.15mL`

Putting values in above equation, we get:

`2* M_1* 50.00mL=1* 0.0973M* 11.15mL`

`M_1=0.0108M`

Now we have to calculate the
`K_(sp)` for
`Ca(OH)_2`

The solubility equilibrium reaction will be:

`Ca(OH)_2\rightleftharpoons Ca^(2+)+2OH^(-)`

The expression for solubility constant for this reaction will be,

`K_(sp)=[Ca^(2+)][OH^(-)]^2`

Let the solubility be, x

x = 0.0108 M

`K_(sp)=(x)* (2x)^2`

`K_(sp)=(0.0108)* (2* 0.0108)^2`

`K_(sp)=5.04* 10^(-6)`

Therefore, the value of
`K_(sp)` is
`5.04* 10^(-6)`