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 11.22 mL of 0.0983 M HCl to reach the end point. Part A Calculate Ksp for Ca(OH)2.

Answer 1

`5.2*10^(-6)`

Step-by-step explanation:

The balanced chemical equation of the reaction is :

Ca(OH)2 + 2HCl → CaCl2 + 2 H20.

Ksp can be calculated by the following formula:

Ksp = [Ca^{2+} ]+ [OH^{2-}].

Moles of HCl = Molarity × Volume of solution ( liters).

Moles of HCl can be calculated by multiplying 0.01122 (liters) ×0.0983

Moles of HCl = 0.0011 or
`1.0*10^(-3)`

The calculation of the concentration of Calcium hydroxide ( as starting with 50 ml) is :

`Ca(OH)_2 = (1/2 * 0.0011)/(0.05 (liters))`

`Ca(OH)_2 =0.011`.

`Ca(OH)_2 = 1.1 * 10^(-2)`.

Ksp = [Ca^{2+} ]+ [2OH^{2-}].

Ksp =
`1.1 * 10^(-2)* (2.2 * 10^(-2))^2`

Ksp =
`5.2*10^(-6)`

Hence, the Ksp of calcium hydroxide is
`5.2*10^(-6)`

Answer 2

Answer: The
`K_(sp)` for calcium hydroxide is
`5.324* 10^(-6)`

Step-by-step explanation:

To calculate the concentration of acid, 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 acid which is
`HCl`

`n_2,M_2\text{ and }V_2` are the n-factor, molarity and volume of base which is
`Ca(OH)_2`

We are given:

`n_1=1\\M_1=0.0983M\\V_1=11.22mL\\n_2=2\\M_2=?M\\V_2=50mL`

Putting values in above equation, we get:

`1* 0.0983* 11.22=2* M_2* 50\\\\M_2=0.011M`

The concentration of
`Ca(OH)_2` comes out to be 0.011 M.

The balanced equilibrium reaction for the ionization of calcium hydroxide follows:

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

The expression for solubility constant for this reaction follows:

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

Putting the values in above equation, we get:

`K_(sp)=(0.011)* (2* 0.11)^2`

`K_(sp)=5.324* 10^(-6)`

Hence, the
`K_(sp)` for calcium hydroxide is
`5.324* 10^(-6)`