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 11.21 mL of 0.0983 M HCl to reach the end point.

Calculate
`K_(sp)` for Ca(OH)₂. Compare your result with that in Appendix D. Do you think the solution was kept at 25°C?

Answer 1

The solubility product of calcium hydroxide is
`5.35* 10^(-6)`.

Step-by-step explanation:

To calculate the concentration of acid, we use the equation given y 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=?0.0983 M\\V_1=11.21 mL\\n_2=2\\M_2=?\\V_2=50.00 mL`

Putting values in above equation, we get:

`1* 0.0983 M* 11.21 mL=2* M_2* 50.00 mL\\\\M_2=0.01102 M`

Molarity of calcium hydroxide solution = 0.01102 M

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

1 mole of calcium hydroxide gives 1 mole of calcium ions and 2 moles of hydroxide ions.

`[Ca^(2+)]=1* 0.01102 M=0.01102 M`

`[OH^(-)]=2* 0.01102 M=0.02204 M`

Solubility product is given by:

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

`=0.01102 M* (0.02204 M)^2=5.35* 10^(-6)`

The solubility product of calcium hydroxide is
`5.35* 10^(-6)`.