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

Answer:

$5.2*10^{-6}$

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 :

$Ca(OH)_2 = \frac{1/2 * 0.0011}{0.05 (liters)}$

$Ca(OH)_2 =0.011$.

$Ca(OH)_2 = 1.1 \times 10^{-2}$.

Ksp =  [Ca^{2+} ]+ [2OH^{2-}].

Ksp = $1.1 \times 10^{-2}* (2.2 \times 10^{-2})^2$

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 $5.324\times 10^{-6}$

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:

$1\times 0.0983\times 11.22=2\times M_2\times 50\\\\M_2=0.011M$

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:

$K_{sp}=(0.011)\times (2\times 0.11)^2$

$K_{sp}=5.324\times 10^{-6}$

Hence, the $K_{sp}$ for calcium hydroxide is $5.324\times 10^{-6}$