calculate the solubility of Ca(OH)2 in a 0.469M CaCl2 solution at 31 degrees C, Ksp of Ca(OH)2 is 4.96 X 10*-6
To calculate the solubility of Ca(OH)2 in a 0.469M CaCl2 solution at 31 degrees C, you need to use the concept of stoichiometry and the Ksp (solubility product constant) of Ca(OH)2.
The solubility product constant (Ksp) is a measure of the equilibrium between a solid substance and its ions in a saturated solution. For Ca(OH)2, the Ksp value is given as 4.96 × 10^-6.
The balanced equation for the dissociation of Ca(OH)2 is:
Ca(OH)2 (s) ⇌ Ca^2+ (aq) + 2OH^- (aq)
Let's assume that 'x' represents the solubility of Ca(OH)2 in moles/L. Therefore, the concentration of Ca^2+ ions is also 'x' moles/L, and the concentration of OH^- ions is '2x' moles/L.
The concentration of Ca^2+ ions is already given in the problem as 0.469M because it is provided by the CaCl2 solution. So, the expression for Ksp at equilibrium is:
Ksp = [Ca^2+] × [OH^-]^2
= (0.469) × (2x)^2
= 4.96 × 10^-6
Now, you can solve this equation to find the value of 'x', which represents the solubility of Ca(OH)2 in the CaCl2 solution at 31 degrees C.
(0.469) × (2x)^2 = 4.96 × 10^-6
Solving this equation will give you the value of 'x', which is the solubility of Ca(OH)2 in moles/L in the given solution.