10. Use the accepted Ksp value of Calcium hydroxide to calculate the following:
a. What is the most calcium hydroxide that can be dissolved in 1.0 Liters of de-ionized water?
b. What is the most calcium hydroxide that can be dissolved in 125.0 ml of de-ionized water?
c. What is the most calcium hydroxide that can be dissolved in 287.0 ml of
de-ionized water ?
.......Ca(OH)2 ==> Ca^2+ + 2OH^-
I......solid.......0........0
C......solid........x.......2x
E......solid.......x........2x
Ksp = (Ca^2+)(OH^-)^2
Substitute the E line into the Ksp expression and solve for x. That will give you the solubility of Ca(OH)2 in mols/L. That is part a in mols/L. If you want grams, that is g = mols x molar mass = ?
part b.
?mols/L x (125/1000) = ? mols/125 mL
part c.
c is a repeat procedure of b.
To calculate the maximum amount of calcium hydroxide (Ca(OH)2) that can be dissolved in a given volume of de-ionized water, we need to use the solubility product constant (Ksp) for calcium hydroxide. The Ksp value represents the equilibrium constant for the dissolution of a sparingly soluble compound.
The balanced equation for the dissolution of calcium hydroxide is:
Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH-(aq)
The Ksp expression for calcium hydroxide is:
Ksp = [Ca2+][OH-]^2
Given that the accepted Ksp value for calcium hydroxide is 5.5 x 10^-6, we can use this value to calculate the maximum solubility of calcium hydroxide.
a. To calculate the maximum amount of calcium hydroxide that can be dissolved in 1.0 liter of de-ionized water:
First, we convert the volume of water from liters to moles of hydroxide ions (OH-). Since the concentration of hydroxide ions is twice that of calcium ions:
1 L of de-ionized water = 1000 mL * 1 mol/L * 2 mol OH-/L = 2000 mmol OH-
Since the stoichiometric coefficient of OH- is 2, this means we have 4000 mmol of OH- ions available.
Using the Ksp expression, we can now calculate the concentration of calcium ions (Ca2+):
Ca2+ = Ksp / [OH-]^2
Ca2+ = (5.5 x 10^-6) / (4000 mmol/L)^2 = 6.875 x 10^-11 mol/L
Since the concentration of Ca2+ is equal to the solubility of calcium hydroxide, i.e., the maximum amount that can be dissolved, the answer is 6.875 x 10^-11 mol/L.
b. To calculate the maximum amount of calcium hydroxide that can be dissolved in 125.0 ml of de-ionized water:
First, convert the volume of water from milliliters to liters: 125.0 mL = 0.125 L.
Using the same calculations as in part a, we get:
Ca2+ = (5.5 x 10^-6) / (200 mmol/L)^2 = 1.375 x 10^-9 mol/L
Since the concentration of Ca2+ is equal to the solubility of calcium hydroxide, the maximum amount that can be dissolved in 125.0 ml of de-ionized water is 1.375 x 10^-9 mol/L.
c. To calculate the maximum amount of calcium hydroxide that can be dissolved in 287.0 ml of de-ionized water:
First, convert the volume of water from milliliters to liters: 287.0 mL = 0.287 L.
Using the same calculations as in part a, we get:
Ca2+ = (5.5 x 10^-6) / (574 mmol/L)^2 = 1.525 x 10^-9 mol/L
So, the maximum amount of calcium hydroxide that can be dissolved in 287.0 ml of de-ionized water is 1.525 x 10^-9 mol/L.