4) Calculate the solubility of calcium hydroxide, Ca(OH)2 (Ksp = 5.5 x 10-5) in grams per liter in:
a. Pure water
b. 0.10 M CaCl2
In pure H2O.
........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 Ksp expression and solve for x = solubility in mols/L and convert to g/L.
In 0.10M CaCl2 which ionizes 100%.
Same as above except CaCl2 must be added in as follows:
.........CaCl2 = Ca^2+ + 2Cl^-
I.........0.1....0.......0
C........-0.1...0.1......0.1
E.........0......0.1.....0.1
Ksp still = (Ca^2+)(OH^-)
(Ca^2+) = 0.1 from CaCl2 and x from Ca(OH)2 for total of 0.1+x
(OH^-) = 2x
Substitute into Ksp expression and solve for x = solubility in M = mols/L and convert to g/L.
grams = mols x molar mass.
0.027
To calculate the solubility of calcium hydroxide (Ca(OH)2) in grams per liter, we need to use the solubility product constant (Ksp) and the given conditions.
The Ksp value for calcium hydroxide is given as 5.5 x 10^-5.
a. Pure water:
In pure water, calcium hydroxide will dissociate into calcium ions (Ca^2+) and hydroxide ions (OH^-). Let's assume the solubility of calcium hydroxide is "x" mol/L.
Ca(OH)2 ⇌ Ca^2+ + 2OH^-
Using the Ksp expression:
Ksp = [Ca^2+][OH^-]^2
5.5 x 10^-5 = (x)(2x)^2
5.5 x 10^-5 = 4x^3
Solving for "x":
(4x^3) = 5.5 x 10^-5
x^3 = 5.5 x 10^-5 / 4
x^3 = 1.375 x 10^-5
Taking the cube root of both sides:
x = (1.375 x 10^-5)^(1/3)
x ≈ 0.034 M
Since the molar mass of calcium hydroxide is 74.093 g/mol, we can convert the solubility from molarity to grams per liter:
0.034 mol/L x 74.093 g/mol = 2.52 g/L
Therefore, the solubility of calcium hydroxide in pure water is approximately 2.52 grams per liter.
b. 0.10 M CaCl2:
When calcium hydroxide is in the presence of 0.10 M CaCl2, it will form calcium chloride (CaCl2) and water (H2O). This will decrease the amount of calcium hydroxide that dissolves in water.
Ca(OH)2 + CaCl2 ⇌ 2CaCl + 2H2O
The presence of calcium chloride will shift the equilibrium to the left, reducing the solubility of calcium hydroxide. As a result, the solubility of calcium hydroxide in 0.10 M CaCl2 will be lower than in pure water.
Unfortunately, without the specific value of the common ion effect constant for calcium hydroxide and calcium chloride, we cannot determine the exact solubility.
To calculate the solubility of calcium hydroxide (Ca(OH)2) in grams per liter, we need to use the solubility product equilibrium expression (Ksp). The Ksp value provided for calcium hydroxide is 5.5 x 10^-5.
a. Solubility of Calcium Hydroxide in Pure Water:
When calcium hydroxide is dissolved in pure water, it completely dissociates into calcium ions (Ca^2+) and hydroxide ions (OH^-).
The balanced equation for the dissociation of calcium hydroxide is:
Ca(OH)2(s) ⟶ Ca^2+(aq) + 2OH^-(aq)
Using the stoichiometry of the balanced equation, we can determine that the molar solubility of calcium hydroxide is 's' in moles per liter.
The Ksp expression for calcium hydroxide is:
Ksp = [Ca^2+][OH^-]^2
Since calcium hydroxide dissociates into Ca^2+ and 2OH^-, the expression becomes:
Ksp = [Ca^2+][OH^-]^2 = s * (2s)^2 = 4s^3
Substituting the given Ksp value of 5.5 x 10^-5:
5.5 x 10^-5 = 4s^3
Now, solve for 's' by taking the cube root and dividing by 4:
s = (5.5 x 10^-5)^(1/3) / 4
Calculating the value:
s ≈ 6.509 x 10^-3 moles per liter
To convert the molar solubility into grams per liter, we need to multiply it by the molar mass of calcium hydroxide.
The molar mass of calcium hydroxide (Ca(OH)2) is:
Ca: 1 atom * 40.08 g/mol = 40.08 g/mol
O: 2 atoms * 16.00 g/mol = 32.00 g/mol
H: 2 atoms * 1.01 g/mol = 2.02 g/mol
Total molar mass = 40.08 + 32.00 + 2.02 = 74.10 g/mol
Multiplying the molar solubility by the molar mass:
Solubility in grams per liter = (6.509 x 10^-3 moles/L) * (74.10 g/mol)
Therefore, the solubility of calcium hydroxide in pure water is approximately 0.482 g/L.
b. Solubility of Calcium Hydroxide in 0.10 M CaCl2:
When calcium hydroxide is dissolved in a solution of 0.10 M CaCl2, the concentration of Ca^2+ ions is already present. This affects the solubility of calcium hydroxide.
The balanced chemical equation for calcium chloride (CaCl2) dissociation is:
CaCl2(s) ⟶ Ca^2+(aq) + 2Cl^-(aq)
To determine the solubility in this situation, we will compare the reaction quotient (Q) to the Ksp value.
Let 'x' represent the solubility of calcium hydroxide in moles per liter.
The concentration of Ca^2+ ions in the solution is already 0.10 M, which means the concentration of Ca^2+ ions in the reaction is 0.10 M + x.
The concentration of OH^- ions is 2x because calcium hydroxide dissociates into 1 Ca^2+ ion and 2 OH^- ions.
The reaction quotient (Q) is given by:
Q = [Ca^2+][OH^-]^2 = (0.10 M + x)(2x)^2 = 4x^3 + 0.4x^2 (since 0.10 M is significantly smaller than x)
Comparing Q to Ksp:
Q ≤ Ksp
4x^3 + 0.4x^2 ≤ 5.5 x 10^-5
Since x is much smaller compared to 0.1, we can make an approximation:
0.4x^2 ≤ 5.5 x 10^-5
0.4x^2 ≈ 5.5 x 10^-5
Solving for x, we get:
x ≈ (5.5 x 10^-5 / 0.4)^(1/2)
Calculating the value:
x ≈ 1.68 x 10^-3 moles per liter
Converting the molar solubility into grams per liter, assuming the same molar mass as before (74.10 g/mol):
Solubility in grams per liter = (1.68 x 10^-3 moles/L) * (74.10 g/mol)
Therefore, the solubility of calcium hydroxide in a 0.10 M CaCl2 solution is approximately 0.124 g/L.