Calculate the solubility (in grams per 1.00 102 mL of solution) of magnesium hydroxide in a solution buffered at pH = 9
the ksp of Mg(oh)2 = 2.06e-13
To calculate the solubility of magnesium hydroxide (Mg(OH)2) in a pH 9 buffered solution, we need to use the concept of the solubility product constant (Ksp). The Ksp of Mg(OH)2 is given as 2.06 × 10^-13.
First, we need to set up the balanced chemical equation for the dissociation of Mg(OH)2:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)
The Ksp expression for this reaction is:
Ksp = [Mg2+][OH-]^2
Since the solution is buffered at pH 9, it means there is a significant amount of OH- ions present. At pH 9, the concentration of OH- is 10^-(14-pOH) = 10^-(14-9) = 10^-5.
Let's assume the molar solubility of Mg(OH)2 is 'x'. Therefore, the concentration of Mg2+ and OH- ions formed is also 'x'.
Substituting these values into the Ksp expression:
2.06 × 10^-13 = x(x)^2
2.06 × 10^-13 = x^3
Now, solve for 'x' by taking the cubic root of both sides:
x = (2.06 × 10^-13)^(1/3)
x ≈ 1.295 × 10^-5 M
Finally, to calculate the solubility in grams per 1.00 × 10^2 mL of solution, we need to convert the molar solubility to grams per volume.
Molar mass of Mg(OH)2 = (24.31 g/mol + 2(16.00 g/mol + 1.01 g/mol)) = 58.33 g/mol
Converting to grams per volume:
Solubility = (1.295 × 10^-5 M) × (58.33 g/mol) × (1.00 × 10^2 mL) × (1 L / 1000 mL)
Solubility ≈ 7.56 × 10^-5 g/1.00 × 10^2 mL
Therefore, the solubility of magnesium hydroxide in a solution buffered at pH 9 is approximately 7.56 × 10^-5 grams per 1.00 × 10^2 mL.
To calculate the solubility of magnesium hydroxide (Mg(OH)2) in a solution buffered at pH = 9, we need to consider the hydrolysis of Mg(OH)2 in water. The hydrolysis reaction can be represented as follows:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)
The solubility of Mg(OH)2 can be determined using the solubility product (Ksp) expression, which is given as:
Ksp = [Mg2+][OH-]^2
Given that the Ksp of Mg(OH)2 is 2.06e-13, we can set up an equilibrium expression:
2.06e-13 = [Mg2+][OH-]^2
Since the solution is buffered at pH = 9, we can assume that the concentration of OH- ions is equal to the concentration of H+ ions, as water undergoes self-ionization to produce equal amounts of H+ and OH- ions.
Using the equation Kw = [H+][OH-], where Kw is the ion product of water (~1.0e-14 at 25°C), and assuming that the solution is close to neutral, we can approximate [H+] as 1.0e-7 M.
Therefore:
[H+] ≈ [OH-] ≈ 1.0e-7 M
Substituting this concentration into the Ksp expression, we have:
2.06e-13 = [Mg2+](1.0e-7)^2
Simplifying, we get:
2.06e-13 = [Mg2+](1.0e-14)
Now, solve for [Mg2+]:
[Mg2+] = (2.06e-13)/(1.0e-14) = 2.06e1
So, the concentration (or solubility) of Mg(OH)2 in the solution is approximately 2.06 × 101 g/L.
To convert this to grams per 1.00 × 102 mL, we need to multiply the solubility by the volume conversion factor:
Solubility (g/1.00 × 102 mL) = 2.06 × 101 g/L × (1.00 × 102 mL / 1 L)
Simplifying, we find:
Solubility (g/1.00 × 102 mL) = 2.06 × 103 g/100 mL
Therefore, the solubility of magnesium hydroxide in the solution buffered at pH = 9 is approximately 2.06 × 103 grams per 1.00 × 102 mL of solution.
........Mg(OH)2 ==> Mg^2+ + 2OH^-
E........x...........x.......2x
Ksp = 2.06E-13 = (Mg^2+)(OH^-)
For (Mg^+) substitute x
For (OH^-)^2 substitute (2x + 1E-5)^2 [note: 2x for the Mg(OH)2 and 1E-5 for the OH^- concn of a soln with pH = 9
Solve for x = solubility in mols/L, then correct that to mols/100 mL, then to g/100 mL.