Calculate the solubility (in grams per 1.00 102 mL of solution) of magnesium hydroxide in a solution buffered at pH = 9. Enter your answer to 2 significant figures.) Ksp = 2.06e-13
Yes, the OH^- is squared.
2.06E-13 = (Mg^2+)(1E-5)^2
Solve for Mg^2+. I see I made a typo and omitted the square of OH on my first response. It is
Ksp = (Mg^2+)(OH^-)^2 etc. Sorry about that.
pH = 9 means pOH = 5 and (OH^-) = 1E-5
........Mg(OH)2 ==> Mg^2+ + 2OH^-
Ksp = (Mg^+)(OH^-) = 2.06E-13
Substitute 1E-5 for OH and solve for Mg. That gives solubility in mols/L. Convert to mols/100 mL, the convert mols to grams.
does the 2 in front of the OH^- get corporate into the Ksp making it (OH)^2??
To calculate the solubility of magnesium hydroxide (Mg(OH)2) in a solution buffered at pH = 9, we can make use of the concept of pH and the solubility product constant (Ksp).
1. First, write the balanced equation for the dissociation of magnesium hydroxide:
Mg(OH)2 ⇌ Mg2+ + 2OH-
We can see that for every one molecule of magnesium hydroxide that dissolves, it produces one magnesium ion (Mg2+) and two hydroxide ions (OH-).
2. Next, recall that the pH of a solution is related to the concentration of hydroxide ions (OH-) in the solution. In this case, we are given that the solution is buffered at pH = 9.
pH is defined as the negative logarithm (base 10) of the concentration of hydroxide ions:
pOH = -log[OH-]
In the case of a pH = 9 solution, we can find the pOH as follows:
pOH = 14 - pH = 14 - 9 = 5
3. Now, convert the pOH value to the concentration of hydroxide ions (OH-) using the formula:
[OH-] = 10^(-pOH)
[OH-] = 10^(-5) = 1.00 × 10^(-5)
4. Since the stoichiometry of the balanced equation tells us that for every two hydroxide ions (OH-), one molecule of magnesium hydroxide (Mg(OH)2) dissolves, the concentration of Mg2+ ions in the solution will be half of the OH- concentration.
[Mg2+] = 1/2 × [OH-] = 1/2 × 1.00 × 10^(-5) = 5.00 × 10^(-6)
5. Finally, we can use the solubility product constant (Ksp) to determine the solubility of Mg(OH)2. The Ksp expression for magnesium hydroxide is:
Ksp = [Mg2+][OH-]^2
Since we have the concentration of Mg2+ and OH-, we can rearrange the equation to solve for the solubility (S) of Mg(OH)2:
S = [Mg2+] = Ksp / [OH-]^2 = 2.06 × 10^(-13) / (1.00 × 10^(-5))^2
S = 2.06 × 10^(-13) / 1.00 × 10^(-10) = 2.06 × 10^(-3) grams per 1.00 × 10^2 mL of solution (to 2 significant figures)
Therefore, the solubility of magnesium hydroxide in a solution buffered at pH = 9 is 2.06 × 10^(-3) grams per 1.00 × 10^2 mL of solution.