he Ksp of mercury(II) hydroxide, Hg(OH)2, is 3.60 × 10-26. Calculate the solubility of this compound in g/L.
I came up with x=2.8*10^-13
I don't think so. Post your work and I'll find the error. And you need to define x.
To calculate the solubility of mercury(II) hydroxide, Hg(OH)2, in g/L using its Ksp value, you will need to follow a few steps.
Step 1: Write the balanced chemical equation for the dissociation of Hg(OH)2.
Hg(OH)2 ⇌ Hg2+ + 2OH-
Step 2: Set up the expression for the solubility product constant, Ksp.
Ksp = [Hg2+][OH-]^2
Step 3: Since the stoichiometric ratio between Hg2+ and OH- is 1:2, and assuming x represents the concentration of Hg2+ and 2x represents the concentration of OH- in the solution, substitute these values into the Ksp expression.
Ksp = (x)(2x)^2 = 3.60 × 10^-26
Step 4: Simplify the equation and solve for x.
4x^3 = 3.60 × 10^-26
Step 5: Take the cubic root of both sides to isolate x.
x = (3.60 × 10^-26)^(1/3)
Calculating this expression should give you the concentration of Hg2+ in the solution.
Step 6: To convert the concentration from moles per liter to grams per liter, you need to multiply by the molar mass of Hg(OH)2.
The molar mass of Hg(OH)2 is:
Atomic mass of Hg = 200.59 g/mol
Atomic mass of O = 16.00 g/mol
Atomic mass of H = 1.01 g/mol
So, molar mass of Hg(OH)2 = (200.59 + 2*16.00 + 2*1.01) g/mol = 234.62 g/mol
Step 7: Multiply your calculated concentration (x) by the molar mass of Hg(OH)2 to get the solubility in grams per liter.
solubility = x * molar mass of Hg(OH)2
By following these steps, you should be able to determine the solubility of mercury(II) hydroxide, Hg(OH)2, in g/L.