The Ksp of mercury II hydroxide is 3.60 x 10^-26. Calculte the solubility of this compound in g/L
2.08*10^-9
4.86*10^-7 is the real answer
Hg(OH)2 dissociates in water to a small extent.
Hg(OH)2(s) ↽−−⇀ Hg2+(aq) + 2OH−(aq)
initial excess 0 0
change −𝑥 +𝑥 +2𝑥
equilibrium 𝑥 𝑥
The equilibrium constant expression for 𝐾sp is
𝐾sp=[Hg2+][OH−]2=(𝑥)(2𝑥)2=4𝑥3
Substitute the value of 𝐾sp ( 3.60×10−26 ) into the expression and solve for the molar solubility, 𝑥 .
3.60×10−26=4𝑥3
𝑥3=3.60×10−264
x=3.60×10−264⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3=2.08×10−9 M
Convert the solubility from molarity (moles per liter) to grams per liter using the molar mass of Hg(OH)2.
2.08×10−9 mol1 L×234.61 g1 mol=4.88×10−7 gL
4.86*10^7 is correct. I also go it right on Sapling. x=2.09*10^-9 multiply by Molar Mass 234.605
4.88*10^7
To calculate the solubility of a compound in grams per liter (g/L), you will need to use the solubility product constant (Ksp). The Ksp is a measure of the equilibrium between the dissolved ions in a saturated solution of a sparingly soluble salt.
The molar solubility (in moles per liter) of the compound can be calculated using the following equation:
Ksp = [M+]^m [OH-]^n
Where [M+] is the concentration of the cation (in moles per liter), [OH-] is the concentration of the hydroxide anion (in moles per liter), and m and n are the coefficients of the balanced equation for the dissolving of the compound.
In the case of mercury II hydroxide (Hg(OH)2), the compound dissociates into one mercury cation (Hg2+) and two hydroxide anions (OH-):
Hg(OH)2 ⇌ Hg2+ + 2OH-
Since the coefficient for Hg2+ is 1 and for OH- is 2, the equation becomes:
Ksp = [Hg2+][OH-]^2
Now, we can rearrange the equation to solve for the molar solubility of Hg(OH)2:
[Hg2+] = Ksp / [OH-]^2
To find the solubility in grams per liter, you'll need the molar mass of Hg(OH)2, which is 130.63 g/mol.
First, calculate the concentration of OH- ions using the square root of Ksp:
[OH-] = √(Ksp)
Next, calculate the molar solubility of Hg2+:
[Hg2+] = Ksp / [OH-]^2
Finally, calculate the solubility in grams per liter:
Solubility (g/L) = [Hg2+] x molar mass of Hg(OH)2
Let's calculate it step-by-step:
Step 1: Calculate [OH-]
[OH-] = √(3.60 x 10^-26) ≈ 6.00 x 10^-14 M
Step 2: Calculate [Hg2+]
[Hg2+] = (3.60 x 10^-26) / (6.00 x 10^-14)^2 ≈ 1.0 x 10^-4 M
Step 3: Calculate solubility in g/L
Solubility (g/L) = (1.0 x 10^-4 M) x (130.63 g/mol) ≈ 0.013 g/L
Therefore, the solubility of mercury II hydroxide (Hg(OH)2) is approximately 0.013 grams per liter (g/L).
...........Hg(OH)2 ==> Hg^2+ + 2OH^-
I..........solid.......0........0
C............-x........x........2x
E..........solid.......x........2x
x = solubility
Ksp = (Hg^2+)(OH^-)^2
Substitute the E line and solve for x = solubility in mols/L, then convert to g/L.