Question

What is the molar solubility of Cr(OH)3 in pure water? The Ksp for chromium (III) hydroxide is 6.3 x 10-31.
Answered by Bot
To calculate the molar solubility of Cr(OH)3 in pure water, we need to use the Ksp expression and solve for the concentration of the compound.

The Ksp expression for Cr(OH)3 is:
Ksp = [Cr3+][OH-]^3

Since Cr(OH)3 dissociates into one Cr3+ ion and three OH- ions, we can write:
Ksp = x * (3x)^3
Ksp = 6.3 x 10^-31

Simplifying the equation:
Ksp = 27x^4
6.3 x 10^-31 = 27x^4

Rearranging the equation:
x^4 = (6.3 x 10^-31) / 27

Taking the fourth root on both sides:
x = (6.3 x 10^-31 / 27)^(1/4)

Calculating the value on the right side of the equation:
x ≈ 9.17 x 10^-8 M

Therefore, the molar solubility of Cr(OH)3 in pure water is approximately 9.17 x 10^-8 M.
Answered by DrBob
Bot, did you make a math error?
Answered by Bot
Apologies for that, I made an error in my calculation. Let's correct it:

Ksp = [Cr3+][OH-]^3
Ksp = x * (3x)^3
Ksp = 27x^4

6.3 x 10^-31 = 27x^4

x^4 = (6.3 x 10^-31) / 27
x^4 ≈ 7.6 x 10^-33

Taking the fourth root on both sides:
x ≈ (7.6 x 10^-33)^(1/4)

Calculating the value on the right side of the equation:
x ≈ 7.85 x 10^-9 M

Therefore, the correct molar solubility of Cr(OH)3 in pure water is approximately 7.85 x 10^-9 M.
Answered by DrBob
I think you made another math error
Answered by Bot
Apologies for the confusion. Let's recalculate the molar solubility correctly:

We have the Ksp expression for Cr(OH)3:
Ksp = [Cr3+][OH-]^3

Substituting the given Ksp value:
6.3 x 10^-31 = [Cr3+][OH-]^3

Since Cr(OH)3 dissociates into one Cr3+ ion and three OH- ions, we can write:
Ksp = x * (3x)^3
6.3 x 10^-31 = x * (27x^3)
6.3 x 10^-31 = 27x^4

Solving for x:
x^4 = (6.3 x 10^-31) / 27
x^4 ≈ 2.33 x 10^-33

Taking the fourth root on both sides:
x ≈ (2.33 x 10^-33)^(1/4)

Calculating the value on the right side of the equation:
x ≈ 2.18 x 10^-9 M

Therefore, the correct molar solubility of Cr(OH)3 in pure water is approximately 2.18 x 10^-9 M.
Answered by DrBob
I think your corrected answer still is incorrect. Bot, check out my response below.
Do you agree or disagree. Starting with your equation of
6.3 x 10^-31 = 27x^4
x^4 = 6.3E-31/27 = 2.33E-32
x = (2.33E-32)^0.25 = 1.24E-8 M
Answered by Bot
Apologies for the error in my previous responses. Yes, I agree with your corrected calculations:

Starting with the equation:
x^4 = 6.3 x 10^-31 / 27
x^4 = 2.33 x 10^-32

Taking the fourth root on both sides:
x ≈ (2.33 x 10^-32)^(1/4)

Calculating the value on the right side of the equation:
x ≈ 1.24 x 10^-8 M

Therefore, the correct molar solubility of Cr(OH)3 in pure water is approximately 1.24 x 10^-8 M. Thank you for catching that error.