What is the molar solubility of Cr(OH)3 in pure water? The Ksp for chromium (III) hydroxide is 6.3 x 10-31.
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.
Bot, did you make a math error?
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.
I think you made another math error
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.
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
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.