The ksp of aluminum hydroxide is 5.6e-17. What is the concentration of hydroxide in a saturated solution?

For the answer I got
[AlOH] <=> [Al] [OH]^2
5.6e-17 x 3x
5.6e-17 = 3x^4
x = 6.6e-5

Is this right? I'm not sure about 3x^4 part...

yep its right

No it isn't right.

Al(OH)3 ==> Al^3+ + 3OH^-
Ksp = (Al^3+)(OH^-) = 5.6E-17
(Al^3+) = x
(OH^-) = 3x
(x)(3x)^3 = 5.6E-17
27x^4 = 5.6E-17
Solve for x, then multiply that by 3 for OH^-

Thank you !!

To calculate the concentration of hydroxide in a saturated solution of aluminum hydroxide (Al(OH)3), we first need to write the balanced equation for the dissociation of aluminum hydroxide:

Al(OH)3(s) ⇌ Al3+(aq) + 3OH-(aq)

From the equation, we can see that the concentration of hydroxide is three times the concentration of aluminum ion. Let's assume the concentration of aluminum ion is represented by x. So, the concentration of hydroxide will be 3x.

Now, we can write the expression for the solubility product constant (Ksp) of aluminum hydroxide:

Ksp = [Al3+][OH-]^3

Since the balanced equation shows that the concentration of hydroxide is three times the concentration of aluminum, we can rewrite the expression for Ksp as:

Ksp = [Al3+][(3x)^3]

Given that the value of Ksp is 5.6e-17, we can substitute this value into the equation:

5.6e-17 = x[27x^3]

Simplifying the equation, we have:

5.6e-17 = 27x^4

To solve for x, we need to isolate it. Divide both sides of the equation by 27:

(5.6e-17) / 27 = x^4

x^4 = 2.074e-18

To find x, we need to take the fourth root of both sides:

x = (2.074e-18)^(1/4)

Calculating this expression, we find:

x ≈ 6.56e-5

Therefore, the concentration of hydroxide in a saturated solution of aluminum hydroxide is approximately 6.56e-5 M.