Calulate the molar solubility of Al(OH)3 in water. (Ksp = 3 x 10^-34)
Al(OH)3(s) ==>Al^+3 + 3OH^-
Ksp = (Al^+3)(OH^-)^3 = 3E-34
Let x = molar solubility of Al(OH)3
then x molar = (Al^+3)
and 3x molar = (OH^-)
Plug in to Ksp and solve for x. Don't forget to cube the 3x.
To calculate the molar solubility of Al(OH)3 in water, we can start by setting up the equation using the given solubility product constant (Ksp) expression:
Ksp = (Al^+3)(OH^-)^3 = 3 x 10^-34
Let's assume that the molar solubility of Al(OH)3 is x. This means that the concentration of Al^+3 would be equal to x, and the concentration of OH^- would be equal to 3x since there are three OH^- ions produced for every one Al^+3 ion.
Substituting these values into the Ksp expression, we get:
(Al^+3)(OH^-)^3 = (x)(3x)^3 = 3x^4 = 3 x 10^-34
To solve for x, we need to isolate it. Dividing both sides of the equation by 3 gives:
x^4 = 10^-34
To eliminate the fourth power, we can take the fourth root of both sides:
x = (10^-34)^(1/4)
Simplifying,
x ≈ 3.98 x 10^-9 M
Therefore, the molar solubility of Al(OH)3 in water is approximately 3.98 x 10^-9 M.
To solve for the molar solubility (x) of Al(OH)3 in water, we will use the given Ksp expression:
Ksp = (Al^+3)(OH^-)^3 = 3E-34
Let's substitute the molar solubility (x) into the equation:
Ksp = (x)(3x)^3
Simplifying the expression:
3E-34 = 27x^4
Now, let's solve for x:
x^4 = (3E-34) / 27
x^4 = 1E-36
To find the value of x, take the fourth root of both sides:
x = (1E-36)^(1/4)
Calculating the value:
x ≈ 0.031
Therefore, the molar solubility of Al(OH)3 in water is approximately 0.031 mol/L.