The ksp for Al(OH)3 is 3.7*-5 at 20 degrees. What is the solubility of Al(OH)3 in grams per litre?
I don't think Ksp is correct. Can't be that large. Also, how advanced is this class?
That is the Ksp I was given
This won't be the correct answer to the problem (but it is the correct answer to the problem that is posted).
.......Al(OH)3 ==> Al^3+ + 3OH^-
I.......solid......0.........0
C.......solid......x.........x
E.......solid......x.........x
Substitute the E line into Ksp and solve for x in mols/L, then convert to grams/L.
BTW, I think that exponent is 10^-15 and not 10^-5
so x=6.08*10-3 and then I use the gram equivalent weight of Al(OH)3 to get grams per Litre? I got 0.474 g per litre, is the correct?
Sorry, I may have misled you due to a typo I didn't put in. That equilibrium should have looked like this;
.......Al(OH)3 ==> Al^3+ + 3OH^-
I.......solid......0.........0
C.......solid......x.........3x
E.......solid......x.........3x
3.7E-5 = (x)(3x)^3
3.7E-5 = 27x^4
x = approx 0.03 but you can do it more accurately.
The answer comes out in mols/L, then you multiply by gram MOLECULAR weight (not gram equivalent weight) to get grams/L. I obtained approx 3 g/L but that was a quickie and you should confirm that. It is an estimate.
To find the solubility of Al(OH)3 in grams per liter, we need to use the solubility product constant (Ksp) and the molar mass of Al(OH)3.
First, let's write the balanced equation for the dissociation of Al(OH)3:
Al(OH)3 (s) ⇌ Al3+ (aq) + 3OH- (aq)
From the balanced equation, we can see that one mole of Al(OH)3 dissociates to give one mole of Al3+ ions and three moles of OH- ions.
The expression for the solubility product constant (Ksp) is given by:
Ksp = [Al3+][OH-]^3
Since the stoichiometry of Al3+ and OH- in the equation is 1:3, we can substitute [OH-]^3 with [Al(OH)3]^3:
Ksp = [Al3+][Al(OH)3]^3
Given the value of Ksp for Al(OH)3 is 3.7 × 10^-5, we can rewrite the equation as:
3.7 × 10^-5 = [Al3+][Al(OH)3]^3
To find the solubility of Al(OH)3, we need to calculate the concentration of Al3+ ions in the solution, as well as the molar mass of Al(OH)3.
The molar mass of Al(OH)3 is:
Molar mass = 26.98 g/mol (for Al) + 3(16.00 g/mol) (for O) + 3(1.01 g/mol) (for H)
Molar mass = 78.00 g/mol
Now, we need to rearrange the equation for Al(OH)3 solubility to solve for [Al3+]:
[Al3+] = (3.7 × 10^-5 / [Al(OH)3]^3)^(1/4)
Next, we need to convert moles of Al3+ to grams of Al(OH)3. This can be done by multiplying the Al3+ concentration by the molar mass of Al(OH)3:
Solubility of Al(OH)3 (g/L) = [Al3+] × Molar mass of Al(OH)3
Once we substitute the values into the equation, we can solve for the solubility of Al(OH)3 in grams per liter.