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?
5 answers
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
.......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.
.......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.
5 answers