Given:

Al^3+ forms a complex ALSO4+ in the presence of sulfate...
Al^3+ + SO4^2- = AlSO4+

with Kstab = 10^3.01

Question: How many ppm Al(total) would be in equilibrium with gibbsite at pH 4 and 25 degrees celsius in the presence of 10^3- m
SO4^2- (consider only this one complex and assume activities = concentrations)

Someone responded and said to look at a phase diagram but the problem is not set up for that...the answer is 3 ppm but need help with the set-up. The textbook offered very little guidance with complexes. Can someone please help. I have another problem similiar to this one but can't do that one either until I can get some feedback on this one...please...

To solve this problem, you need to use the concept of chemical equilibrium and the equation you provided. Here's a step-by-step guide on how to set up the problem:

Step 1: Write the chemical reaction equation for the formation of the complex:
Al^3+ + SO4^2- = AlSO4+

Step 2: Write the expression for the equilibrium constant (Kstab):
Kstab = [AlSO4+] / [Al^3+][SO4^2-]

Step 3: Use the given value of Kstab (10^3.01) to set up the equation:
10^3.01 = [AlSO4+] / [Al^3+][SO4^2-]

Step 4: Determine the concentration of Al^3+ and SO4^2- by converting the given molarity to concentration (assuming they are the same due to activity = concentration):
[Al^3+] = 10^-3 M
[SO4^2-] = 10^-3 M

Step 5: Substitute the concentrations into the equation:
10^3.01 = [AlSO4+] / (10^-3)(10^-3)
10^3.01 = [AlSO4+] / 10^-6
[AlSO4+] = (10^3.01)(10^-6)

Step 6: Calculate the concentration of AlSO4+:
[AlSO4+] = 10^3.01 × 10^-6
[AlSO4+] = 10^3.01-6
[AlSO4+] = 10^-2.99 M

Step 7: Convert the concentration of AlSO4+ to ppm:
1 ppm = 1 mg/L = 1 mg/1000 mL
So, [AlSO4+] (ppm) = 10^-2.99 M * (26.98 g/mol) * (1000 mg/g)

Calculating the above value gives [AlSO4+] (ppm) ≈ 30.36 ppm.

Therefore, in equilibrium with gibbsite at pH 4 and 25 degrees Celsius in the presence of 10^-3 M SO4^2-, the concentration of aluminum as Al(total) would be approximately 30.36 ppm.