Al^3+ forms a complex ALSO4+ in the presence of sulfate...
Al^3+ + SO4^2- = AlSO4+
with Kstab = 10^3.01
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)
Am having a tough time with complexes, hope you can help. Thanks.
To calculate the concentration of Al^3+ in equilibrium with gibbsite at pH 4 and 25 degrees Celsius in the presence of 10^-3 M SO4^2-, you need to consider the equilibrium constant (Kstab) and the speciation of aluminum in solution.
Step 1: Write down the balanced chemical equation for the reaction between Al^3+ and SO4^2- to form AlSO4+:
Al^3+ + SO4^2- = AlSO4+
Step 2: Write down the expression for the equilibrium constant (Kstab):
Kstab = [AlSO4+]/[Al^3+][SO4^2-]
Given that Kstab = 10^3.01, the ratio [AlSO4+]/[Al^3+][SO4^2-] is equal to 10^3.01.
Step 3: Assume that the concentration of AlSO4+ is equal to x, and the concentration of Al^3+ is also equal to x, since AlSO4+ is a 1:1 complex.
Now, the concentration of [SO4^2-] is given as 10^-3 M. Therefore, the expression becomes:
Kstab = x/(x * 10^-3)
Simplifying the equation:
10^3.01 = 1/(10^-3)
10^3.01 = 10^3
Since the exponents are equal, x = 10^-3 M.
Step 4: Convert the concentration to parts per million (ppm):
ppm = (10^-3 M) * (10^6 ppm/1 M) = 1000 ppm
Therefore, the concentration of Al(total) in equilibrium with gibbsite in the presence of 10^-3 M SO4^2- at pH 4 and 25 degrees Celsius is 1000 ppm.