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)
To find the number of ppm (parts per million) of Al(total) in equilibrium with gibbsite, we need to use the information given in the question and apply the principles of complex formation and equilibrium chemistry.
First, let's understand the given information and what it represents:
- Al^3+ forms a complex AlSO4+ in the presence of sulfate (SO4^2-): This means that Al^3+ ions bind with SO4^2- ions to form AlSO4+ complexes.
- Kstab = 10^3.01: This is the stability constant for the formation of the AlSO4+ complex. It represents the degree to which the complex is formed in equilibrium. Higher values of Kstab indicate higher formation of the complex.
To solve the problem, we can set up an equilibrium expression for the formation of the AlSO4+ complex:
[AlSO4+] = [Al^3+][SO4^2-]/Kstab
In this expression, [AlSO4+] represents the concentration (or activity, since activities are assumed to be equal to concentrations) of AlSO4+ complex, [Al^3+] represents the concentration of Al^3+ ions, [SO4^2-] represents the concentration of SO4^2- ions, and Kstab is the stability constant.
Now, let's apply this equilibrium expression to find the concentration of Al^3+ ions in equilibrium with gibbsite.
In the presence of 10^-3 M SO4^2- (as given in the question), we can assume that the concentration/activity of SO4^2- ions is equal to their initial concentration, which is 10^-3 M.
Let's assume the concentration of Al^3+ ions in equilibrium is x M. Therefore, the concentration of AlSO4+ complex would also be x M, considering their stoichiometry as 1:1.
Using the equilibrium expression, we can substitute the given values and solve for the concentration of Al^3+ ions in equilibrium:
x = (Al^3+)(SO4^2-)/Kstab
x = (x)(10^-3)/(10^3.01)
Simplifying the equation:
x^2 = 10^-3 * 10^(-3.01)
x^2 = 10^(-6.01)
x = 10^(-6.01/2) (taking the square root of both sides)
x ≈ 10^(-3.006)
Since the concentration of Al^3+ ions is in M (moles per liter), we can convert this value to ppm:
ppm = (10^(-3.006)) * (10^6)
ppm ≈ 10^2.994
ppm ≈ 786
Therefore, approximately 786 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-.