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 determine the concentration of Al(total) in equilibrium with gibbsite at pH 4 and 25 degrees Celsius, we need to use the given equilibrium constant (Kstab).
First, we need to understand the species present for Al in solution at pH 4. In this case, gibbsite (Al(OH)3) is the dominant species. At pH 4, the hydrolysis of Al^3+ occurs, and it forms Al(OH)^2+ and Al(OH)2+. However, for simplicity, we will consider Al(OH)3 as the dominant species.
The equation for the formation of the complex AlSO4+ is:
Al^3+ + SO4^2- ⇌ AlSO4+
The equilibrium constant, Kstab, has a value of 10^3.01, which means the reaction favors the formation of the complex.
Now, let's define the concentration of Al(total) as [Al(total)].
Since AlSO4+ is the only complex considered, we can assume the concentration of AlSO4+ is equal to [Al(total)]. Thus, [AlSO4+] = [Al(total)].
The equation for the equilibrium expression is:
Kstab = [AlSO4+]/([Al^3+] * [SO4^2-])
At equilibrium, the concentrations can be related as:
[Al(total)] = [Al^3+] (since [Al(total)] = [AlSO4+])
[SO4^2-] = 10^-3 M (given)
Substituting the known values into the equilibrium expression, we get:
10^3.01 = [Al(total)] / ([Al^3+] * 10^-3)
Rearranging the equation, we have:
[Al(total)] = 10^3.01 * [Al^3+] * 10^-3
Simplifying further, we find:
[Al(total)] = 10^0.01 * [Al^3+]
Now, we need to find the concentration of Al^3+ in equilibrium with gibbsite at pH 4. For this, we can use a solubility equilibrium expression for gibbsite:
Al(OH)3 ⇌ Al^3+ + 3OH^-
Ksp = [Al^3+] * [OH^-]^3
At pH 4, the concentration of OH^- can be determined from the pH:
pOH = 14 - pH = 14 - 4 = 10
[OH^-] = 10^-pOH = 10^-10
Substituting the value of [OH^-] into the solubility equilibrium expression:
Ksp = [Al^3+] * (10^-10)^3
Ksp = [Al^3+] * 10^-30
Rearranging the equation, we get:
[Al^3+] = Ksp / 10^-30
Substituting the value of Ksp and simplifying, we find:
[Al^3+] = 10^-30 / 10^-3
Further simplifying:
[Al^3+] = 10^-27
Now, we can substitute this value into the equation for [Al(total)]:
[Al(total)] = 10^0.01 * [Al^3+]
[Al(total)] = 10^0.01 * 10^-27
Simplifying further, we find:
[Al(total)] = 10^-26 M
To convert this concentration to ppm, we can use the fact that 1 ppm is equal to 1 mg/L or 10^-6 g/L. The molar mass of Al is approximately 27 g/mol.
[Al(total)] = 10^-26 M
= 10^-26 mol/L
= 10^-26 * 27 g/L
= 2.7 * 10^-25 g/L
= 2.7 * 10^-22 mg/L
= 2.7 * 10^-22 ppm
Therefore, the concentration of Al(total) in equilibrium with gibbsite at pH 4 and 25 degrees Celsius in the presence of 10^-3 M SO4^2- is approximately 2.7 * 10^-22 ppm.