"Calculate the pH of 0.10 mol/L aqueous solution of aluminum sulfate."
I am also given the ka value for the aluminum ion (9.8x10^-6)...but I don't know why I need it. I'm not even sure if I need to use it. Help please?
Oops! Sorry, I figured it out. But I still need help...I'm not clear on the process.
Does it give you the equation for the Al hydrated ion. Something like
Al(H2O)6+3 ==> Al(H2O)5(OH^-)^+2 + H^+
So you st up and ICE chart and solve for H^+ and convert to pH.
Sorry, I have a question within a question. I think the ka value for the aluminum ion is used for when we have a solution like AlCl3, right? When you dissociate it, the Cl ion has a very large ka value...so we use the ka for Al instead? I hope I'm making sense. :)
No, it doesn't. But I have it in my notes. Ok, I think I understand how to approach this. Thanks!
Is the initial concentration for Al(H20)6 also 0.10 mol/L?
You're right. It doesn't make sense. Cl^- is a Bronsted-Lowry base BUT a very weak one. It won't even displace H from HOH. The acidity of AlCl3 solutions and Al2(SO4)3 solutions is due to the hydration of the Al^+3 ion and the Ka given in the problem is the one for which I wrote the ionization above although it may have shown H2O as part of the reaction. You work those problem just like the Ka for a weak acid, HA. Note: Gardener's put Al2(SO4)3 in their soil to increase the acidity (lower the pH of the soil) in order to accommodate plants that require an acid soil to grow. Plants like hydrangea (the flowers) actually turn color, pink for acid and blue for basic.
Not if the solution is Al2(SO4)3. A 0.1 M solution of Al2(SO4)3 is 0.2 M in Al(H2O)6^+3
Ohhh! Ok! Thank you for that! I missed a class, and so I'm relying on my notes to teach me...which isn't easy. Interesting little fact btw.
Oh you're right! Thanks.
To calculate the pH of an aqueous solution of aluminum sulfate, we need to understand that the compound dissociates in water, producing aluminum ions (Al3+) and sulfate ions (SO4^2-). The aluminum ion can hydrolyze in water, resulting in the generation of H+ ions and forming an acidic solution. The Ka value for the aluminum ion (Al3+) indicates the degree of acidity it produces when it reacts with water.
The hydrolysis equation for aluminum ion (Al3+) can be represented as follows:
Al3+ + H2O ⇌ Al(OH)2+ + H+
The Ka value tells us the extent to which the reaction shifts to the right and provides information about the concentration of H+ ions produced. In this case, the Ka value for the aluminum ion is given as 9.8x10^-6. This value will be useful in determining the concentration of H+ ions in the solution.
To find the pH of the solution, we need to consider both the hydrolysis of aluminum ions and the auto-ionization of water. The auto-ionization of water can be represented by the following equation:
H2O ⇌ H+ + OH-
Here's the step-by-step process to calculate the pH:
1. Write the balanced equation for the hydrolysis of aluminum ions:
Al3+ + H2O ⇌ Al(OH)2+ + H+
2. Write the expression for the equilibrium constant (Ka) using the equilibrium concentrations of the species involved:
Ka = [Al(OH)2+][H+] / [Al3+]
3. Assume x as the concentration of Al3+ and x as the subsequent concentrations of Al(OH)2+ and H+.
4. Write the equilibrium expression in terms of x:
Ka = (x)(x) / (0.10 - x)
Note: The initial concentration of Al3+ is assumed to be 0.10 M, and the concentration of Al(OH)2+ can be assumed to be approximately equal to x since x will be small in comparison.
5. Solve the quadratic equation for x to find the concentration of H+ ions.
6. Convert the concentration of H+ ions to pH using the formula:
pH = -log[H+]
By following these steps and using the given Ka value for aluminum ions, you will be able to calculate the pH of the 0.10 mol/L aqueous solution of aluminum sulfate.