# Chemistry

posted by Stuck

"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?

1. Stuck

Oops! Sorry, I figured it out. But I still need help...I'm not clear on the process.

2. DrBob222

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.

3. Stuck

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. :)

4. Stuck

No, it doesn't. But I have it in my notes. Ok, I think I understand how to approach this. Thanks!

5. Stuck

Is the initial concentration for Al(H20)6 also 0.10 mol/L?

6. DrBob222

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.

7. DrBob222

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

8. Stuck

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.

9. Stuck

Oh you're right! Thanks.

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