[big]

A solution containing 1.0 M Al³⁺ and 1.0 M Cu²⁺ is buffered to a pH of 4.00. What is the ratio of the ions, Cu²⁺/Al³⁺, left in solution?

K_sp of Cu(OH)₂=2.20*10^-20
K_sp of Al(OH)₃=3.00*10^-33

Here's what I tried, but I got the wrong answer:
So at pH=4, [OH]^-=10^-10M.
So if you set 2.20*10^-20=(1.0-s)(10^-10-2s)², s for Cu²⁺=1.242*10^-10
And if you set 3.00*10^-33=(1.0-s)(10^-10-3s)³, s for Al³⁺=2.853*10^-11

Then when I divided the first s by the second, I got 4.35, but the answer is 333, so is there somewhere where I went wrong?

[Borek]

2.20*10^-20=(1.0-s)(10^-10-2s)²

You are overthinking. Solution is buffered, which means final concentration of OH^- is 10^-4 M - it doesn't change during precipitation.

[LackOfFuel]

What is the pH of the unbuffered solution?

[big]

2.20*10^-20=(1.0-s)(10^-10-2s)²

You are overthinking. Solution is buffered, which means final concentration of OH^- is 10^-4 M - it doesn't change during precipitation.

So I redid it, and this time I did:

K_sp of Cu(OH)₂=2.20*10^-20
K_sp of Al(OH)₃=3.00*10^-33

At pH=4, [OH]^-=10^-10M.
2.20*10^-20=[Cu²⁺](10^-10)², and s for Cu²⁺=2.2
3.00*10^-33=[Al³⁺](10^-10)³, s for Al³⁺=.003

2.2/0.003=733 though, not 333. Did I do something wrong again or did they probably mean 733 instead of 333?

What is the pH of the unbuffered solution?

How would the pH of the unbuffered solution help?

[Borek]

s for Cu²⁺=2.2

2.2M? In 1.0M solution?

[big]

s for Cu²⁺=2.2

2.2M? In 1.0M solution?

Oh! So you do 1/0.003=333, which is the answer! Wow, that completely caught me off guard, but thanks so much! 

[LackOfFuel]

What is the pH of the unbuffered solution?

How would the pH of the unbuffered solution help?

It wouldn't, it seems I've been clumsy on reading the text.