Question

Determine the pH of a 0.260M solution of oxalic acid.

Answer 1

pH = -log[\(H^+\)]

Answer 2

So Ka isn't necessary?

Answer 3

Yes, set up the expression of course for your acid. Remember its \(K_a =\large \frac{[product]}{[reactant]}\)
can you set up your equation?

Answer 4

So, like this?:
\[H_{2}C _{2}O _{4}+H _{2}O \leftarrow \rightarrow HC _{2}O ^{-} _{4}+H _{3}O\]

\[K _{a}=[HC _{2}O ^{-} _{4}][H _{3}O]/[H_{2}C _{2}O _{4}][H _{2}O]\]

Answer 5

But then how do we utilize that? Can you walk me through please?

Answer 6

For any acid, you generally have: \(HA \rightarrow H^+ + A^-\) so you'd have: \(\large \frac{[H^+][A^-]}{[HA]}\) DOES THAT make sens?

Answer 7

No.
Is what I have incorrecT?

Answer 8

\(\LARGE\frac{[H^+][HC_2O_4^-]}{H_2C_2O_4}\)

Answer 9

Hmm, okay. But the question says oxalic acid SOLUTION, so would the reactants not just be oxalic acid and water?

Answer 10

Hello?

Answer 11

Hi?

Answer 12

you don't include water in equilibrium expressions

Answer 13

you have to make an ICE table to figure the extend of the dissociation of the acid.
When you you do that you'll find the value for H3O+, then just plug that into the formula abb0t wrote earlier

Answer 14

Could you assist me with that please? I'm really struggling with what values to put in the ICE table

Answer 15

yeah no problem, what do you have so far?