anonymous
  • anonymous
Determine the pH of a 0.260M solution of oxalic acid.
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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abb0t
  • abb0t
pH = -log[\(H^+\)]
anonymous
  • anonymous
So Ka isn't necessary?
abb0t
  • abb0t
Yes, set up the expression of course for your acid. Remember its \(K_a =\large \frac{[product]}{[reactant]}\) can you set up your equation?

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anonymous
  • anonymous
So, like this?: \[H_{2}C _{2}O _{4}+H _{2}O \leftarrow \rightarrow HC _{2}O ^{-} _{4}+H _{3}O\] \[K _{a}=[HC _{2}O ^{-} _{4}][H _{3}O]/[H_{2}C _{2}O _{4}][H _{2}O]\]
anonymous
  • anonymous
But then how do we utilize that? Can you walk me through please?
abb0t
  • abb0t
For any acid, you generally have: \(HA \rightarrow H^+ + A^-\) so you'd have: \(\large \frac{[H^+][A^-]}{[HA]}\) DOES THAT make sens?
anonymous
  • anonymous
No. Is what I have incorrecT?
abb0t
  • abb0t
\(\LARGE\frac{[H^+][HC_2O_4^-]}{H_2C_2O_4}\)
anonymous
  • anonymous
Hmm, okay. But the question says oxalic acid SOLUTION, so would the reactants not just be oxalic acid and water?
anonymous
  • anonymous
Hello?
anonymous
  • anonymous
Hi?
aaronq
  • aaronq
you don't include water in equilibrium expressions
aaronq
  • aaronq
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
anonymous
  • anonymous
Could you assist me with that please? I'm really struggling with what values to put in the ICE table
aaronq
  • aaronq
yeah no problem, what do you have so far?
anonymous
  • anonymous
|dw:1370562575805:dw|
anonymous
  • anonymous
|dw:1370562666036:dw|
aaronq
  • aaronq
|dw:1370562731390:dw|
aaronq
  • aaronq
initially you had zero of the products, so their initial values are zero for both. now since in the balanced equation all the coefficients were 1, you have a 1x change for each, negative and positive dependent on whether they are reactants or products, respectively. can you finish the table now?
anonymous
  • anonymous
Ah, okay! I'll try, and then can you confirm?
anonymous
  • anonymous
|dw:1370563017714:dw|
anonymous
  • anonymous
Hi?
anonymous
  • anonymous
ICE table needed indeed, good work.
anonymous
  • anonymous
So then where do I go?
aaronq
  • aaronq
you plug the equilibrium values into the equilibrium expression and solve for x
anonymous
  • anonymous
X being [H+], correct?
aaronq
  • aaronq
yes
anonymous
  • anonymous
Sweet, thanks everyone!!!
aaronq
  • aaronq
no problem !
abb0t
  • abb0t
Sorry, I went off for a bit to do some work. But I see @aaronq helped already.

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