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Analytical Chem Question

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user098123
Posts: 6072 - Karma: 18
What is the pH of a solution that is prepared by dissolving 9.24 grams of salicylic acid (138.12 g/mol) and 7.03 grams of sodium salicylate (160.10 g/mol) in water and diluting to 500 mL. The Ka for salicylic acid is 1.05×10-3.

Can anyone help me solve this?
Feb 8 2007 8:54PM
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sopher
Posts: 760 - Karma: 15
pH = pKa + log[acid]/[base]
Feb 8 2007 8:56PM
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philipc
Posts: 6651 - Karma: 1,286
im horrible at chem but a mol is 6.02x10^23
Feb 8 2007 8:56PM
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skierdudeguy
Posts: 6574 - Karma: 81
and planks constant is 6.626x10^34
Feb 8 2007 9:00PM
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freezed
Posts: 12973 - Karma: 500
write the balanced equation...do an ICE table, solve for x. then it'll be easy to see how much acid is left, and therefore the pH.
Feb 8 2007 9:09PM
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niels.
Posts: 1282 - Karma: 14
^what he said

no i really have no clue. sounds fun though
Feb 8 2007 9:14PM
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user098123
Posts: 6072 - Karma: 18
not exactly sure what that is
Feb 8 2007 9:24PM
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freezed
Posts: 12973 - Karma: 500
this page will probably help you http://www.chem.ubc.ca/courseware/pH/section1/content.html
Feb 8 2007 9:27PM
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user098123
Posts: 6072 - Karma: 18
Alright how about this one? Determine the pH of an aqueous solution that is 3.98×10^-3 M in NaOCl.

I looked up the Ka of HOCl, because I believe that is what it dissociates into. I then use that Ka value (3.0x10^-8) and plug it into the equation Ka = X^2/(3.98x10^-3 - X) and then....I'm not sure where from that. -log to get the pH? Dunno if that is right tho.
Feb 8 2007 9:55PM
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