Analytical Chem Question

[user098123]

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?

 

[sopher]

pH = pKa + log[acid]/[base]

 

[philipc]

im horrible at chem but a mol is 6.02x10^23

 

[skierdudeguy]

and planks constant is 6.626x10^34

 

[freezed]

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.

 

[niels.]

^what he said

no i really have no clue. sounds fun though

 

[user098123]

not exactly sure what that is

 

[freezed]

this page will probably help you http://www.chem.ubc.ca/courseware/pH/section1/content.html

 

[user098123]

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.