Chemistry
posted by Jessica .
Aspirin is a weak acid with a Ka of 3.0x10^5. Find the pH of a solution by dissolving .65g of aspirin in water and diluting it to 50mL. You can use RCOOH to represent aspirin. Molecular weight of aspirin is 180g/mol

moles aspirin = grams/molar mass = 0.65/180 = approximately 0.004 mole.
.........RCOOH ==> H^+ + RCOO^
initial....0.004....0......0
change.....x......x........x
equil...0.004x.....x........x
Ka = (H^+)(RCOO^)/(RCOOH)
Ka = (x)(x)/(0.004x)
What is x. Solve for x = (H^+), then convert to H.
pH = log (H^+). 
The answer that was in my answer key was 2.83 but I got 3.48. Is there something wrong in the calculations?

2.83 is right (and I obtained 3.46 using 0.004, also). Apparently you took my "apparoximately 0.004) and assumed that was gospel. I had to stick something there and didn't want to calculate it so I guessed at it and wrote approximately. I assumed you would substitute the correct value although I didn't say you should confirm my approximate number (but you should have). The molarity of the aspirin is 0.0722. (NOTE: 0.65g x (1 mol/180g) x (1000 mL/50 mL)

For the ICE box, you are supposed to use the molarity of the aspirin. I did not realize that. Thank you so much for your help!