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

Answer 1

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...0.004-x.....x........x

Ka = (H^+)(RCOO^-)/(RCOOH)
Ka = (x)(x)/(0.004-x)
What is x. Solve for x = (H^+), then convert to H.
pH = -log (H^+).

Answer 2

The answer that was in my answer key was 2.83 but I got 3.48. Is there something wrong in the calculations?

Answer 3

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)

Answer 4

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!

Answer 5

as we know that pH=-log(H^+)
we firstly need to know the concentration of H^+ after dissociating the aspirin.
We should therefore calculate the number of moles of aspirin. Then we should use the ICE table to determine the concentrations of H^+ and the associated conjugate base. The volume in litres may be helpful if we want to get the concentration of aspirin by molarity... remember that we need this in the ICE table.