Question

Acetylsalicyclic Acid, HC9H704, is the active component is aspirin. Two extra strength aspirin tablets, each containing 500 mg of acetylsalicylic acid, are dissolved in 325 ml of water. What is the pH of the solution? (Assume Ka=3.3x10^-4)

Answer 1

Let's call aspirin, HA
HA ==> H^+ + A^-

Ka = (H^+)(A^-)/(HA)
(H^+) = x
(A^-) = x
(HA) = 500 mg x 2 tablets = 1000 mg (1 gram and moles is 1g/molar mass with M = mols/0.325L.
Solve for x and convert to pH.

Answer 2

so where does the Ka=3.3x10^-4 come in? and im lost with what you put in parethesis in ur explanation.

Answer 3

<b>Note the FIRST equation I wrote has Ka. Substitute 3.3E-4 there.</b>

Ka = (H^+)(A^-)/(HA)
(H^+) = x <b>Let x = (H^+)</b>
(A^-) = x <b>then x = (A^-)</b>

(HA) = 500 mg x 2 tablets = 1000 mg (1 gram and moles is 1g/molar mass with M = mols/0.325L.
<b>You must substitute something for HA in the denominator of the Ka expression. The problem tells you that two (2) 500 mg tablets were placed in 325 mL solution. 2*500 mg = 1000 mg and that is 1.00 grams. How many moles is that. It is 1.00g/molar mass aspirin--you do the addition to find molar mass. Molarity is what goes into the equation, you have moles in 325 mL so that is moles/0.325L = (HA).<b>
Solve for x and convert to pH. </b></b>

Answer 4

does (HA)=58.695. my math was 1g/180.6 * 1/.325L

Answer 5

moles aspirin = 1.00g/180.16 = ??
M = moles/L = ??/0.325L = ??

Answer 6

x=2.37x10^-3
pH=2.63
is this correct?