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)

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.

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

Note the FIRST equation I wrote has Ka. Substitute 3.3E-4 there.

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

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

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

moles aspirin = 1.00g/180.16 = ??

M = moles/L = ??/0.325L = ??

x=2.37x10^-3

pH=2.63
is this correct?

To find the pH of the solution, we need to first calculate the concentration of acetylsalicylic acid (ASA) in the solution.

Given that two extra strength aspirin tablets are dissolved in 325 ml of water, and each tablet contains 500 mg (0.5 g) of ASA, we can calculate the total mass of ASA:

Total mass of ASA = (2 tablets) * (0.5 g/tablet) = 1 g

Next, let's calculate the concentration of ASA in the solution:

Concentration of ASA = (mass of ASA) / (volume of solution)
= (1 g) / (325 mL)
= (1 g) / (0.325 L)
= 3.08 g/L

Now, we need to convert the concentration from grams per liter (g/L) to moles per liter (M). The molar mass of ASA is 180.16 g/mol.

Concentration of ASA in moles per liter = (concentration of ASA in g/L) / (molar mass of ASA)
= (3.08 g/L) / (180.16 g/mol)
= 0.0171 mol/L

Now, let's write the dissociation equation for ASA in water:

HC9H7O4 (s) + H2O (l) ⇄ H3O+ (aq) + C9H7O4- (aq)

The equilibrium constant for this reaction is denoted as Ka and its value is given as 3.3 x 10^-4.

Using the equation for Ka, which relates the concentrations of the products and reactants at equilibrium, we have:

Ka = [H3O+] / [C9H7O4-]

Since the initial concentration of H3O+ is equal to the initial concentration of ASA, we can let [H3O+] = x and [C9H7O4-] = x.

Therefore, we have:

Ka = x / x
x = Ka

Now, we can substitute the value of Ka into our equation:

x = 3.3 x 10^-4

Finally, we can calculate the pH of the solution:

pH = -log[H3O+]
pH = -log(x)
pH = -log(3.3 x 10^-4)

Using a calculator, we can find that the pH of the solution is approximately 3.48.