A typical aspirin tablet contains 325 mg of acetylsalicylic acid, hc9h7o4. calculate the ph of a solution that is prepared by dissolving five aspirin tablet(s) in one cup (237 ml) of solution. assume the aspirin tablets are pure acetylsalicylic acid, ka = 3.3 ✕ 10-4.

Respuesta :

Acetylsalicylic acid (ASA) is a compound containing a carboxylic acid functionality that can act as an acid and give up a proton. A simple equation to represent this type of acid/base reaction can be written as follows, where HA = ASA.

HA + H2O --> A- + H3O+

We can now determine the initial concentration of our HA. We have 325 mg ASA per tablet, and we have 5 tablets, therefore:

325 mg ASA/tablet x 5 tablets = 1625 mg ASA

We can convert this into the initial concentration by figuring out how many moles of ASA are present and dividing it by the total volume of solution, which is 237 mL.

1625 mg/ 180.157 g/mol = 9.02 mmoles ASA

9.02 mmoles / 237 mL = 0.0381 mmoles/mL or 0.0381 mol/L.

With the initial concentration, we now record the changes in concentration and determine what the concentrations will be at equilibrium. This is referred to as an ICE table (initial, change, equilibrium).

        HA     +    H2O    -->     A-     +     H3O+
i     0.0381                             0              0
c       -x                                 +x             +x
e    0.0381 - x                       x                x

We were given the Ka value which is the acid dissociation constant of ASA. We write the formula for this value which is as follows:

Ka = [A-][H3O+]/[HA] = 3.3 x 10^-4
Ka = [x][x]/[0.0381-x] = 3.3 x 10^-4

At this point we can ignore the change to the intial concentration as a value of 10^-4 is negligible. So our simplified equation is now:

x^2/0.0381 = 3.3 x 10^-4
x^2 = 1.3 x 10^-5
x = 0.0035 = [A-] = [H3O+]

We have now solved for the concentration of hydronium ions, which is what is required in order to calculate the pH or a solution. To determine the pH, we use the formula pH = -log[H3O+]:

pH = -log[0.0035]
pH = 2.5

The pH of the solution is 2.5.