A typical aspirin tablet contains 324 mg of aspirin (acetylsalicylic acid, C9H8O

Question

A typical aspirin tablet contains 324 mg of aspirin (acetylsalicylic acid, C9H8O4 (Figure 1) ), a monoprotic acid having Ka=3.0×104.

If you dissolve four aspirin tablets in a 200 mL glass of water, what is the pH of the solution?

What is the percent dissociation?

...I dont just want the answer, I need help understanding how to get to the solution. I am stuck with the equillibrium constant. I know that following that portion I will use that answer for calculating the dissociation I just need some help. Thank you so much advance

Explanation / Answer

Calculate molar concentration of asprin solution:

Molar mass = C9H8O4 = 12.011*9 + 1.008*8+15.99*4 = 180.123g/mol
You have 2*324mg = 648mg
moles = 0.648/180.123 = 3.59*10^-3 mol
Moles = 3.59*10^-3moles in 300ml solution, this is :
3.59*10^-3*1000/200 = 0.017 moles per litre or 0.017M solution of acid

Ka = x² / ( 0.017M -x)

substitute the value Ka and solve the quadratic equation for x.

x²/ ( 0.017M -x) = 3.00*10^-4

X² = (0.017M - X) * 3.00*10^-4

X² = 5.1*10^-6M - 3.00*10^-4X

X² + 0.0003X - 0.0000051M = 0

X = 0.001167M

acid is monoprotic, so [H+] = 0.001167
pH = -log [H+]
pH = -log 0.001167
pH = 2.94

% dissociation: solution was 0.012M

Dissociated to produce 0.001167M H+

% dissociation = 0.001167/0.012*100 = 9.72% dissociated.

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