Problem: The active ingredient in aspirin is acetylsalicylic acid (HC9H7O4), a monoprotic acid with a Ka of 3.3x10-4 at 25°C .What is the pH of a solution obtained by dissolving two extra-strength aspirin tablets, containing 510 mg of acetylsalicylic acid each, in 330 mL of water?

FREE Expert Solution

Equilibrium reaction:        HC9H7O4(aq) + H2O(l)  C9H7O4-(aq) + H3O+(aq) 


Step 1: Calculate the concentration of the solution.

molarity(M)=moles soluteL solution


molar mass HC9H7O4 = 180.154 g/mol

mass HC9H7O4 = 510 mg x 2 = 1020 mg

moles HC9H7O4=1020 mg×10-3 g1 mg×1 mol180.154 g

moles HC9H7O4 = 5.6618x10–3 mol


[HC9H7O4]=5.6618×10-3 mol330 mL×10-3 L1 mL[HC9H7O4]=5.6618×10-3 mol0.330 L

[HC9H7O4] = 0.0171 M


Step 2: Construct an ICE chart for the reaction.


Step 3: Write the Ka expression.

Ka=productsreactantsKa=[C9H7O4-][H3O+]HC9H7O4

Solids and liquids are not included in the expression



Step 4: Calculate the equilibrium concentrations.

Ka=xx0.0171-x3.3×10-4=x20.0171-x

[HC9H7O4]initialKa500

3.3×10-4 (0.0171-x)=x20.0171-x(0.0171-x)5.6618×10-6-(3.3×10-4)x=x20=x2+(3.3×10-4)x-5.6618×10-6


x=-b±b2-4ac2a

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Problem Details

The active ingredient in aspirin is acetylsalicylic acid (HC9H7O4), a monoprotic acid with a Ka of 3.3x10-4 at 25°C .

What is the pH of a solution obtained by dissolving two extra-strength aspirin tablets, containing 510 mg of acetylsalicylic acid each, in 330 mL of water?

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Weak Acids concept. If you need more Weak Acids practice, you can also practice Weak Acids practice problems.

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