Part A Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a...

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Part A
Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoproticacid. To determine its acid-dissociation constant, a studentdissolved 2.00 g of aspirin in 0.600 L of water and measured thepH. What was the Ka value calculated by the student if thepH of the solution was 2.62?
Express your answer numerically using two significantfigures.
Answer I got: 3.16 x 10^-4 (incorrect) this is the reason why:You used the initial concentration of aspirin in the Ka expression.Instead, you need to use the equilibrium concentration. Considerthat the amount of aspirin that reacted is equal to the amount ofH3O+ produced. How much is left over?

Answer & Explanation Solved by verified expert

3.6 Ratings (421 Votes)

Mass of aspirin in the solution = 2.00 g

Moles of aspirin = ( Mass / Molar mass) = ( 2.00 g / 180.157 g/ mol) = 0.011 mol

Volume of the solution = 0.600 L

Molarity of the solution = ( Moles / Volume) = ( 0.011/ 0.600 ) M = 0.018 M

pH = 2.62

-log [H+] = pH

-log [H+] = 2.62

[H+] = 10^-2.62

[ H+] = 0.0024

[H+] = [H3O+] = 0.0024 M

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â C9H8O4(aq)Â Â Â Â Â Â Â +Â Â Â Â Â H2O(l) ---------> H3O+ (aq)Â Â Â Â Â Â +Â Â Â Â Â Â Â Â Â Â C9H7O4-(aq)

Â Â Â Â Â Â Â Â Â Â Â InitialÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.018 MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

Â Â Â Â Â Â Â Â Â Â Â ChangeÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â - 0.0024Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â + 0.0024Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.0024

Â Â Â Â Â Â Â Â Â EquilibriumÂ Â Â Â Â Â Â Â Â Â Â Â Â Â 0.0156 MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.0024 MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.0024 M

KaÂ Â =Â Â Â [ H3O+] x [C9H7O4-] / [C9H8O4]

Ka = [0.0024] x [ 0.0024] / [ 0.0156] = 3.7 x 10^-4

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