Aspirin (acetylsalicylic acid, C_9H_8O_4) is a weak monoprotic acid. To determin

Question

Aspirin (acetylsalicylic acid, C_9H_8O_4) is a weak monoprotic acid. To determine its acid dissociation constant, a student dissolved 2.00 g of aspirin in 0.600 L of water and measured the pH. What was the K_a value calculate by the student if the pH of the solution was 2.60? Express your answer numerically using two significant figures. K_a = _______ A 0.100 M solution of ethylamine (C_2H_5NH_2) has a pH of 11.87. Calculate the K_b for ethylamine. Express your answer numerically using two significant figures. k_b = _______

Explanation / Answer

A)

apply

Ka = [H+][A-]/[HA]

pH = 2.6

[H+] = 10^-2.6 = 0.0025118 M

M = mol/V

mol = mass/MW = 2/180.157 = 0.011101 mol of aspirin

M = 0.011101/0.6 = 0.0185 M

so...

in equilibrium,w e need to account for the dissovled HA i..e that which formed H+ and A-

[HA] = 0.0185- 0.0025118 = 0.0159882

Ka = (0.0025118 *0.0025118 ) /(0.0159882) = 3.94612*10^-4

B)

for Kb

Base + H2O <--> BaseH+ + OH-

Kb = [BH+][OH-]/[B]

pOH = 14-pH = 14-11.87 = 2.13

[OH-] = 10^-pOH = 10^-2.13 = 0.007413

so

[BH+] = [OH-] = 0.007413

then

B = 0.1-0.007413 = 0.092587

then

Kb = (0.007413 )(0.007413 ) / 0.092587

KB = 5.935235*10^-4

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