For the dissociation reaction of a weak acid in water, HA(aq) + H2O(l) H3O+(aq)
ID: 602040 • Letter: F
Question
For the dissociation reaction of a weak acid in water, HA(aq) + H2O(l) H3O+(aq) + A-(aq) the equilibrium constant is the acid-dissociation constant, Ka, and takes the form Ka = [H3O+][A-]/[HA] Weak bases accept a proton from water to give the conjugate acid and OH- ions: B(aq) + H2O(l) BH+(aq) + OH-(aq) The equilibrium constant Kb is called the base-dissociation constant and can be found by the formula Kb = [BH+][OH-]/[B] Aspirin (acetylsalicylic acid. C9H8O4) 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 Ka value calculated by the student if the pH of the solution was 2.62? Express your answer numerically using two significant figures. Ka= My Answers Give Up Review Part A 0.100 M solution of ethylamine (C2H5NH2) has a pH of 11.87. Calculate the Kb for ethylamine. Express your answer numerically using two significant figures. Kb = My Answers Give Up Review PartExplanation / Answer
moles of aspirin =2/164=.0122mol
Molarity . . . . . .HA + H2O <==> H3O+ + A-
At Equil. . . . .0.0122-x . . . . . . . . . . . .x . . . . .x
pH = 2.62
[H3O+] = 10^-pH = .003988 = [H3O+] = [A-]
[HA] = 0.100 - x = 0.0122 - 0.003988 = 0.00821
Ka = [H3O+][A-] / [HA] = (0.003988)(0.003988) / (0.00821 = 1.94x 10^-3
in the same way calculate ka
then by using formula
ka x kb = kw
kb = kw / kb
kb= 1 x 10^-14 /Ka
calculate kb