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Kindly answer All the questions.

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Kindly answer All the questions.

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Answer #1

2.00 g C9H8O4 x (1 mole C9H8O4 / 180 g C9H8O4) = 0.0111 moles C9H8O4

0.0111 moles C9H8O4 / 0.600 L = 0.0185 M C9H8O4

Molarity . . . . .C9H8O4 + H2O <==> H3O+ + C9H7O4-
Initial . . . . . . . .0.0185 . . . . . . . . . . . .0 . . . . . .0
Change . . . . . . . .-x . . . . . . . . . . . . . .x . . . . . .x
At Equil. . . . .0.0185 - x . . . . . . . . . . .x . . . . . .x

Ka = [H3O+][C9H7O4-] / [C9H8O4] = x^2 / (0.0185 - x)

pH = 2.60; [H3O+] = 10^-pH = 10^-2.60 = 0.0025 = x

Ka = (0.0025)^2 / (0.0185 - 0.0025) = 3.9 x 10^-4

======================================...

Molarity . . . . . .C2H5NH2 + H2O <==> C2H5NH3+ + OH-
Initial . . . . . . . . . .0.100 . . . . . . . . . . . . . . .0 . . . . . .0
Change . . . . . . . . . .-x . . . . . . . . . . . . . . . .x . . . . . .x
At Equil. . . . . .0.100 - x . . . . . . . . . . . . . . .x . . . . . .x

pH = 11.87
pOH = 14.00 - pH = 14.00 - 11.87 = 2.13
[OH-] = 10^-pOH = 10^-2.13 = 0.0074 = x

Kb = [C2H5NH3+][OH-] / [C2H5NH2] = x^2 / (0.100 - x) = (0.0074)^2 / (0.100 - 0.0074) = 5.9 x 10^-4

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