Calculate the ph of a 0.17 m solution of c6h5nh3no3 (kb for c6h5nh2 = 3.8 x 10-10). record your ph value to 2 decimal places.

Question

Belen Bridges

Chemistry

12 August, 04:47

Calculate the ph of a 0.17 m solution of c6h5nh3no3 (kb for c6h5nh2 = 3.8 x 10-10). record your ph value to 2 decimal places.

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Answers (1)

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Duncan · Answer 1

Mass of the solution = 0.17m

Kb for C6H5NH2 = 3.8 x 10^-10

We know Ka for C6H5NH2 = 1.78x10^-11

We have Kw = Ka x Kb = > Ka = Kw / Kb

=> (C2H5NH2) (H3O^+) / (C2H5NH3^+) = > 1.78x10^-11 = K^2 / 0.17

K^2 = 3 x 10^-12 = > K = 1.73 x 10^-6.

pH = - log (Kw (H3O^+)) = - log (1.73 x 10^-6) = 5.76