Question

Calculate the pH at 25 °C of a 0.16 M solution of ethylammonium bromide (C,HNH,Br). Note that ethylamine (C,H,NH) is a weak base with a p K, of 3.19 Round your answer to 1 decimal place. pH X

Answer 1

The hydrolysis that occurs is:

C2H5NH3 + + H2O = C2H5NH2 + H3O +

Kb and Ka are calculated:

Kb = 10 ^ -pKb = 10 ^ -3.19 = 6.45x10 ^ -4

Ka = 10 ^ -14 / 6.45x10 ^ -4 = 1.55x10 ^ -11

You have the expression of Ka:

Ka = [C2H5NH2] * [H3O +] / [C2H5NH3 +]

1.55x10 ^ -11 = X ^ 2 / 0.16

It clears X = 1.57x10 ^ -6 M

The pH is calculated:

pH = - log 1.57x10 ^ -6 = 5.80

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