Question

a) What is the pH of a 0.108 M monoprotic acid whose K_a is 6.097 ×10⁻³?

b) Calculate the pH of a 0.61 M methylamine solution.

Answer 1

a) The acid partially dissociates as follows:

HA + H2O = H3O + + A-

You have the expression of Ka:

Ka = [H3O +] * [A-] / [HA]

6.097x10 ^ -3 = X ^ 2 / 0.108

It clears X = 0.026 M

PH is calculated:

pH = - log 0.026 = 1.59

2) You have the expression of Kb:

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

3.7x10 ^ -4 = X ^ 2 / 0.61

It clears X = 0.015 M

POH and pH are calculated:

pOH = - log 0.015 = 1.82

pH = 14 - 1.82 = 12.18

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