need help with calculating the pH of the following...

Question

Determine the pH for the following solutions. Show your work to support your answer:

a. 0.227M HCl    

 b. 0.186M NH₃      

c. 0.088M KHCO₃

Answer 1

pH = -log [H⁺]

a). HCl (strong acid) ==> H⁺ + Cl^-

0.227 M HCl => 0.227 M H⁺

pH = -log 0.227

pH = 0.64

b), NH3 (weak base)

NH₃ + H₂O => NH₄⁺ + OH^-

To determine pH, we need to know either Kb for NH3 or Ka for NH4+

Using Kb for NH3 = 1.8x10^-5 we can solve for [OH-]

Kb = 1.8x10^-5 = [NH4+][OH-] / [NH3]

1.8x10^-5 = (x)(x) / 0.186 - x (assume x is small relative to 0.186 and ignore it in the denominator)

x² =3.3x10^-6

x = [OH-] = 1.8x10^-3 M (note: this is insignificant relative to 0.186 to above assumption was valid)

pOH = -log [OH-] = 2.74

pH = 14 - pOH = 14 - 2.74

pH = 11.3

c). KHCO₃ (both a weak acid and a weak base; amphoteric)

This is more complex than what can be explained here. It involves 3 equilibria.

H₂CO₃ <==> H⁺ + HCO₃^- Ka1 = [H⁺][HCO₃^-] / [H₂CO₃

HCO₃^- <==> H⁺ + CO₃^2- Ka2 = [H⁺][CO₃]^2- / HCO₃^-

H₂O <==> H⁺ + OH^- Kw = [H⁺][OH^-] = 1x10^-14

This will ultimately involve using a polynomial. Perhaps the person posing the question expected you to solve it in a more simplistic manner, but I'm not sure what that might be.