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Consider a titration of 24.2 mL of 0.0878M solution of aqueous ammonia [Kb(NH3) = 1.8e-5] with a 0.115M of hydrochloric acid.Calculate:
a) the pH of the aqueous ammonia solution before the titration
b) the pH of the solution at half-equivalence point
c) the pH of the solution at the equivalence point
For part "a" I have:[OH-] = √(1.8e-5 x 0.115)
For part "b" I have:
pOH = pKb
pOH = -log(1.8e-5)
then I subtract pOH from 14 to get pH
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