Determine the pH at the equivalence (stoichiometric) point in the titration of 30 mL of 0.16 M NH3(aq) with 0.19 M HCl(aq).

Question

Determine the pH at the equivalence (stoichiometric) point in the titration of 30 mL of 0.16 M NH₃(aq) with 0.19 M HCl(aq). The K_b of NH₃ is 1.8 x 10^-5.

 The K_b of NH₃ is 1.8 x 10^-5.

The answer is 5.16.

Please show the steps for the answer, thank you :)

Answer 1

This problem involves titrating a weak base (NH₃) with a strong acid (HCl). When titrated to completion (stoichiometric equivalence) all of the NH₃ will be converted to NH₄⁺ which will then be hydrolyzed to produce H₃O⁺ (see below). Thus, the pH at equivalence will be acidic (<7)

NH₃ + H₊ ===> NH₄⁺

moles NH₃ present = 0.030 L x 0.16 mol/L = 0.0048 moles NH₃

moles HCl needed for stoichiometric equivalence = 0.0048 moles HCl (1:1 mole ratio in balanced equation)

volume HCl needed: (x L)(0.19 mol/L) = 0.0048 moles and x = 0.0253 L = 25.3 mls

moles NH₄⁺ formed = 0.0048 moles NH₄⁺

Final volume at equivalence = 30 ml + 25.3 ml = 55.3 mls = 0.0553 L

FINAL [NH₄⁺] = 0.0048 moles/0.05531 L = 0.08679 M NH₄⁺

So, to this point, in summary, we have completely neutralized the NH₃ and formed NH₄⁺. The NH₄⁺ will now be hydrolyzed by the water present as follows:

NH₄⁺ + H₂O <===>NH₃ + H₃O⁺

0.0868.....0..............0.........0.......Initial

-x...........................+x.........+x.....Change

0.0868-x................x............x......Equilibrium

From the provided value of Kb, we can find the corresponding value of Ka (for NH₄⁺):

KaKb = Kw

Ka = Kw/Kb = 1x10^-14/1.8x10^-5 = 5.6x10^-10 = Ka for NH₄⁺ acting as an acid

Ka = 5.6x10^-10 = [NH₃][H₃O⁺]/[NH₄⁺] = (x)(x)/0.0868 - x (since Ka is so small we can assume that x is small relative to 0.0868 and neglect it to simplify the math)

5.6x10^-10 = x²/0.0868

x² = 4.86x10^-11

x = 6.97x10^-6 = [H⁺]

pH = -log [H⁺]

pH = -log 6.97x10^-6 = 5.16