Find the pH of the equivalence point and the volume(ml) of 0.125M HCl needed to reach the point in titrations of: a) 65.5ml of 0.234M NH3.b) 21.8 mL of 1.11 M CH3NH2.

a. 122.6 mL / pH = 4.03

b. 193.6 mL / pH = 5.73

Explanation:

In the equivalence point we know: mmoles acid = mmoles base- And the pH in a titration between a weak base and a strong base, is acid, at the equivalence point. For the volume, we can replace the equation with the data given.

a. 0.125M . volume of acid = 65.5 mL . 0.234M

Volume of acid = (65.5 . 0.234) / 0.125 = 122.6 mL

Total volume at the equivalence point = 188.1 mL

b. 0.125M . volume of acid = 21.8 mL . 1.11 M

Volume of acid = (21.8 .  1.11) / 0.125 = 193.6 mL

Let's calculate the pH. In the equilavence point we have a neutralization reaction.

a. NH₃ + HCl → NH₄Cl

All the mmoles of protons (65.5 mL . 0.234M) react to ammonia, to obtain ammonium.

New concentration is: 15.32 mmoles / 188.1 mL = 0.0814 M

This is the [NH₄⁺] to determine the pH in the acid base equilibrium.

NH₄⁺  +  H₂O  ⇄  NH₃  +  H₃O⁺      Ka

Expression for Ka = [NH₃]  . [H₃O⁺] / [NH₄⁺]

5.6×10⁻¹⁰ = x² / (15.32 - x)

(We can avoid the quadratic equation 'cause Ka is so small)

√(5.6×10⁻¹⁰ . 15.32) = x → [H₃O⁺] = 9.26×10⁻⁵

pH = - log [H₃O⁺] → 4.03

b. CH₃NH₂ + HCl → CH₃NH₃Cl

All the mmoles of protons (21.8 mL . 1.11M) react to methylamine, to obtain methylammonium.

New concentration is: 24.2 mmoles / 193.6 mL = 0.125 M

This is the [CH₃NH₃⁺] to determine the pH in the acid base equilibrium.

CH₃NH₃⁺  +  H₂O  ⇄  CH₃NH₂  +  H₃O⁺      Ka

Expression for Ka = [CH₃NH₂]  . [H₃O⁺] / [CH₃NH₃⁺]

2.7×10⁻¹¹ = x² / (0.125 - x)

(We can avoid the quadratic equation 'cause Ka is so small)

√(2.7×10⁻¹¹ . 0.125) = x → [H₃O⁺] = 1.84×10⁻⁶

pH = - log [H₃O⁺] → 5.73