Respuesta :
Answer:
(1) Before the addition of any HBr, the pH is 12.02
(2) After adding 12.0 mL of HBr, the pH is 10.86
(3) At the titration midpoint, the pH is 10.73
(4) At the equivalence point, the pH is 5.79
(5) After adding 45.1 mL of HBr, the pH is 1.18
Explanation:
First of all, we have a weak base:
- 0 mL of HBr is added
(CH₃)₂NH + H₂O ⇄ (CH₃)₂NH₂⁺ + OH⁻ Kb = 5.4×10⁻⁴
0.289 - x x x
Kb = x² / 0.289-x
Kb . 0.289 - Kbx - x²
1.56×10⁻⁴ - 5.4×10⁻⁴x - x²
After the quadratic equation is solved x = 0.01222 → [OH⁻]
- log [OH⁻] = pOH → 1.91
pH = 12.02 (14 - pOH)
- After adding 12 mL of HBr
We determine the mmoles of H⁺, we add:
0.286 M . 12 mL = 3.432 mmol
We determine the mmoles of base⁻, we have
27.9 mL . 0.289 M = 8.0631 mmol
When the base, react to the protons, we have the protonated base plus water (neutralization reaction)
(CH₃)₂NH + H₃O⁺ ⇄ (CH₃)₂NH₂⁺ + H₂O
8.0631 mm 3.432 mm -
4.6311 mm 3.432 mm
We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.
[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M
[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M
We have just made a buffer.
pH = pKa + log (CH₃)₂NH / (CH₃)₂NH₂⁺
pH = 10.73 + log (0.116/0.0860) = 10.86
- Equivalence point
mmoles of base = mmoles of acid
Let's find out the volume
0.289 M . 27.9 mL = 0.286 M . volume
volume in Eq. point = 28.2 mL
(CH₃)₂NH + H₃O⁺ ⇄ (CH₃)₂NH₂⁺ + H₂O
8.0631 mm 8.0631mm -
8.0631 mm
We do not have base and protons, we only have the conjugate acid
We calculate the new concentration:
mmoles of conjugated acid / Total volume (initial + eq. point)
[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL) = 0.144 M
(CH₃)₂NH₂⁺ + H₂O ⇄ (CH₃)₂NH + H₃O⁻ Ka = 1.85×10⁻¹¹
0.144 - x x x
[H₃O⁺] = √ (Ka . 0.144) → 1.63×10⁻⁶ M
pH = - log [H₃O⁺] = 5.79
- Titration midpoint (28.2 mL/2)
This is the point where we add, the half of acid. (14.1 mL)
This is still a buffer area.
mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)
mmoles of base = 8.0631 mmol - 4.0326 mmol
[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M
[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M
pH = pKa + log (0.096M / 0.096 M)
pH = 10.73 + log 1 = 10.73
Both concentrations are the same, so pH = pKa. This is the maximum buffering capacity.
- When we add 45.1 mL of HBr
mmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol
mmoles of base = 8.0631 mmoles
This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.
(CH₃)₂NH + H₃O⁺ ⇄ (CH₃)₂NH₂⁺ + H₂O
8.0631 mm 12.8986 mm -
- 4.8355 mm
[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)
[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M
- log [H₃O⁺] = pH
- log 0.0662 = 1.18 → pH
