Answer :
To determine the pH at the equivalence point in the titration of methylamine ([tex]\(CH_3NH_2\)[/tex]) with hydrochloric acid (HCl), you can follow these steps:
1. Calculate Moles of [tex]\(CH_3NH_2\)[/tex]:
- Volume of [tex]\(CH_3NH_2\)[/tex] solution = 39.3 mL = 0.0393 L
- Molarity of [tex]\(CH_3NH_2\)[/tex] = 0.259 M
- Moles of [tex]\(CH_3NH_2\)[/tex] = Volume (L) × Molarity (mol/L) = 0.0393 L × 0.259 mol/L = 0.0101817 moles
2. Equivalence Point:
- At the equivalence point, the moles of [tex]\(CH_3NH_2\)[/tex] will equal the moles of HCl added.
- Molarity of HCl = 0.166 M
- Volume of HCl needed = Moles of [tex]\(CH_3NH_2\)[/tex] / Molarity of HCl = 0.0101817 moles / 0.166 M = 0.0613175 L or 61.3175 mL
3. Concentration of Conjugate Acid ([tex]\(CH_3NH_3^+\)[/tex]) at Equivalence Point:
- At the equivalence point, all [tex]\(CH_3NH_2\)[/tex] is converted to [tex]\(CH_3NH_3^+\)[/tex].
- Total volume = Volume of [tex]\(CH_3NH_2\)[/tex] + Volume of HCl = 39.3 mL + 61.3175 mL = 100.6175 mL or 0.1006175 L
- Concentration of [tex]\(CH_3NH_3^+\)[/tex] = Moles of [tex]\(CH_3NH_2\)[/tex] / Total Volume (L) = 0.0101817 moles / 0.1006175 L = 0.101162 M
4. Calculate [tex]\(K_a\)[/tex] of the Conjugate Acid:
- Given [tex]\(K_b\)[/tex] of [tex]\(CH_3NH_2\)[/tex] is [tex]\(3.6 \times 10^{-4}\)[/tex].
- [tex]\(K_w\)[/tex] (ion product of water) = [tex]\(1.0 \times 10^{-14}\)[/tex].
- [tex]\(K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{3.6 \times 10^{-4}} = 2.77778 \times 10^{-11}\)[/tex].
5. Calculate [H[tex]\(^+\)[/tex]] Concentration:
- Use the formula [H[tex]\(^+\)[/tex]] = [tex]\(\sqrt{K_a \times [CH_3NH_3^+]}\)[/tex].
- [H[tex]\(^+\)[/tex]] = [tex]\(\sqrt{2.77778 \times 10^{-11} \times 0.101162}\)[/tex] = [tex]\(1.67632 \times 10^{-6}\)[/tex] M
6. Calculate the pH:
- pH = [tex]\(-\log_{10}([H^+])\)[/tex].
- pH = [tex]\(-\log_{10}(1.67632 \times 10^{-6}) \approx 5.776\)[/tex].
Therefore, the pH at the equivalence point in this titration is approximately 5.776.
1. Calculate Moles of [tex]\(CH_3NH_2\)[/tex]:
- Volume of [tex]\(CH_3NH_2\)[/tex] solution = 39.3 mL = 0.0393 L
- Molarity of [tex]\(CH_3NH_2\)[/tex] = 0.259 M
- Moles of [tex]\(CH_3NH_2\)[/tex] = Volume (L) × Molarity (mol/L) = 0.0393 L × 0.259 mol/L = 0.0101817 moles
2. Equivalence Point:
- At the equivalence point, the moles of [tex]\(CH_3NH_2\)[/tex] will equal the moles of HCl added.
- Molarity of HCl = 0.166 M
- Volume of HCl needed = Moles of [tex]\(CH_3NH_2\)[/tex] / Molarity of HCl = 0.0101817 moles / 0.166 M = 0.0613175 L or 61.3175 mL
3. Concentration of Conjugate Acid ([tex]\(CH_3NH_3^+\)[/tex]) at Equivalence Point:
- At the equivalence point, all [tex]\(CH_3NH_2\)[/tex] is converted to [tex]\(CH_3NH_3^+\)[/tex].
- Total volume = Volume of [tex]\(CH_3NH_2\)[/tex] + Volume of HCl = 39.3 mL + 61.3175 mL = 100.6175 mL or 0.1006175 L
- Concentration of [tex]\(CH_3NH_3^+\)[/tex] = Moles of [tex]\(CH_3NH_2\)[/tex] / Total Volume (L) = 0.0101817 moles / 0.1006175 L = 0.101162 M
4. Calculate [tex]\(K_a\)[/tex] of the Conjugate Acid:
- Given [tex]\(K_b\)[/tex] of [tex]\(CH_3NH_2\)[/tex] is [tex]\(3.6 \times 10^{-4}\)[/tex].
- [tex]\(K_w\)[/tex] (ion product of water) = [tex]\(1.0 \times 10^{-14}\)[/tex].
- [tex]\(K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{3.6 \times 10^{-4}} = 2.77778 \times 10^{-11}\)[/tex].
5. Calculate [H[tex]\(^+\)[/tex]] Concentration:
- Use the formula [H[tex]\(^+\)[/tex]] = [tex]\(\sqrt{K_a \times [CH_3NH_3^+]}\)[/tex].
- [H[tex]\(^+\)[/tex]] = [tex]\(\sqrt{2.77778 \times 10^{-11} \times 0.101162}\)[/tex] = [tex]\(1.67632 \times 10^{-6}\)[/tex] M
6. Calculate the pH:
- pH = [tex]\(-\log_{10}([H^+])\)[/tex].
- pH = [tex]\(-\log_{10}(1.67632 \times 10^{-6}) \approx 5.776\)[/tex].
Therefore, the pH at the equivalence point in this titration is approximately 5.776.
