Question

Calculate the pH for each case in the titration of 50ml of 0.220 with 0.22.

Answer 1

The pH for the titration of 50 mL of 0.220 M HCl with 0.22 M NaOH can be calculated using the Henderson-Hasselbalch equation. At the equivalence point, where moles of acid equal moles of base, the pH is 7. In the titration of a strong acid (HCl) with a strong base (NaOH), the pH will change rapidly near the equivalence point.

Step-by-step explanation:

In the given titration, HCl is a strong acid, and NaOH is a strong base. The stoichiometry of the reaction is 1:1, meaning that one mole of HCl reacts with one mole of NaOH at the equivalence point. At the equivalence point, the solution contains only water and the salt formed by the reaction, which is NaCl in this case. The pH at the equivalence point is 7, as it represents a neutral solution.

Near the equivalence point, the pH changes rapidly. This is because small additions of NaOH significantly alter the concentration of H+ ions. To calculate the pH in the buffer region before the equivalence point, the Henderson-Hasselbalch equation is used:

`\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]`

In this case, since HCl is a strong acid, the concentration of its conjugate base (A⁻ ) is negligible. Therefore, the pH is mainly determined by the concentration of HCl and NaOH before the equivalence point. The rapid change in pH occurs around the equivalence point, making the titration curve steep.