Answer :
The pH of a titration between 0.1 M HCl and 0.1 M NaOH changes through several stages: starting from pH 1.00, passing a pH close to 7.00 at the equivalence point, and becoming basic with excess NaOH. The pH is slightly higher than 4.74 when 0.1 M NaOH is added to an unbuffered solution with a pH of 4.74.
Calculating pH During a Titration
To calculate the pH of the solution during the titration of 0.1 M HCl with 0.1 M NaOH, we'll go through several stages:
- At the beginning: No NaOH added, the pH is that of 0.1 M HCl, which is 1.00 since pH = -log[H3O+], and [H3O+] = 0.1 M.
- 50 ml of 0.1 M NaOH added: This is the halfway point to the equivalence point, where [H3O+] = 0.05 M, so pH = -log(0.05) ≈ 1.30.
- 98 ml of 0.1 M NaOH added: Just before the equivalence point, with pH slightly above 1.30 due to the excess of NaOH.
- 99.9 ml of 0.1 M NaOH added: Almost at equivalence, the pH will be close to 7 as the solution is nearly neutralized.
- 100 ml of 0.1 M NaOH added: At the equivalence point, the pH should be 7.00 since HCl and NaOH neutralize each other completely and the resultant solution is of water and NaCl, which are neutral.
- 100.1 ml of 0.1 M NaOH added: Now NaOH is in excess, the pH will be above 7, determined by the concentration of the remaining NaOH.
Regarding the pH after 1.0 mL of 0.1 M NaOH is added to a 100 mL unbuffered solution with pH 4.74, the resulting pH will be slightly higher than 4.74 as the added NaOH will neutralize a small fraction of the H+ ions.
