Final answer:
The initial pH of a 25.00 mL sample of 0.350 M NaOH before titration with 0.750 M HNO₃ is approximately 13.544, assuming complete dissociation of the strong base NaOH.
Step-by-step explanation:
The initial pH before any titrant is added in a titration of a 25.00 mL sample of 0.350 M NaOH with 0.750 M HNO₃ can be calculated assuming that NaOH is a strong base that completely dissociates in solution. Therefore, the initial concentration of OH- ions is equal to the concentration of NaOH.
The pOH of the solution can be calculated using the formula pOH = -log[OH-], and the pH can be found using the relationship pH + pOH = 14 at 25 °C. For 0.350 M NaOH, the pOH is -log(0.350), which is approximately 0.456. Since pH + pOH = 14, the initial pH is thus 14 - 0.456 = 13.544.