Final answer:
The pH at 10.0 mL of added base in the titration of 0.100 M HNO3 with 0.200 M NaOH is 12.60.
Step-by-step explanation:
In a strong acid-strong base titration, the pH at different volumes of added base can be calculated using the equation pH = 14 - pOH, where pOH = -log[OH-].
For the given titration of 50.0 mL of 0.100 M HNO3 with 0.200 M NaOH, the pH at 10.0 mL of added base can be calculated by finding the concentration of OH- at that point and then calculating the pOH and pH.
Since the concentration of NaOH is 0.200 M, the volume of added base at 10.0 mL is (0.2 M NaOH * 10.0 mL) / 50.0 mL = 0.04 M. Therefore, the pOH at 10.0 mL of added base is -log(0.04) = 1.40. The pH is then calculated as 14 - 1.40 = 12.60.