Final answer:
To calculate the initial pH of a 0.280 M LiOH solution, find the pOH using the concentration of OH- and then subtract it from 14.
Step-by-step explanation:
The question involves calculating the initial pH of a Lithium Hydroxide (LiOH) solution before the titration begins.
Since the solution is of a strong base (LiOH), we can directly find the concentration of OH- ions and use it to calculate the pH.
The initial molarity of LiOH is provided as 0.280 M.
The formula for pH in the case of strong bases is pH = 14 - pOH, where pOH = -log[OH-].
In this case, since no HI has been added yet (titrant volume = 0 mL), the concentration of OH- ions is equal to the concentration of LiOH, which is 0.280 M. Thus:
pOH = -log(0.280) and then
pH = 14 - pOH