Calculate the pH during the titration of 25.00 mL of 0.1000 M LiOH(aq) with 0.1000 M HI(aq) after 20.92 mL of the acid have been added.

pH = 11.95

Explanation:

The reaction of LiOH with HI -Reaction of strongs bases and acids- is:

LiOH + HI → H₂O + LiI

The reaction of LiOH and HI is 1:1

The moles of LiOH and HI are:

Moles LiOH:

0.025L * (0.100mol / L) = 0.0025moles LiOH

Moles HI:

0.02092L * (0.100mol / L) = 0.002092moles HI

As moles of LiOH > Moles HI, after the addition of the acid, some LiOH remains, that is:

0.0025 moles - 0.002092 moles = 4.08x10⁻⁴ moles of LiOH

These moles are in 25mL + 20.92mL = 45.92mL = 0.04592L:

4.08x10⁻⁴ moles of LiOH / 0.04592L = 8.885x10⁻³M of LiOH = [OH⁻]

As pOH = -log[OH⁻]

pOH = 2.05

And pH = 14 - pOH

pH = 11.95

Molarity is moles of solute over liters of solution. idk if that is what you are talking about, but you can convert moles of solute to grams of solute using dimensional analyzes by using molar mass found on the periodic table.

H tricky one let me think and get back to u