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

r just means there's a number of carbon in teh chain. for example rcooh means that after x number of carbon, the chain ends with cooh.

explanation:

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