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

linda: 3 ; laura: 15

step-by-step explanation:

note that linda + laura's age is 18, and that linda is 12 years younger than her sister.

let "linda" = x , "laura" = y:

x + y = 18

x = y - 12

first, plug in y - 12 for x in the first equation:

(y - 12) + y = 18

simplify.

2y - 12 = 18

isolate the variable (y). note the equal sign, what you do to one side, you do to the other. do the opposite of pemdas. first add, then divide.

2y - 12 (+12) = 18 (+12)

2y = 18 + 12

2y = 30

(2y)/2 = (30)/2

y = 30/2

y = 15

next, plug in 15 for y in the second equation (the other equation would work too).

x = y - 12

x = (15) - 12

x = 3

linda's age is 3, laura's age is 15.

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