Answer :
Answer:
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
The pH of the solution is 10.61
Data;
- volume of LiOH = 25.0mL= 0.025L
- conc. of LiOH = 0.1M
- volume of LiOH = 20.92mL =0.02092L
- conc. of HI = 0.10M
pH of the solution
To calculate the pH of this solution, we have to find the excess moles of LiOH that reacted in the solution since it is a very strong base compared to HI.
The number of moles of LiOH can be found by multiplying the volume by the concentration of the base.
[tex]n = C*V \\n = 0.1*0.025 = 0.0025moles[/tex]
The number of moles of HI is
[tex]n = C*V\\n = 0.1*0.02092 = 0.002092moles[/tex]
The excess moles after reacting is
[tex]0.0025 - 0.002092 = 0.000408 moles[/tex]
Since this is the concentration of the base [OH-]
[tex][OH^-]=0.00408M[/tex]
The pOH of the solution is calculated as
[tex]pOH = -log[OH^-]\\pOH= -log[0.000408] = 3.39\\[/tex]
but pH + pOH = 14
[tex]pH + pOH = 14\\pH = 14 - pOH\\pH = 14 - 3.39\\pH = 10.61[/tex]
The pH of the solution is 10.61
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