1) A 60.0 mL sample of 0.18 M HClO 4 is titrated with 0.27 M LiOH. Determine the
ID: 994642 • Letter: 1
Question
1) A 60.0 mL sample of 0.18 M HClO4 is titrated with 0.27 M LiOH. Determine the pH of the solution before the addition of any LiOH.
-> [H+] = 0.18
--> pH = -log[H+]
---> pH = -log(0.18M) = 0.74 pH (correct)
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2) Determine the pH of the solution after the addition of 15.0 mL LiOH.
-> mol [OH-] after tritration
--> (0.27M x 0.06L) - (0.18M x 0.06L) = 0.0054 mol
-> concentration [OH-] = mol / total volume = (0.0054 / 0.06L + 0.015L) = 0.072 M [OH-]
-> pOH = -log(0.072M) = 1.14 -> pH = 14 - 1.14 = 12.85 pH (incorrect)
Answers listed on MC: 0.57 [incorrect], 1.57, 1.74, 1.05, 0.74 ? which is it?
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3) Determine the pH of the solution after the addition of 40.0 mL LiOH.
Answers listed on MC: 12.11 [incorrect], 5.42, 7.00, 1.52, 1.07 ? which is it?
Explanation / Answer
1.
before any addition
pH = -log(H+)
[H+] = 0.18
pH = -log(0.18) = 0.74472
2.
mmol of acid = MV = 60*0.18 = 10.8
mmol of base = MV = 0.27*15 = 4.05
mmol of acid left = 10.8-4.05 = 6.75
[H+] = mmol/V = 6.75/(60+15) = 0.09
pH = -log(0.09) = 1.04575
3.
after 40 mL of base
mmol of acid = MV = 60*0.18 = 10.8
mmol of base = MV = 0.27*40= 10.8
this is complete neutralization
so pH = 7
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