A Chemistry student is asked to predict the pH of a potassium hydroxide solution when 1.82 g of potassium hydroxide is dissolved in 250 mL of water. He shows the steps of his calculations as follows: 1KOH == n 1.82 g 56.11 g/mol 0.032 43...mol KOH n 0.03243...mol = Скон = 0.000 129 7...mol/L KOH V 250 mL [KOH(aq)] = [OH(aq)] because it is a 1:1 ratio and 100% dissociation [OH (aq)] = 0.000 129 7...mol/L pH = -log[OH(aq)] = -log(0.000 129 7...mol/L) = 3.8869... pH = 3.897 Identify the three mistakes that the student made in his calculations. Correct the mistakes and provide the correct answer.

Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter14: Acids And Bases
Section: Chapter Questions
Problem 127QRT
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A Chemistry student is asked to predict the pH of a potassium hydroxide solution when
1.82 g of potassium hydroxide is dissolved in 250 mL of water. He shows the steps of his
calculations as follows:
n 1KOH ==
1.82 g
56.11 g/mol
0.032 43...mol KOH
n
0.03243...mol
=
CKOH=
0.000 129 7...mol/L KOH
V
250 mL
[KOH(aq)] = [OH(aq)] because it is a 1:1 ratio and 100% dissociation
[OH(aq)] = 0.000 129 7...mol/L
pH = -log[OH(aq)] = -log(0.000 129 7...mol/L) = 3.8869...
pH = 3.897
Identify the three mistakes that the student made in his calculations. Correct the mistakes and
provide the correct answer.
Transcribed Image Text:A Chemistry student is asked to predict the pH of a potassium hydroxide solution when 1.82 g of potassium hydroxide is dissolved in 250 mL of water. He shows the steps of his calculations as follows: n 1KOH == 1.82 g 56.11 g/mol 0.032 43...mol KOH n 0.03243...mol = CKOH= 0.000 129 7...mol/L KOH V 250 mL [KOH(aq)] = [OH(aq)] because it is a 1:1 ratio and 100% dissociation [OH(aq)] = 0.000 129 7...mol/L pH = -log[OH(aq)] = -log(0.000 129 7...mol/L) = 3.8869... pH = 3.897 Identify the three mistakes that the student made in his calculations. Correct the mistakes and provide the correct answer.
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