The following information is for questions 1 and 2. Consider the titration of 25.0 mL of 0.100 MHCI with 0.100 M NaOH. Calculate the pH after adding the following amounts of the NaOH solution: 1. 22.0 mL a) 1.27 b) 1.00 2.19 d) 3.52 e) 1.92
The following information is for questions 1 and 2. Consider the titration of 25.0 mL of 0.100 MHCI with 0.100 M NaOH. Calculate the pH after adding the following amounts of the NaOH solution: 1. 22.0 mL a) 1.27 b) 1.00 2.19 d) 3.52 e) 1.92
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter17: Principles Of Chemical Reactivity: Other Aspects Of Aqueous Equilibria
Section17.3: Acid-base Titrations
Problem 17.6CYU: The titration of 0.100 M acetic acid with 0.100 M NaOH is described in the text. What is the pH of...
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can someone explain what I did wrong. the correct answer is circled for number one, but I got 2.19 as the poh, not the pH
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NaOH + HC) ⇒ NaCl + H₂0
Dinitial moles of NaOH: 7500 × 0.1 = 0.0025 mol
1000
(a) moles of NaOH added: 22 x 0.1 = 0.0022 mal
1000
excess moles of NaOH:
G
0.0025-0.0022 = 0.0003 mol
Total volume = 25 mL + 22 = 47mL
0.047 L
-3
COH-] = exass mole - 0.0003 6013 X10
total val
0.047
рон
pOH = -log (OH-] = 10g [613X10²³] = 2.19
-p+) = 14-pOH = 14-2.19 - 11.81
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Transcribed Image Text:-
NaOH + HC) ⇒ NaCl + H₂0
Dinitial moles of NaOH: 7500 × 0.1 = 0.0025 mol
1000
(a) moles of NaOH added: 22 x 0.1 = 0.0022 mal
1000
excess moles of NaOH:
G
0.0025-0.0022 = 0.0003 mol
Total volume = 25 mL + 22 = 47mL
0.047 L
-3
COH-] = exass mole - 0.0003 6013 X10
total val
0.047
рон
pOH = -log (OH-] = 10g [613X10²³] = 2.19
-p+) = 14-pOH = 14-2.19 - 11.81
XID
Transcribed Image Text:The following information is for questions 1 and 2.
Consider the titration of 25.0 mL of 0.100 MHCI with 0.100 M NaOH. Calculate the
pH after adding the following amounts of the NaOH solution:
1. 22.0 mL
a) 1.27
b) 1.00
2.19
d) 3.52
e) 1.92
2. 25.0 mL
a) 2.30
b) 1.00
c) 2.60
d) 1.30
7.00
lor folubility of BaF2 in pure water. (Ksp for BaF2=2.45 x 10)
577²
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