Exercise Construct a titration curve by plotting pH vs. VNaOH added.HCl(aq) + NaOH(aq)H₂O(1) + NaCl(aq)Volume NaOH (mL) [H'] pH0.005.0010.0015.0020.0025.0030.0035.0040.0045.0048.0050.0052.0055.0060.0065.0070.0075.0080.0085.0090951001×107Hints:The pH of the solution is calculated using the following equations:[H]=(a) Initial conditions (VNaOH = 0), [H] = [HCI](b) After initial addition of NaOH and before the end point:[H]=MHCI VHCI-MVHCI + V NaOH1x10-14XV.titrant: NaOH, 0.1M(c) At the end point (VNaOH = 50 mL), all the HCI is being consumed and the acidityof the solution comes from auto ionization of water which is:[H"]=1x107(d) After the end point:Moanalyte: HC1, 0.1 M, 50 ml.NaOH *V NaOHNaOffV +V₂C²NaOH

This is an acid-base titration. The table for graph construction is as follows

Answered: Construct a titration curve by plotting…

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Chemistry

Construct a titration curve by plotting pH vs. VNªOH added.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

Question

Exercise Construct a titration curve by plotting pH vs. VNaOH added.
HCl(aq) + NaOH(aq)
H₂O(1) + NaCl(aq)
Volume NaOH (mL) [H'] pH
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
40.00
45.00
48.00
50.00
52.00
55.00
60.00
65.00
70.00
75.00
80.00
85.00
90
95
100
1×107
Hints:
The pH of the solution is calculated using the following equations:
[H]=
(a) Initial conditions (VNaOH = 0), [H] = [HCI]
(b) After initial addition of NaOH and before the end point:
[H]=
MHCI VHCI-M
VHCI + V NaOH
1x10-14
XV.
titrant: NaOH, 0.1M
(c) At the end point (VNaOH = 50 mL), all the HCI is being consumed and the acidity
of the solution comes from auto ionization of water which is:
[H"]=1x107
(d) After the end point:
Mo
analyte: HC1, 0.1 M, 50 ml.
NaOH *V NaOH
NaOff
V +V₂C²
NaOH

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