A titration is carried out for 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M of a strong base NaOH the titration curve is as shown. Calculate the pH at these volumes of added base solution:(a) 0.00 mL(b) 12.50 mL(c) 25.00 mL(d) 37.50 mL Titration of Strong AcidTitration of Weak Acid14141313121211111010Equivalencepoint pH, 8.7298.Equivalencepoint pH, 7.006.643210 5 10 15 20 25 30 35 40 45 50Volume of 0.100M NAOH added (mL)0 5 10 15 20 25 30 35 40 45 50Volume of 0.100 M NaOH added (mL)(a)(b)Figure 14.21 (a) The titration curve for the titration of 25.00 mL of 0.100 M HCI (strong acid) with 0.100 M NAOH(strong base) has an equivalence point of 7.00 pH. (b) The titration curve for the titration of 25.00 mL of 0.100 M HCI(strong acid) with 0.100 M NaOH (strong base) has an equivalence point of 8.72 pH.(O LO

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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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A titration is carried out for 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M of a strong base NaOH the titration curve is as shown. Calculate the pH at these volumes of added base solution:
(a) 0.00 mL
(b) 12.50 mL
(c) 25.00 mL
(d) 37.50 mL

Titration of Strong Acid
Titration of Weak Acid
14
14
13
13
12
12
11
11
10
10
Equivalence
point pH, 8.72
9
8.
Equivalence
point pH, 7.00
6.
6
4
3
2
1
0 5 10 15 20 25 30 35 40 45 50
Volume of 0.100M NAOH added (mL)
0 5 10 15 20 25 30 35 40 45 50
Volume of 0.100 M NaOH added (mL)
(a)
(b)
Figure 14.21 (a) The titration curve for the titration of 25.00 mL of 0.100 M HCI (strong acid) with 0.100 M NAOH
(strong base) has an equivalence point of 7.00 pH. (b) The titration curve for the titration of 25.00 mL of 0.100 M HCI
(strong acid) with 0.100 M NaOH (strong base) has an equivalence point of 8.72 pH.
(O LO

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