B. Calculate the pH after 17.50 mL of NAOH is added to 35.00 mL of acetic acid using the concentration assigned in Beyond Labz: • First assume the acid-base neutralization reaction is irreversible and goes to completion: HC2H;O2 (ag) + OH° (ag) - H2O (1) + C;H;O;" (ag) Moles before reaction Change in moles Moles remaining after reaction • Then equilibrium is established HC;H;O2 (ag) 2H* (ag) + C;H;O2 (ag) Initial Concentration (M) Change in Concentration (M) Equilibrium Concentration (M) E Equilibrium Concentration, Calculated (M) Show your work here or replicate this on a separate piece of paper. Number and label each calculation, include units, and show the math performed to get your answer. Report answers with correct units and significant figures.
B. Calculate the pH after 17.50 mL of NAOH is added to 35.00 mL of acetic acid using the concentration assigned in Beyond Labz: • First assume the acid-base neutralization reaction is irreversible and goes to completion: HC2H;O2 (ag) + OH° (ag) - H2O (1) + C;H;O;" (ag) Moles before reaction Change in moles Moles remaining after reaction • Then equilibrium is established HC;H;O2 (ag) 2H* (ag) + C;H;O2 (ag) Initial Concentration (M) Change in Concentration (M) Equilibrium Concentration (M) E Equilibrium Concentration, Calculated (M) Show your work here or replicate this on a separate piece of paper. Number and label each calculation, include units, and show the math performed to get your answer. Report answers with correct units and significant figures.
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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Question
Acetic Acid = 0.1033 M
Ka = 1.8 x 10-5
Sodium hydroxide = 0.1104 M
The first picture are the data, if needed
second picture are the ones needs to be solved
![HC2H3O2
H+
+
C2H302-
Initial
0.1033
Change
Equilibrium
1. Initial concentration [HC2H3O2]o = 0.1033 M
-X
+X
+X
0.1033-x
X
[H*][C2H307]
2. Ka symbolic expression: Ka
[HC2H302]
x2
3. Ka numerical equation with x: 1.8 x 10-5
0.1033-x
4. Quadratic formula method
a. quadratic equation: x2 + 1.8x10-5x – 1.86x10-6 = 0
%3D
-1.8x10-5+V (1.8x10-5)2-4(1)(-1.86×10-6)
b. quadratic formula: x =
2(1)
x = 1.35x10-3 M; -1.37x103 M (only positive root is used)
5. Assumption (5% rule) method
a. equation for x: x =
V(1.8 x 10-5)(0.1033) = 1.36 × 10-3M
1.36x10-3 M
b. % ionization =
x 100 = 1.32%
0.1033 M
6. Equilibrium concentrations
a. [HC2H3O2leg = 0.1033 – 1.35x10-3 = 0.1020 M
%3D
%3D
b. [H*]eg = 1.35x10-3 M
Jeq
c. [C2H3O2]eg = 1.35x10-3 M
7. pH = -log[H*] = -log(1.35x10 3) = 2.87](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F378a66cd-977d-4b08-83be-b0cd331b8abe%2Fab3c6acf-f131-4fcc-9773-ccae6bc9137c%2F9e5zgdb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:HC2H3O2
H+
+
C2H302-
Initial
0.1033
Change
Equilibrium
1. Initial concentration [HC2H3O2]o = 0.1033 M
-X
+X
+X
0.1033-x
X
[H*][C2H307]
2. Ka symbolic expression: Ka
[HC2H302]
x2
3. Ka numerical equation with x: 1.8 x 10-5
0.1033-x
4. Quadratic formula method
a. quadratic equation: x2 + 1.8x10-5x – 1.86x10-6 = 0
%3D
-1.8x10-5+V (1.8x10-5)2-4(1)(-1.86×10-6)
b. quadratic formula: x =
2(1)
x = 1.35x10-3 M; -1.37x103 M (only positive root is used)
5. Assumption (5% rule) method
a. equation for x: x =
V(1.8 x 10-5)(0.1033) = 1.36 × 10-3M
1.36x10-3 M
b. % ionization =
x 100 = 1.32%
0.1033 M
6. Equilibrium concentrations
a. [HC2H3O2leg = 0.1033 – 1.35x10-3 = 0.1020 M
%3D
%3D
b. [H*]eg = 1.35x10-3 M
Jeq
c. [C2H3O2]eg = 1.35x10-3 M
7. pH = -log[H*] = -log(1.35x10 3) = 2.87
Transcribed Image Text:B. Calculate the pH after 17.50 mL of N2OH is added to 35.00 mL of acetic acid using the concentration
assigned in Beyond Labz:
• First assume the acid-base neutralization reaction is irreversible and goes to completion:
田
HC;H;O2 (ag)
+ Он (ag)
→ H2O ()
+ C;H3O2 (aq)
Moles before reaction
Change in moles
Moles remaining after reaction
Then equilibrium is established
HC;H;O2 (ag)
2H* (ag)
+ C;H;O2 (aq)
I
Initial Concentration (M)
Change in Concentration (M)
Equilibrium Concentration (M)
E
Equilibrium Concentration,
Calculated (M)
Show your work here or replicate this on a separate piece of paper. Number and label each calculation, include
units, and show the math performed to get your answer. Report answers with correct units and significant
figures.
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