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 ireversible and goes to completion: HCH3O2 (ag) + OH" (ag) → H2O () + C2H3O2 (ag) Moles before reaction Change in moles Moles remaining after reaction

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Original concentration of Acetic Acid was 0.1033 M and NaOH was 0.1104 M

there is part a done, need to do part B

A. Calculate the pH of 35.00 mL of acetic acid using the concentration assigned in Beyond Labz:
HC2H3O2 (aa)
2 H* (aa)
+ C2H3O2 (aa)
I
Initial Concentration (M) 0.1033
Change in Concentration (M)
+x
+x
-X-
Equilibrium Concentration (M) 0.1033 - x
E
Equilibrium Concentration,
Calculated (M)
0.1020
1.36x10-3
1.36x10-3
1. Initial concentration [HC2H3O2]0=0.1033 M
2. Ka symbolic expression: [H*][ C2H3O2 ]
[HC2H3O2]
(x)(x)
3. Ka numerical equation with x: 1.8 x 10-S =
0.1033-x
4. Quadratic Formula Method
a. Quadratic equation: x2+ 1.8 x 10-S – 1.86 x 10-6 = 0
-1.8 x 10A-5t/(-1.8x 10)-4(1)(1.86 x 10A-6)
b.Quadratic formula: x =
2(1)
5. Assumption (5% rule) Method
(1.8x10)(0.1033)
a. Equation for x: x =
= 1.36x10-°M
b.% ionization = (1.36x10^-3M)(0.1033M)x100 = 1.3%
6. Equilibrium concentrations
a. [HC2H3O2lm =0.1033 - 1.36x103 = 0.1020M
b.[H* Jeg =1.36x10-3 M
c.[C2H3O2] =1.36x103 M
7. pH = - log(1.36x10-3 ) = 2.8
Transcribed Image Text:A. Calculate the pH of 35.00 mL of acetic acid using the concentration assigned in Beyond Labz: HC2H3O2 (aa) 2 H* (aa) + C2H3O2 (aa) I Initial Concentration (M) 0.1033 Change in Concentration (M) +x +x -X- Equilibrium Concentration (M) 0.1033 - x E Equilibrium Concentration, Calculated (M) 0.1020 1.36x10-3 1.36x10-3 1. Initial concentration [HC2H3O2]0=0.1033 M 2. Ka symbolic expression: [H*][ C2H3O2 ] [HC2H3O2] (x)(x) 3. Ka numerical equation with x: 1.8 x 10-S = 0.1033-x 4. Quadratic Formula Method a. Quadratic equation: x2+ 1.8 x 10-S – 1.86 x 10-6 = 0 -1.8 x 10A-5t/(-1.8x 10)-4(1)(1.86 x 10A-6) b.Quadratic formula: x = 2(1) 5. Assumption (5% rule) Method (1.8x10)(0.1033) a. Equation for x: x = = 1.36x10-°M b.% ionization = (1.36x10^-3M)(0.1033M)x100 = 1.3% 6. Equilibrium concentrations a. [HC2H3O2lm =0.1033 - 1.36x103 = 0.1020M b.[H* Jeg =1.36x10-3 M c.[C2H3O2] =1.36x103 M 7. pH = - log(1.36x10-3 ) = 2.8
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:
HC2H3O2 (aa)
+ ОH (ag)
→H2O ()
+ C2H3O2 (aa)
Moles before reaction
Change in moles
Moles remaining after reaction
Then equilibrium is established
HCH3O2 (aa)
2H* (ag)
+ C2H3O2 (aa)
I
Initial Concentration (M)
Change in Concentration (M)
Equilibrium Concentration (M)
E
Equilibrium Concentration,
Calculated (M)
Transcribed Image Text: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: HC2H3O2 (aa) + ОH (ag) →H2O () + C2H3O2 (aa) Moles before reaction Change in moles Moles remaining after reaction Then equilibrium is established HCH3O2 (aa) 2H* (ag) + C2H3O2 (aa) I Initial Concentration (M) Change in Concentration (M) Equilibrium Concentration (M) E Equilibrium Concentration, Calculated (M)
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