A) The Ka for acetic acid is 1.8X105. If the initial concentration of acetic acid is 0.012 M. Calculate the [H₂O*] coming ONLY from acetic acid using an ice table. Show all steps of calculation. Reaction stage [CH₂COOH] (M) [CH3COO] (M) [H₂O¹] (M) 0 0.012 Initial Change Equilibrium K. (acetic acid) = 1.8x10-5 [H₂O*] = 0 B) On question 1B you calculated the [H3O+] coming only from autoproteolysis of water. That value is . How does this number compare with the [H3O+] coming only from acetic acid as calculated above? 10-7 C) Calculate pH of this 0.012 M acetic acid in water considering the hydronium generated from the acid as well as the solvent. Where pH = -log[H3O*]. D) Calculate pH of the solution containing 0.012 M acetic acid in water considering the hydronium generated ONLY from the acid. Where pH = -log[H3O*]. E) Most common pH meters can only read two decimal places. Based on this information, would you for all practical purpose consider the hydronium concentration coming from water when considering pH of a strong acid? Explain.
A) The Ka for acetic acid is 1.8X105. If the initial concentration of acetic acid is 0.012 M. Calculate the [H₂O*] coming ONLY from acetic acid using an ice table. Show all steps of calculation. Reaction stage [CH₂COOH] (M) [CH3COO] (M) [H₂O¹] (M) 0 0.012 Initial Change Equilibrium K. (acetic acid) = 1.8x10-5 [H₂O*] = 0 B) On question 1B you calculated the [H3O+] coming only from autoproteolysis of water. That value is . How does this number compare with the [H3O+] coming only from acetic acid as calculated above? 10-7 C) Calculate pH of this 0.012 M acetic acid in water considering the hydronium generated from the acid as well as the solvent. Where pH = -log[H3O*]. D) Calculate pH of the solution containing 0.012 M acetic acid in water considering the hydronium generated ONLY from the acid. Where pH = -log[H3O*]. E) Most common pH meters can only read two decimal places. Based on this information, would you for all practical purpose consider the hydronium concentration coming from water when considering pH of a strong acid? Explain.
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Question
Please help with A, B, C, D, and E
![CH3COOH (I) + H₂O (1)→ H₂O* (aq) + CH3COO (aq)
K. (acetic acid) =
A) The Ka for acetic acid is 1.8X105. If the initial concentration of acetic acid is 0.012 M. Calculate
the [H₂O*] coming ONLY from acetic acid using an ice table. Show all steps of calculation.
Reaction stage
[CH₂COOH] (M)
[CH3COO] (M)
[H₂O¹] (M)
0
Initial
0.012
Change
Equilibrium
K. (acetic acid) = 1.8x10-5
[H3O+] =
0
B) On question 1B you calculated the [H3O+] coming only from autoproteolysis of water. That value
is
-1
10°
. How does this number compare with the [H3O+] coming only from acetic
acid as calculated above?
C) Calculate pH of this 0.012 M acetic acid in water considering the hydronium generated from the
acid as well as the solvent. Where pH = -log[H3O*].
D) Calculate pH of the solution containing 0.012 M acetic acid in water considering the hydronium
generated ONLY from the acid. Where pH = -log[H3O+].
E) Most common pH meters can only read two decimal places. Based on this information, would you
for all practical purpose consider the hydronium concentration coming from water when
considering pH of a strong acid? Explain.
3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa3b13a71-cd62-4da6-9e1d-f686bcc45db9%2F654482df-08db-4c25-8d9f-81f7a85ac024%2Fw8jscpk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:CH3COOH (I) + H₂O (1)→ H₂O* (aq) + CH3COO (aq)
K. (acetic acid) =
A) The Ka for acetic acid is 1.8X105. If the initial concentration of acetic acid is 0.012 M. Calculate
the [H₂O*] coming ONLY from acetic acid using an ice table. Show all steps of calculation.
Reaction stage
[CH₂COOH] (M)
[CH3COO] (M)
[H₂O¹] (M)
0
Initial
0.012
Change
Equilibrium
K. (acetic acid) = 1.8x10-5
[H3O+] =
0
B) On question 1B you calculated the [H3O+] coming only from autoproteolysis of water. That value
is
-1
10°
. How does this number compare with the [H3O+] coming only from acetic
acid as calculated above?
C) Calculate pH of this 0.012 M acetic acid in water considering the hydronium generated from the
acid as well as the solvent. Where pH = -log[H3O*].
D) Calculate pH of the solution containing 0.012 M acetic acid in water considering the hydronium
generated ONLY from the acid. Where pH = -log[H3O+].
E) Most common pH meters can only read two decimal places. Based on this information, would you
for all practical purpose consider the hydronium concentration coming from water when
considering pH of a strong acid? Explain.
3
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Step 1: Interpretation of the problem
VIEWStep 2: Part A) Calculation of [H3O+] from ICE table
VIEWStep 3: Part B) Comparison of [H3O+] of water and acetic acid solution
VIEWStep 4: Part C) Calculating pH using [H3O+] generated from acid as well as solvent
VIEWStep 5: Part D) Calculating pH using [H3O+] generated from acid only
VIEWStep 6: Part E) pH of a strong acid
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