The 100. mL of buffer solution is 0.05 M in C2H3O2- and 0.10 M in HC2H3O2. What is the resulting pH when 20.0 mL of 0.1 M HCl is added to the buffer? Ka of HC2H3O2 = 1.8×10-5 pH after strong acid added = [?] pH of Soln w/ Acid Enter
The 100. mL of buffer solution is 0.05 M in C2H3O2- and 0.10 M in HC2H3O2. What is the resulting pH when 20.0 mL of 0.1 M HCl is added to the buffer? Ka of HC2H3O2 = 1.8×10-5 pH after strong acid added = [?] pH of Soln w/ Acid Enter
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![### Buffer Solution pH Calculation
**Problem Statement:**
100 mL of buffer solution contains 0.05 M in C₂H₃O₂⁻ and 0.10 M in HC₂H₃O₂. What is the resulting pH when 20.0 mL of 0.1 M HCl is added to the buffer?
1. **Given Data:**
- Buffer Solution Volume: 100 mL
- Concentration of C₂H₃O₂⁻: 0.05 M
- Concentration of HC₂H₃O₂: 0.10 M
- Added HCl Volume: 20.0 mL
- Concentration of HCl: 0.1 M
- Acid Dissociation Constant (Kₐ) for HC₂H₃O₂: \( 1.8 \times 10^{-5} \)
2. **Calculation Goal:**
Determine the resulting pH after the addition of the strong acid (HCl) to the buffer solution.
### Steps to Calculate pH After Adding Strong Acid:
1. **Calculate the Initial Moles:**
- Moles of C₂H₃O₂⁻: \( 0.05 \times 0.1 \) = 0.005 moles
- Moles of HC₂H₃O₂: \( 0.10 \times 0.1 \) = 0.010 moles
2. **Calculate the Moles of HCl Added:**
- Moles of HCl: \( 0.1 \times 0.02 \) = 0.002 moles
3. **Reaction of HCl with C₂H₃O₂⁻:**
HCl will react with C₂H₃O₂⁻ to form HC₂H₃O₂.
\[ C₂H₃O₂⁻ + HCl \rightarrow HC₂H₃O₂ \]
- Moles of C₂H₃O₂⁻ Left: 0.005 - 0.002 = 0.003 moles
- Moles of HC₂H₃O₂ Formed: 0.010 + 0.002 = 0.012 moles
4. **Calculate the New Concentr](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F74ff972a-42b6-478c-a012-297da9a07acf%2F5a17d719-5828-470b-93b1-ace33c2a7604%2F4ovj8ce_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Buffer Solution pH Calculation
**Problem Statement:**
100 mL of buffer solution contains 0.05 M in C₂H₃O₂⁻ and 0.10 M in HC₂H₃O₂. What is the resulting pH when 20.0 mL of 0.1 M HCl is added to the buffer?
1. **Given Data:**
- Buffer Solution Volume: 100 mL
- Concentration of C₂H₃O₂⁻: 0.05 M
- Concentration of HC₂H₃O₂: 0.10 M
- Added HCl Volume: 20.0 mL
- Concentration of HCl: 0.1 M
- Acid Dissociation Constant (Kₐ) for HC₂H₃O₂: \( 1.8 \times 10^{-5} \)
2. **Calculation Goal:**
Determine the resulting pH after the addition of the strong acid (HCl) to the buffer solution.
### Steps to Calculate pH After Adding Strong Acid:
1. **Calculate the Initial Moles:**
- Moles of C₂H₃O₂⁻: \( 0.05 \times 0.1 \) = 0.005 moles
- Moles of HC₂H₃O₂: \( 0.10 \times 0.1 \) = 0.010 moles
2. **Calculate the Moles of HCl Added:**
- Moles of HCl: \( 0.1 \times 0.02 \) = 0.002 moles
3. **Reaction of HCl with C₂H₃O₂⁻:**
HCl will react with C₂H₃O₂⁻ to form HC₂H₃O₂.
\[ C₂H₃O₂⁻ + HCl \rightarrow HC₂H₃O₂ \]
- Moles of C₂H₃O₂⁻ Left: 0.005 - 0.002 = 0.003 moles
- Moles of HC₂H₃O₂ Formed: 0.010 + 0.002 = 0.012 moles
4. **Calculate the New Concentr
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