How would you prepare 100 mL of a 0.15 M acetate buffer at pH 5.0 using sodium acetate trihydrate crystalline (CH3COONa 3H₂O, MW = 136 g/mol) and a solution of 0.2 M HCl (pKa for acetic acid is 4.76)?
How would you prepare 100 mL of a 0.15 M acetate buffer at pH 5.0 using sodium acetate trihydrate crystalline (CH3COONa 3H₂O, MW = 136 g/mol) and a solution of 0.2 M HCl (pKa for acetic acid is 4.76)?
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![**Question:**
How would you prepare 100 mL of a 0.15 M acetate buffer at pH 5.0 using sodium acetate trihydrate crystalline (CH₃COONa·3H₂O, MW = 136 g/mol) and a solution of 0.2 M HCl (pKa for acetic acid is 4.76)?
**Explanation:**
To prepare the buffer solution, follow these steps:
1. **Calculate the ratio of [A-] (acetate) to [HA] (acetic acid) using the Henderson-Hasselbalch equation:**
\[
\text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
Given:
- pH = 5.0
- pKa = 4.76
Rearrange the equation to solve for the ratio:
\[
5.0 = 4.76 + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
\[
5.0 - 4.76 = \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
\[
0.24 = \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
\[
\frac{[\text{A}^-]}{[\text{HA}]} = 10^{0.24} \approx 1.737
\]
2. **Determine the total moles of buffer components:**
Since we need 100 mL of 0.15 M buffer:
\[
\text{Total moles} = 0.15 \text{ M} \times 0.1 \text{ L} = 0.015 \text{ moles}
\]
3. **Set up equations to find the moles of acetic acid (HA) and acetate (A-):**
Let \( x \) be the moles of HA (acetic acid), then:
\[
[\text{A}^-] = 1.737 \cdot](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc62819af-1f81-4d59-bfb3-1ca1702855ed%2F4d922054-54f7-4c45-bccd-cd31c5d3afb3%2F7e6ko7b_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
How would you prepare 100 mL of a 0.15 M acetate buffer at pH 5.0 using sodium acetate trihydrate crystalline (CH₃COONa·3H₂O, MW = 136 g/mol) and a solution of 0.2 M HCl (pKa for acetic acid is 4.76)?
**Explanation:**
To prepare the buffer solution, follow these steps:
1. **Calculate the ratio of [A-] (acetate) to [HA] (acetic acid) using the Henderson-Hasselbalch equation:**
\[
\text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
Given:
- pH = 5.0
- pKa = 4.76
Rearrange the equation to solve for the ratio:
\[
5.0 = 4.76 + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
\[
5.0 - 4.76 = \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
\[
0.24 = \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
\[
\frac{[\text{A}^-]}{[\text{HA}]} = 10^{0.24} \approx 1.737
\]
2. **Determine the total moles of buffer components:**
Since we need 100 mL of 0.15 M buffer:
\[
\text{Total moles} = 0.15 \text{ M} \times 0.1 \text{ L} = 0.015 \text{ moles}
\]
3. **Set up equations to find the moles of acetic acid (HA) and acetate (A-):**
Let \( x \) be the moles of HA (acetic acid), then:
\[
[\text{A}^-] = 1.737 \cdot
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