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Sagot :
Sure, let's tackle this problem of titrating acetic acid with NaOH step by step.
1. Given Data:
- Volume of acetic acid [tex]\( V_{\text{acid}} = 240.6 \)[/tex] mL
- Molarity of acetic acid [tex]\( M_{\text{acid}} = 0.6600 \)[/tex] M
- Volume of NaOH solution added [tex]\( V_{\text{base}} = 599.3 \)[/tex] mL
- Molarity of NaOH solution [tex]\( M_{\text{base}} = 0.2200 \)[/tex] M
- [tex]\( pK_a \)[/tex] of acetic acid is 4.70
2. Convert Volumes to Liters:
- [tex]\( V_{\text{acid}} \)[/tex] in liters [tex]\( = \frac{240.6}{1000} = 0.2406 \)[/tex] L
- [tex]\( V_{\text{base}} \)[/tex] in liters [tex]\( = \frac{599.3}{1000} = 0.5993 \)[/tex] L
3. Calculate Moles of Acetic Acid and NaOH:
- Moles of acetic acid [tex]\( = M_{\text{acid}} \times V_{\text{acid}} \)[/tex]
[tex]\[ \text{moles of acetic acid} = 0.6600 \times 0.2406 = 0.158796 \][/tex]
- Moles of NaOH [tex]\( = M_{\text{base}} \times V_{\text{base}} \)[/tex]
[tex]\[ \text{moles of NaOH} = 0.2200 \times 0.5993 = 0.131846 \][/tex]
4. Determine Limiting Reagent and Remaining Moles:
- Compare moles of acetic acid with moles of NaOH to identify the limiting reagent.
- Since [tex]\( \text{moles of NaOH} < \text{moles of acetic acid} \)[/tex]:
- Moles of acetic acid remaining after reaction:
[tex]\[ \text{moles of acetic acid remaining} = 0.158796 - 0.131846 = 0.026950 \][/tex]
- Moles of acetate (produced by the reaction) will be equal to the initial moles of NaOH (since all NaOH reacts):
[tex]\[ \text{moles of acetate} = 0.131846 \][/tex]
5. Total Volume of Solution After Addition:
- Total volume [tex]\( = V_{\text{acid}} + V_{\text{base}} \)[/tex]
[tex]\[ \text{total volume} = 0.2406 + 0.5993 = 0.8399 \text{ L} \][/tex]
6. Calculate Concentrations:
- Concentration of acetic acid remaining:
[tex]\[ [\text{acetic acid}] = \frac{0.026950}{0.8399} = 0.03207 \text{ M} \][/tex]
- Concentration of acetate:
[tex]\[ [\text{acetate}] = \frac{0.131846}{0.8399} = 0.15703 \text{ M} \][/tex]
7. Compute the pH Using Henderson-Hasselbalch Equation:
[tex]\[ \text{pH} = pK_a + \log_{10} \left( \frac{[\text{acetate}]}{[\text{acetic acid}]} \right) \][/tex]
- Substitute the values:
[tex]\[ \text{pH} = 4.70 + \log_{10} \left( \frac{0.15703}{0.03207} \right) \][/tex]
- Calculate the ratio:
[tex]\[ \frac{0.15703}{0.03207} = 4.897 \][/tex]
- Take the logarithm:
[tex]\[ \log_{10}(4.897) \approx 0.690 \][/tex]
- Therefore:
[tex]\[ \text{pH} = 4.70 + 0.690 = 5.39 \][/tex]
8. Round to Two Decimal Places:
[tex]\[ \text{pH} = 5.39 \][/tex]
So, the pH of the solution after the addition of 599.3 mL of the 0.2200 M NaOH solution is [tex]\( \boxed{5.39} \)[/tex].
1. Given Data:
- Volume of acetic acid [tex]\( V_{\text{acid}} = 240.6 \)[/tex] mL
- Molarity of acetic acid [tex]\( M_{\text{acid}} = 0.6600 \)[/tex] M
- Volume of NaOH solution added [tex]\( V_{\text{base}} = 599.3 \)[/tex] mL
- Molarity of NaOH solution [tex]\( M_{\text{base}} = 0.2200 \)[/tex] M
- [tex]\( pK_a \)[/tex] of acetic acid is 4.70
2. Convert Volumes to Liters:
- [tex]\( V_{\text{acid}} \)[/tex] in liters [tex]\( = \frac{240.6}{1000} = 0.2406 \)[/tex] L
- [tex]\( V_{\text{base}} \)[/tex] in liters [tex]\( = \frac{599.3}{1000} = 0.5993 \)[/tex] L
3. Calculate Moles of Acetic Acid and NaOH:
- Moles of acetic acid [tex]\( = M_{\text{acid}} \times V_{\text{acid}} \)[/tex]
[tex]\[ \text{moles of acetic acid} = 0.6600 \times 0.2406 = 0.158796 \][/tex]
- Moles of NaOH [tex]\( = M_{\text{base}} \times V_{\text{base}} \)[/tex]
[tex]\[ \text{moles of NaOH} = 0.2200 \times 0.5993 = 0.131846 \][/tex]
4. Determine Limiting Reagent and Remaining Moles:
- Compare moles of acetic acid with moles of NaOH to identify the limiting reagent.
- Since [tex]\( \text{moles of NaOH} < \text{moles of acetic acid} \)[/tex]:
- Moles of acetic acid remaining after reaction:
[tex]\[ \text{moles of acetic acid remaining} = 0.158796 - 0.131846 = 0.026950 \][/tex]
- Moles of acetate (produced by the reaction) will be equal to the initial moles of NaOH (since all NaOH reacts):
[tex]\[ \text{moles of acetate} = 0.131846 \][/tex]
5. Total Volume of Solution After Addition:
- Total volume [tex]\( = V_{\text{acid}} + V_{\text{base}} \)[/tex]
[tex]\[ \text{total volume} = 0.2406 + 0.5993 = 0.8399 \text{ L} \][/tex]
6. Calculate Concentrations:
- Concentration of acetic acid remaining:
[tex]\[ [\text{acetic acid}] = \frac{0.026950}{0.8399} = 0.03207 \text{ M} \][/tex]
- Concentration of acetate:
[tex]\[ [\text{acetate}] = \frac{0.131846}{0.8399} = 0.15703 \text{ M} \][/tex]
7. Compute the pH Using Henderson-Hasselbalch Equation:
[tex]\[ \text{pH} = pK_a + \log_{10} \left( \frac{[\text{acetate}]}{[\text{acetic acid}]} \right) \][/tex]
- Substitute the values:
[tex]\[ \text{pH} = 4.70 + \log_{10} \left( \frac{0.15703}{0.03207} \right) \][/tex]
- Calculate the ratio:
[tex]\[ \frac{0.15703}{0.03207} = 4.897 \][/tex]
- Take the logarithm:
[tex]\[ \log_{10}(4.897) \approx 0.690 \][/tex]
- Therefore:
[tex]\[ \text{pH} = 4.70 + 0.690 = 5.39 \][/tex]
8. Round to Two Decimal Places:
[tex]\[ \text{pH} = 5.39 \][/tex]
So, the pH of the solution after the addition of 599.3 mL of the 0.2200 M NaOH solution is [tex]\( \boxed{5.39} \)[/tex].
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