Understanding the Mixture
In this scenario, we are mixing 500 ml of 0.1 M acetic acid (CH3COOH) with 250 ml of 0.1 M sodium hydroxide (NaOH). This solution will undergo a neutralization reaction.
Calculating Moles of Reactants
- Moles of acetic acid:
0.1 M × 0.5 L = 0.05 moles
- Moles of sodium hydroxide:
0.1 M × 0.25 L = 0.025 moles
Neutralization Reaction
The balanced equation for the reaction is:
CH3COOH + NaOH → CH3COONa + H2O
- Acetic acid reacts with sodium hydroxide in a 1:1 ratio.
- Since we have 0.05 moles of acetic acid and 0.025 moles of NaOH, the NaOH will be the limiting reagent.
Determining Remaining Species
- After the reaction:
- Moles of CH3COOH remaining = 0.05 - 0.025 = 0.025 moles
- Moles of CH3COONa formed = 0.025 moles
Final Concentrations
The total volume after mixing is:
500 ml + 250 ml = 750 ml = 0.75 L
- Concentration of CH3COOH:
0.025 moles / 0.75 L = 0.0333 M
- Concentration of CH3COONa:
0.025 moles / 0.75 L = 0.0333 M
Calculating pH using Henderson-Hasselbalch Equation
Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
- pKa of acetic acid: 4.47
- [A-] = [CH3COONa] = 0.0333 M
- [HA] = [CH3COOH] = 0.0333 M
pH = 4.47 + log(0.0333/0.0333)
pH = 4.47 + log(1)
pH = 4.47
Conclusion
The pH of the resulting mixture is 4.47, indicating a buffer solution due to the presence of both acetic acid and its conjugate base, acetate.