Chapter 14: Problem 59
Consider the titration of \(100.0 \mathrm{mL}\) of \(0.200 M\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) by \(0.100 M\) KOH. Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of KOH have been added. a. \(0.0 \mathrm{mL}\) b. \(50.0 \mathrm{mL}\) c. \(100.0 \mathrm{mL}\) d. \(40.0 \mathrm{mL}\) e. \(50.0 \mathrm{mL}\) f. \(100.0 \mathrm{mL}\)
Short Answer
Step by step solution
Find initial moles of acetic acid
Case a: 0.0 mL of KOH added
Case b: 50.0 mL of KOH added
Case c: 100.0 mL of KOH added
Case d: 40.0 mL of KOH added
Case e: 50.0 mL of KOH added
Case f: 100.0 mL of KOH added
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Acetic Acid
Henderson-Hasselbalch Equation
pH Calculation
KOH Titration
- Initially, adding a small amount of KOH results in a slight increase in acetate ions, starting the buffer action of the acetic acid.
- As more KOH is added, more acetic acid is neutralized, increasing the solution's pH towards neutrality or above, depending on the volume of KOH added.
- The point at which equal moles of base and acid have reacted marks the equivalence point, where the amount of remaining acetic acid equals the product formed, acetate ion.