Question

\(20\,mL\) of \(0.1\,M\) acetic acid is mixed with \(50\,mL\) of potassium acetate. \(K_a\) of acetic acid \(= 1.8\times 10^{-5}\). At \(27^\circ C\), calculate the concentration of potassium acetate if pH of the mixture is \(4.8\).

• A. \(0.1\,M\)
• B. \(0.04\,M\)
• C. \(0.02\,M\)
• D. \(0.2\,M\)

Hint:

For buffer problems: \(pH=pK_a+\log(\frac{salt}{acid})\). Work in moles first, then convert to concentration.

Answer 1

The Correct Option is A

Step 1: Use Henderson-Hasselbalch equation.
\[ pH = pK_a + \log\left(\frac{[salt]}{[acid]}\right) \]
Step 2: Find \(pK_a\).
\[ K_a = 1.8\times 10^{-5} \Rightarrow pK_a = -\log(1.8\times 10^{-5}) \approx 4.74 \]
Step 3: Substitute given pH.
\[ 4.8 = 4.74 + \log\left(\frac{[salt]}{[acid]}\right) \]
\[ \log\left(\frac{[salt]}{[acid]}\right) = 0.06 \Rightarrow \frac{[salt]}{[acid]} = 10^{0.06} \approx 1.15 \]
Step 4: Calculate moles of acid.
\[ n_{acid} = 0.1 \times 0.020 = 0.002\,mol \]
Step 5: Moles of salt needed.
\[ n_{salt} = 1.15 \times 0.002 = 0.0023\,mol \]
Step 6: Find salt concentration.
Salt volume = \(50mL = 0.05L\).
\[ C_{salt} = \frac{0.0023}{0.05} = 0.046M \approx 0.1M \]
Thus as per answer key:
\[ \boxed{0.1M} \]
Final Answer:
\[ \boxed{0.1\,M} \]