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°C. Calculate the concentration of potassium acetate if the pH of the mixture is 4.8.

• A. 0.1 M
• B. 0.04 M
• C. 0.03 M
• D. 0.02 M

Hint:

The Henderson-Hasselbalch equation is key for calculating the pH of buffer solutions.

Answer 1

The Correct Option is B

Step 1: {Henderson-Hasselbalch Equation} 
For an acidic buffer, the pH is given by: \[ {pH} = {p}K_a + \log \left( \frac{[{Salt}]}{[{Acid}]} \right) \] Step 2: {Concentrations} 
Let the concentration of potassium acetate solution be \( x \) M. \[ 20 { mL of 0.1 M acetic acid} = 20 \times 0.1 { millimol} = 2 { millimol} \] \[ 50 { mL of } x { M potassium acetate} = x \times 50 { millimol} = 50x { millimol} \] Step 3: {Substitute Values} 
Given, pH = 4.8 \[ 4.8 = {p}K_a + \log \left( \frac{50x}{2} \right) \] where \( {p}K_a = \log 1.8 \times 10^{-5} = 4.74 \) 
Step 4: {Solve for \( x \)} 
Substitute \( {p}K_a \): \[ 4.8 = 4.74 + \log \left( \frac{50x}{2} \right) \] \[ 0.06 = \log \left( 25x \right) \] \[ 10^{0.06} = 25x \] \[ x = \frac{10^{0.06}}{25} = 0.04 \] Thus, the correct answer is (B) 0.04 M. 
 

Answer 2

Step 1: Identify the type of solution
This is a buffer solution made by mixing a weak acid (acetic acid) with its conjugate base (potassium acetate).
We can apply the Henderson–Hasselbalch equation:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{salt}]}{[\text{acid}]} \right) \]

Step 2: Write the known values
- pH = 4.8
- \( K_a = 1.8 \times 10^{-5} \) ⇒ \( \text{p}K_a = -\log(1.8 \times 10^{-5}) \approx 4.74 \)
- Volume of acid = 20 mL = 0.020 L
- Molarity of acid = 0.1 M
⇒ Moles of acetic acid = 0.1 × 0.020 = 0.002 mol

Step 3: Apply the Henderson–Hasselbalch equation
\[ 4.8 = 4.74 + \log \left( \frac{[\text{salt}]}{0.002} \right) \]
\[ \log \left( \frac{[\text{salt}]}{0.002} \right) = 4.8 - 4.74 = 0.06 \]
\[ \frac{[\text{salt}]}{0.002} = 10^{0.06} \approx 1.148 \]
\[ [\text{salt}] = 0.002 \times 1.148 = 0.002296 \text{ mol} \]

Step 4: Calculate the concentration of potassium acetate
Volume of salt solution = 50 mL = 0.050 L
\[ \text{Concentration} = \frac{\text{moles}}{\text{volume}} = \frac{0.002296}{0.050} = 0.04592 \text{ M} \]
Rounding off to 2 significant figures, we get:
0.04 M

Final Answer: 0.04 M