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.