Question

8 g of NaOH is dissolved in 1.0 L solution containing one mole of acetic acid and one mole of sodium acetate. The pH value of the resulting solution is: \[ pK_a=4.74 \]

• A. \(4.91\)
• B. \(3.91\)
• C. \(5.91\)
• D. \(2.91\)

Hint:

When strong base is added to an acidic buffer, acid decreases while salt increases by the same number of moles before applying the Henderson equation.

Answer 1

The Correct Option is A

Concept: This is a buffer solution. Henderson–Hasselbalch equation: \[ pH=pK_a+\log\frac{[\text{salt}]}{[\text{acid}]} \]

Step 1: Calculate moles of NaOH added. \[ n = \frac{8}{40} = 0.2\;mol \]

Step 2: Neutralization reaction. \[ CH_3COOH+NaOH \rightarrow CH_3COONa+H_2O \] Initial: \[ \text{Acid}=1 \] \[ \text{Salt}=1 \] After reaction: \[ \text{Acid}=1-0.2=0.8 \] \[ \text{Salt}=1+0.2=1.2 \]

Step 3: Apply Henderson equation. \[ pH = 4.74+\log\frac{1.2}{0.8} \] \[ = 4.74+\log(1.5) \] \[ = 4.74+0.176 \] \[ = 4.916 \] \[ \boxed{4.91} \]