This question is about acidic solutions - AQA - A-Level Chemistry - Question 2 - 2017 - Paper 1

To find the pH of the buffer solution after adding NaOH:

Initial concentrations:

• ( [CH_3COOH] = 0.260 , \text{mol dm}^{-3} )
• ( [CH_3COO^-] = 0.121 , \text{mol dm}^{-3} )

Amount of NaOH added:

• 7.00 \times 10^{-3} mol in 500 cm$^3$ means the concentrations change as follows:

• Moles of NaOH = 0.00700

• ( [CH_3COOH] ) decreases by 0.007 to become ( 0.260 - 0.007 = 0.253 , \text{mol dm}^{-3} )

• ( [CH_3COO^-] ) increases by 0.007 to become ( 0.121 + 0.007 = 0.128 , \text{mol dm}^{-3} )

Now applying the Henderson-Hasselbalch equation:

$\text{pH} = pK_a + \log \left( \frac{[CH_3COO^-]}{[CH_3COOH]} \right)$ Where, ( pK_a = -\log(1.74 \times 10^{-5}) \approx 4.76 $$

Substituting the values:

$\text{pH} = 4.76 + \log \left( \frac{0.128}{0.253} \right)$

Calculating: ( \log(0.505) \approx -0.295 ) gives:

$\text{pH} \approx 4.76 - 0.295 \approx 4.46$

Thus, the pH of the buffer solution after adding sodium hydroxide is approximately 4.46.