Question

Calculate the pH of a solution obtain by mixing 10 ml of 1 N sodium acetate and 50 ml of 2 N acetic acid ($K_a = 1.8 \times 10^{-5}$).

Answer 1

3.75

Answer 2

To calculate the pH of the solution obtained by mixing sodium acetate and acetic acid, we recognize that this mixture forms an acidic buffer solution. The pH of an acidic buffer can be calculated using the Henderson-Hasselbalch equation:

$pH = pK_a + \log \frac{[Salt]}{[Acid]}$

First, let's determine the millimoles of each component. Normality (N) is equivalent to Molarity (M) for both acetic acid ($CH_3COOH$) and sodium acetate ($CH_3COONa$) because their n-factors are 1.

1. Millimoles of Sodium Acetate (Salt):
   Volume = 10 mL
   Normality = 1 N
   Millimoles of $CH_3COONa = \text{Normality} \times \text{Volume (mL)}$
   Millimoles of $CH_3COONa = 1 \text{ N} \times 10 \text{ mL} = 10 \text{ millimoles}$

2. Millimoles of Acetic Acid (Acid):
   Volume = 50 mL
   Normality = 2 N
   Millimoles of $CH_3COOH = \text{Normality} \times \text{Volume (mL)}$
   Millimoles of $CH_3COOH = 2 \text{ N} \times 50 \text{ mL} = 100 \text{ millimoles}$

3. Calculate $pK_a$:
   Given $K_a = 1.8 \times 10^{-5}$
   $pK_a = -\log K_a$
   $pK_a = -\log (1.8 \times 10^{-5})$
   $pK_a = -(\log 1.8 + \log 10^{-5})$
   $pK_a = -(\log 1.8 - 5)$
   $pK_a = 5 - \log 1.8$
   Using $\log 1.8 \approx 0.255$:
   $pK_a = 5 - 0.255 = 4.745$

4. Calculate pH using the Henderson-Hasselbalch equation:
   Since the total volume is the same for both the salt and the acid in the final mixture, we can use the millimoles directly in the ratio:
   $pH = pK_a + \log \frac{\text{Millimoles of Salt}}{\text{Millimoles of Acid}}$
   $pH = 4.745 + \log \frac{10}{100}$
   $pH = 4.745 + \log (0.1)$
   $pH = 4.745 + (-1)$
   $pH = 3.745$

The pH of the solution is approximately 3.75.