What would be the pH of a solution obtained by mixing 5 g of acetic acid and 7.5 g of sodium acetate and making the volume equal to 500 mL? ($$K_a = 1.75 \times 10^{-5}$$, pK$$_a$$ = 4.76)

To find the pH of the solution obtained by mixing 5 g of acetic acid and 7.5 g of sodium acetate and making the volume up to 500 mL, we recognize that this mixture forms a buffer solution. A buffer solution consists of a weak acid and its conjugate base, and its pH can be calculated using the Henderson-Hasselbalch equation:

$$\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{salt}]}{[\text{acid}]}\right)$$

Given that the pKa of acetic acid is 4.76, we need to find the concentrations of sodium acetate (the salt) and acetic acid (the acid) in the solution.

First, we calculate the molar masses:

Molar mass of acetic acid (CH3COOH) = 12 × 2 + 1 × 4 + 16 × 2 = 24 + 4 + 32 = 60 g/mol.

Molar mass of sodium acetate (CH3COONa) = 12 × 2 + 1 × 3 + 16 × 2 + 23 = 24 + 3 + 32 + 23 = 82 g/mol.

Next, we find the number of moles:

Moles of acetic acid = mass / molar mass = 5 g / 60 g/mol = 5/60 = 1/12 mol.

Moles of sodium acetate = mass / molar mass = 7.5 g / 82 g/mol = 7.5 / 82 = 75 / 820. Simplifying by dividing numerator and denominator by 5: 75 ÷ 5 = 15, 820 ÷ 5 = 164, so 15/164 mol.

The total volume of the solution is 500 mL, which is 0.5 L. Now, we find the concentrations:

Concentration of acetic acid, [acid] = moles / volume = (1/12) mol / 0.5 L = (1/12) × (1 / 0.5) = (1/12) × 2 = 2/12 = 1/6 M.

Concentration of sodium acetate, [salt] = moles / volume = (15/164) mol / 0.5 L = (15/164) × (1 / 0.5) = (15/164) × 2 = 30/164 M. Simplifying by dividing numerator and denominator by 2: 30 ÷ 2 = 15, 164 ÷ 2 = 82, so 15/82 M.

Now, we substitute these values into the Henderson-Hasselbalch equation:

$$\text{pH} = 4.76 + \log_{10}\left(\frac{15/82}{1/6}\right)$$

First, simplify the ratio inside the logarithm:

$$\frac{[\text{salt}]}{[\text{acid}]} = \frac{15/82}{1/6} = \frac{15}{82} \times \frac{6}{1} = \frac{15 \times 6}{82 \times 1} = \frac{90}{82}$$

Simplify 90/82 by dividing both numerator and denominator by 2: 90 ÷ 2 = 45, 82 ÷ 2 = 41, so the ratio is 45/41.

Now, compute the logarithm:

$$\log_{10}\left(\frac{45}{41}\right) = \log_{10}(45) - \log_{10}(41)$$

Using approximate values: log10(45) ≈ 1.6532, log10(41) ≈ 1.6128, so:

$$\log_{10}\left(\frac{45}{41}\right) \approx 1.6532 - 1.6128 = 0.0404$$

Therefore, the pH is:

$$\text{pH} \approx 4.76 + 0.0404 = 4.8004$$

This value is approximately 4.80, which is greater than 4.76 and less than 5.0.

Now, comparing with the options:

A. pH = 4.70 - This is not equal to 4.80.

B. pH < 4.70 - But 4.80 is greater than 4.70.

C. pH of solution will be equal to pH of acetic acid - The pH of pure acetic acid would be much lower (around 2-3 for typical concentrations), so this is incorrect.

D. 4.76 < pH < 5.0 - Since 4.76 < 4.80 < 5.0, this is correct.

Hence, the correct answer is Option D.