What is the final pH of a buffer solution made by mixing 10 mL of 0.6% Acetic Acid (CH3COOH; MW = 60.05; pKa = 4.8) with 50 mL of a 1.36% Sodium Acetate (CH3COONa.3H2O; MW = 136.1) solution?pH = pka + log[salt]/[acid]4.1b.4.5c.5.15d.5.5e.9.8

To solve this problem, we need to use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the base (in this case, the acetate ion from sodium acetate) and [HA] is the concentration of the acid (acetic acid).

First, we need to calculate the molar concentrations of the acetic acid and sodium acetate.

For acetic acid: The volume is 10 mL or 0.01 L. The weight/volume percent concentration is 0.6%, which means there are 0.6 g of acetic acid in 100 mL of solution. So, in 10 mL of solution, there are 0.006 g of acetic acid. The molar mass of acetic acid is 60.05 g/mol. So, the molar concentration of acetic acid is (0.006 g) / (60.05 g/mol) / 0.01 L = 0.01 M.

For sodium acetate: The volume is 50 mL or 0.05 L. The weight/volume percent concentration is 1.36%, which means there are 1.36 g of sodium acetate in 100 mL of solution. So, in 50 mL of solution, there are 0.68 g of sodium acetate. The molar mass of sodium acetate is 136.1 g/mol. So, the molar concentration of sodium acetate is (0.68 g) / (136.1 g/mol) / 0.05 L = 0.1 M.

Now we can plug these values into the Henderson-Hasselbalch equation: pH = 4.8 + log(0.1/0.01) = 4.8 + 1 = 5.8.

So, the final pH of the buffer solution is 5.8. None of the options provided (4.1, 4.5, 5.15, 5.5, 9.8) match this result.