Question

Below what pH will free phenobarbital begin to separate from asolution having an initial concentration of 4 g of sodiumphenobarbital per 100 mL at 25°. The molar solubility ofphenobarbital is 0.0050 and the pKa is 7.41 at 25°C. The molecularweight of sodium phenobarbital is 254​Evaluate x^2 - x - 1 + x for x = 2.

1. 0
2. 1
3. 2
4. 012

Answer 1

The pH of a solution with 0.15 mol benzoic acid and 0.25 mol sodium benzoate is approximately 4.42. To prepare a 0.1 M benzoic acid/sodium benzoate buffer at pH 4.1 for 250 mL, approximately 1.34 g of benzoic acid and 2.01 g of sodium benzoate are required.

Step-by-step explanation:

To calculate the pH of a solution containing 0.15 mol of benzoic acid and 0.25 mol of sodium benzoate, we use the Henderson-Hasselbalch equation:

pH = pKa + log([base]/[acid])

Given that the pKa for benzoic acid is 4.2, we plug in the values:

pH = 4.2 + log(0.25/0.15) = 4.2 + log(1.67) ≈ 4.42

For the preparation of 250 mL of a 0.1 M buffer at pH 4.1 of benzoic acid and sodium benzoate:

Using the same equation and rearranging for the acid/base ratio:

ratio = 10^(pH - pKa) = 10^(4.1 - 4.2) ≈ 0.7943

If the total concentration of the buffer is 0.1 M, then we can assign x to [acid] and 0.1 - x to [base]. Using the ratio, x/(0.1 - x) = 0.7943, we find that x ≈ 0.044 mol/L for the acid. For 250 mL, the mass of benzoic acid is 0.044 mol/L * 0.250 L * 122.12 g/mol (molar mass of benzoic acid) ≈ 1.34 g. Similarly, the mass of sodium benzoate is 0.056 mol/L * 0.250 L * 144.11 g/mol (molar mass of sodium benzoate) ≈ 2.01 g.