Questions: Ka of an Unknown Acid Othello titrates 25.0 mL out of 50.0 mL of a weak acid using NaOH. After combining the two halves, the measured pH of the system is 7.46. Which acid did he titrate? Phosphoric Acid, Ka=7.52 x 10^-3 Hydrosulfuric Acid, Ka=9.1 x 10^-8 Sulfurous Acid, K1=1.54 x 10^-2 Hypochlorous Acid, Ks=3.5 x 10^-8

Solution Steps

Othello titrates 25.0 mL of a weak acid with NaOH and then combines it with another 25.0 mL of the same acid. The pH of the resulting solution is 7.46. We need to determine which acid was titrated based on the given \( \mathrm{K}_a \) values.

The Henderson-Hasselbalch equation is given by:

\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]

Since the solution is at the half-equivalence point (equal volumes of acid and its conjugate base), the ratio \(\frac{[\text{A}^-]}{[\text{HA}]}\) is 1. Therefore, the equation simplifies to:

\[ \text{pH} = \text{pK}_a \]

Given that the pH of the solution is 7.46, we can equate this to the pK_a of the acid:

\[ \text{pK}_a = 7.46 \]

Convert the given \( \mathrm{K}_a \) values to pK_a values:

• Phosphoric Acid: \( \mathrm{pK}_a = -\log(7.52 \times 10^{-3}) \approx 2.123 \)
• Hydrosulfuric Acid: \( \mathrm{pK}_a = -\log(9.1 \times 10^{-8}) \approx 7.041 \)
• Sulfurous Acid: \( \mathrm{pK}_a = -\log(1.54 \times 10^{-2}) \approx 1.812 \)
• Hypochlorous Acid: \( \mathrm{pK}_a = -\log(3.5 \times 10^{-8}) \approx 7.456 \)

Final Answer