When a 20.0 mL sample of a 0.483 M aqueous nitrous acid solution is titrated with a 0.430 M aqueous potassium hydroxide solution, what is the pH at the midpoint in the titration? pH =

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**Titration of Nitrous Acid with Potassium Hydroxide**

When a 20.0 mL sample of a 0.483 M aqueous nitrous acid solution is titrated with a 0.430 M aqueous potassium hydroxide solution, what is the pH at the midpoint in the titration?

**Calculation:**

pH = [Insert calculated pH value here based on details or class notes]

**Explanation:**

In the titration of a weak acid, such as nitrous acid, with a strong base, such as potassium hydroxide, the pH at the midpoint equals the pKa of the acid. This is because, at the midpoint, the concentrations of the acid and its conjugate base are equal, simplifying the Henderson-Hasselbalch equation to:

\[ \text{pH} = \text{pKa} + \log \left(\frac{[\text{Base}]}{[\text{Acid}]}\right) = \text{pKa} \]

Nitrous acid (\( \text{HNO}_2\)) is a weak acid and the \( \text{KOH} \) is a strong base that will completely dissociate in solution.

For educational purposes, this concept can demonstrate the utility of the Henderson-Hasselbalch equation in calculating the pH of buffer solutions and understanding acid-base equilibria.
Transcribed Image Text:**Titration of Nitrous Acid with Potassium Hydroxide** When a 20.0 mL sample of a 0.483 M aqueous nitrous acid solution is titrated with a 0.430 M aqueous potassium hydroxide solution, what is the pH at the midpoint in the titration? **Calculation:** pH = [Insert calculated pH value here based on details or class notes] **Explanation:** In the titration of a weak acid, such as nitrous acid, with a strong base, such as potassium hydroxide, the pH at the midpoint equals the pKa of the acid. This is because, at the midpoint, the concentrations of the acid and its conjugate base are equal, simplifying the Henderson-Hasselbalch equation to: \[ \text{pH} = \text{pKa} + \log \left(\frac{[\text{Base}]}{[\text{Acid}]}\right) = \text{pKa} \] Nitrous acid (\( \text{HNO}_2\)) is a weak acid and the \( \text{KOH} \) is a strong base that will completely dissociate in solution. For educational purposes, this concept can demonstrate the utility of the Henderson-Hasselbalch equation in calculating the pH of buffer solutions and understanding acid-base equilibria.
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