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Let's break down the problem step by step:

1. **Given Reaction**:
   \[
   \text{Cl}_2 + \text{H}_2\text{O} \rightarrow \text{HClO} + \text{H}^+ + \text{Cl}^-
   \]
   Chlorine gas reacts with water to form hypochlorous acid (HClO) and hydrochloric acid (HCl).

2. **Buffer Solution**:
   - The solution contains 0.170 moles of HClO and 0.170 moles of HCl.
   - 0.2 moles of NaOH is added to this solution. NaOH reacts with HCl, as NaOH neutralizes the acid (HCl).

3. **Reaction Between NaOH and HCl**:
   Since NaOH reacts completely with HCl, the moles of NaOH (0.2 moles) will neutralize an equal amount of HCl (0.170 moles of HCl). After the reaction:
   \[
   0.2 - 0.170 = 0.030 \text{ moles of NaOH remain}.
   \]
   All 0.170 moles of HCl are consumed, leaving no HCl, and 0.030 moles of NaOH remain unreacted.

4. **Buffer Composition After Reaction**:
   - After neutralization, all the remaining NaOH will react with HClO (since HClO is a weak acid), converting it to its conjugate base (ClO⁻).
   - So, 0.030 moles of HClO will react with 0.030 moles of NaOH to form 0.030 moles of ClO⁻, and 0.170 moles of HClO are reduced by 0.030 moles.

   Final buffer composition:
   - Moles of HClO = \(0.170 - 0.030 = 0.140 \, \text{moles}\).
   - Moles of ClO⁻ = 0.030 moles.

5. **Using the Henderson-Hasselbalch Equation**:
   \[
   \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
   \]
   Where:
   - \(\text{pKa} = -\log \text{Ka} = -\log(2.9 \times 10^{-8})\)
   - \([\text{A}^-]\) = concentration of \( \text{ClO}^- \)
   - \([\text{HA}]\) = concentration of \( \text{HClO} \)

   Let's calculate each part.The pH of the buffer solution is approximately **6.87**.

Would you like a detailed explanation or have any further questions?

Here are 5 related questions you can explore:
1. How does the buffer capacity change with different amounts of NaOH?
2. What happens to the pH if the volume of the solution increases?
3. How is the pKa related to the strength of an acid?
4. Why does the addition of NaOH affect the ratio of HClO to ClO⁻?
5. What would happen if more NaOH were added beyond the neutralization point?

**Tip:** In buffer solutions, the closer the amounts of the weak acid and its conjugate base, the better the buffer resists changes in pH.

Cl2 is dissolved in water to produce 1000 cm³ of 0.170 moles HClO and 0.170 moles HCl. A buffer solution was prepared by adding 0.2 moles of NaOH to this solution. Calculate the pH of the buffer. Ka = 2.9 x 10⁻⁸ mol dm⁻³.

Learn how to calculate the pH of a buffer solution when 0.2 moles of NaOH are added to a mixture of 0.170 moles of HClO and HCl. Using the Henderson-Hasselbalch equation and Ka value of 2.9 x 10⁻⁸ mol dm⁻³, this detailed guide explains the steps involved in determining the final pH.

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