(b) When chlorine dissolves in water the following reaction occurs.
CL(g) + H,O(l) → HC10(aq) + H'(aq) + CL(aq)
When solutions of chlorine are used for water purification, the pH of the solution of chlorine is kept near to pH 7 by the addition of a base.
Chlorine is dissolved in water to produce 1000 cm' of a solution containing 0.170 mol of HCIO and 0.170 mol of HCL
A buffer solution is then prepared by adding 0.200 mol of NaOHs) to this solution. The NaOH reacts initially with the HCL
Calculate the pH of the buffer solution.
[HCIO is a weak acid with K, = 2.9 × 10 *moldm*]

### Step 1: Understanding the Problem

We are tasked with calculating the pH of a buffer solution after NaOH is added to a solution containing \( \text{HClO} \) and \( \text{HCl} \). The key steps involve:
1. The neutralization of \( \text{HCl} \) by \( \text{NaOH} \), followed by
2. The buffer effect of \( \text{HClO} \) and its conjugate base after the reaction.

Given data### Step 1: Understanding the Problem

We are tasked with calculating the pH of a buffer solution after NaOH is added to a solution containing \( \text{HClO} \) and \( \text{HCl} \). The key steps involve:
1. The neutralization of \( \text{HCl} \) by \( \text{NaOH} \), followed by
2. The buffer effect of \( \text{HClO} \) and its conjugate base after the reaction.

Given 
- Volume of solution: 1000 cm³ = 1 dm³
- Moles of \( \text{HClO} \) and \( \text{HCl} \): 0.170 mol each
- Moles of \( \text{NaOH} \): 0.200 mol
- \( K_a \) of \( \text{HClO} \) = \( 2.9 \times 10^{-8} \, \text{mol/dm}^3 \)

### Step 2: Reaction Between \( \text{NaOH} \) and \( \text{HCl} \)

First, \( \text{NaOH} \) reacts with \( \text{HCl} \), as NaOH is a strong base and HCl is a strong acid:

\[
\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}
\]

Moles of \( \text{NaOH} \) added = 0.200 mol  
Moles of \( \text{HCl} \) present = 0.170 mol

Since \( \text{NaOH} \) is present in excess, all \( \text{HCl} \) will be neutralized:
\[
0.170 \, \text{mol} \, \text{HCl} \, \text{neutralized} \Rightarrow \text{Remaining NaOH} = 0.200 - 0.170 = 0.030 \, \text{mol}
\]

### Step 3: Reaction Between \( \text{NaOH} \) and \( \text{HClO} \)

The remaining \( \text{NaOH} \) (0.030 mol) will react with \( \text{HClO} \) to form its conjugate base \( \text{ClO}^- \):

\[
\text{HClO} + \text{NaOH} \rightarrow \text{NaClO} + \text{H}_2\text{O}
\]

Moles of \( \text{HClO} \) remaining after this reaction = \( 0.170 - 0.030 = 0.140 \, \text{mol} \)

Moles of \( \text{ClO}^- \) produced = 0.030 mol

### Step 4: Calculate the pH Using the Henderson-Hasselbalch Equation

The buffer solution contains \( \text{HClO} \) (weak acid) and \( \text{ClO}^- \) (its conjugate base). We can calculate the pH using the Henderson-Hasselbalch equation:

\[
\text{pH} = \text{p}K_a + \log \left( \frac{[\text{ClO}^-]}{[\text{HClO}]} \right)
\]

Where:
- \( \text{p}K_a = -\log K_a = -\log (2.9 \times 10^{-8}) \)
- \( [\text{ClO}^-] = \frac{0.030 \, \text{mol}}{1 \, \text{dm}^3} = 0.030 \, \text{mol/dm}^3 \)
- \( [\text{HClO}] = \frac{0.140 \, \text{mol}}{1 \, \text{dm}^3} = 0.140 \, \text{mol/dm}^3 \)

### Step 5: Solve for pH

First, calculate \( \text{p}K_a \):

\[
\text{p}K_a = -\log(2.9 \times 10^{-8}) = 7.537
\]

Now apply the Henderson-Hasselbalch equation:

\[
\text{pH} = 7.537 + \log \left( \frac{0.030}{0.140} \right)
\]
\[
\text{pH} = 7.537 + \log(0.214)
\]
\[
\log(0.214) \approx -0.670
\]
\[
\text{pH} = 7.537 - 0.670 = 6.867
\]

### Final Answer:
The pH of the buffer solution is approximately **6.87**.

---

Would you like further clarification on any step? Here are 5 related questions to explore this concept:

1. What is the significance of a buffer solution in chemical reactions?
2. How would the pH change if more \( \text{NaOH} \) was added?
3. How does the Henderson-Hasselbalch equation work for buffer solutions?
4. Why is \( \text{HCl} \) fully neutralized by \( \text{NaOH} \) in this reaction?
5. What role does \( K_a \) play in determining the pH of weak acids?

**Tip:** Remember, strong acids and bases fully dissociate in water, while weak acids only partially dissociate!

When chlorine dissolves in water the following reaction occurs. CL(g) + H,O(l) → HCIO(aq) + H'(aq) + CL(aq). When solutions of chlorine are used for water purification, the pH of the solution of chlorine is kept near to pH 7 by the addition of a base. Chlorine is dissolved in water to produce 1000 cm' of a solution containing 0.170 mol of HCIO and 0.170 mol of HCL. A buffer solution is then prepared by adding 0.200 mol of NaOHs) to this solution. The NaOH reacts initially with the HCL. Calculate the pH of the buffer solution. [HCIO is a weak acid with K, = 2.9 × 10 *moldm*]

This problem involves calculating the pH of a buffer solution after adding NaOH to a solution of HClO and HCl. Learn how to apply the Henderson-Hasselbalch equation, understand buffer reactions, and compute the final pH with given moles of reactants. Suitable for high school students in Grades 11-12.

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