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.

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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!