Calculate `[Cl^(Theta)], [Na^(o+)], [H^(o+)], [overset(Theta)OH]`, and the `pH` of resulting solution obtained by mixing `50mL` of `0.6M HCl` and `50mL` of `0.3M NaOH`.

Explanation

To solve the problem of calculating the concentrations of chloride ions \([Cl^-]\), sodium ions \([Na^+]\), hydrogen ions \([H^+]\), hydroxide ions \([\overset{\sim}{OH}^-]\), and the pH of the resulting solution after mixing \(50 \, \text{mL}\) of \(0.6 \, \text{M} \, HCl\) and \(50 \, \text{mL}\) of \(0.3 \, \text{M} \, NaOH\), we can follow these steps:

### Step 1: Calculate the moles of HCl and NaOH

1. **Calculate moles of HCl:**
\[ \text{Moles of } HCl = \text{Volume (L)} \times \text{Concentration (M)} = 0.050 \, \text{L} \times 0.6 \, \text{M} = 0.030 \, \text{moles} \]

2. **Calculate moles of NaOH:**
\[ \text{Moles of } NaOH = \text{Volume (L)} \times \text{Concentration (M)} = 0.050 \, \text{L} \times 0.3 \, \text{M} = 0.015 \, \text{moles} \]

### Step 2: Determine the limiting reactant and the remaining moles after the reaction

3. **Reaction between HCl and NaOH:**
\[ HCl + NaOH \rightarrow NaCl + H_2O \]
- Since \(0.015 \, \text{moles}\) of NaOH is less than \(0.030 \, \text{moles}\) of HCl, NaOH is the limiting reactant.

4. **Calculate remaining moles of HCl:**
\[ \text{Remaining moles of } HCl = 0.030 - 0.015 = 0.015 \, \text{moles} \]

5. **Calculate moles of NaOH remaining:**
\[ \text{Moles of } NaOH = 0 \, \text{(completely reacted)} \]

6. **Calculate moles of NaCl formed:**
\[ \text{Moles of } NaCl = \text{Moles of } NaOH \text{ reacted} = 0.015 \, \text{moles} \]

### Step 3: Calculate the total volume of the solution

7. **Total volume of the solution:**
\[ \text{Total Volume} = 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]

### Step 4: Calculate the concentrations of ions

8. **Concentration of chloride ions \([Cl^-]\):**
\[ [Cl^-] = \frac{\text{Moles of } Cl^-}{\text{Total Volume}} = \frac{0.015 \, \text{moles}}{0.1 \, \text{L}} = 0.15 \, \text{M} \]

9. **Concentration of sodium ions \([Na^+]\):**
\[ [Na^+] = \frac{\text{Moles of } Na^+}{\text{Total Volume}} = \frac{0.015 \, \text{moles}}{0.1 \, \text{L}} = 0.15 \, \text{M} \]

10. **Concentration of hydrogen ions \([H^+]\):**
\[ [H^+] = \frac{\text{Remaining moles of } HCl}{\text{Total Volume}} = \frac{0.015 \, \text{moles}}{0.1 \, \text{L}} = 0.15 \, \text{M} \]

11. **Concentration of hydroxide ions \([\overset{\sim}{OH}^-]\):**
\[ [\overset{\sim}{OH}^-] = \frac{K_w}{[H^+]} = \frac{1.0 \times 10^{-14}}{0.15} \approx 6.67 \times 10^{-14} \, \text{M} \]

### Step 5: Calculate the pH of the solution

12. **Calculate pH:**
\[ pH = -\log[H^+] = -\log(0.15) \approx 0.82 \]

### Final Results
- \([Cl^-] = 0.15 \, \text{M}\)
- \([Na^+] = 0.15 \, \text{M}\)
- \([H^+] = 0.15 \, \text{M}\)
- \([\overset{\sim}{OH}^-] \approx 6.67 \times 10^{-14} \, \text{M}\)
- \(pH \approx 0.82\)