What will be the pH of a solution obtained by mixing 800 mL of 0.05 M NaOH and 200 mL of 0.1 M HCl assuming complete dissociation of the acid and the base ?

Explanation

To find the pH of the solution obtained by mixing 800 mL of 0.05 M NaOH and 200 mL of 0.1 M HCl, we can follow these steps:

### Step 1: Calculate the number of moles of NaOH and HCl
1. **For NaOH:**
- Volume = 800 mL = 0.800 L
- Concentration = 0.05 M
- Moles of NaOH = Concentration × Volume = 0.05 mol/L × 0.800 L = 0.04 moles = 40 millimoles (mmol)

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