100 mL of 0.2 N HCI is added to 100 mL of 0.18 N NaOH and whole volume is made one litre. What will be the pH of resulting solution.

Answer

To find the pH of the resulting solution after mixing 100 mL of 0.2 N HCl with 100 mL of 0.18 N NaOH and making the total volume 1 liter, we can follow these steps:

### Step 1: Calculate the moles of HCl
1. **Normality of HCl (0.2 N)** means it has 0.2 equivalents of H+ per liter.
2. The volume of HCl used is 100 mL or 0.1 L.
3. Moles of HCl = Normality × Volume = 0.2 N × 0.1 L = 0.02 equivalents.
4. Since HCl is a strong acid, it completely dissociates to give 0.02 moles of H+ ions.

### Step 2: Calculate the moles of NaOH
1. **Normality of NaOH (0.18 N)** means it has 0.18 equivalents of OH- per liter.
2. The volume of NaOH used is also 100 mL or 0.1 L.
3. Moles of NaOH = Normality × Volume = 0.18 N × 0.1 L = 0.018 equivalents.
4. Since NaOH is a strong base, it completely dissociates to give 0.018 moles of OH- ions.

### Step 3: Determine the limiting reagent and the remaining moles after neutralization
1. The reaction between HCl and NaOH is a 1:1 reaction:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
2. Initially, we have 0.02 moles of H+ from HCl and 0.018 moles of OH- from NaOH.
3. After the reaction, NaOH will completely react with HCl:
- H+ remaining = 0.02 - 0.018 = 0.002 moles (or 2 millimoles).
- OH- will be completely consumed.

### Step 4: Calculate the concentration of H+ ions in the final solution
1. The total volume of the solution is made to 1 L (1000 mL).
2. Concentration of H+ ions = Moles of H+ remaining / Total volume = 0.002 moles / 1 L = 0.002 M.

### Step 5: Calculate the pH of the solution
1. pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
2. Substituting the concentration of H+:
\[ \text{pH} = -\log(0.002) = -\log(2 \times 10^{-3}) = -(\log(2) + \log(10^{-3})) = -(\log(2) - 3) \]
3. Using the approximation \(\log(2) \approx 0.301\):
\[ \text{pH} = 3 - 0.301 = 2.699 \approx 2.7 \]

### Final Answer:
The pH of the resulting solution is approximately **2.7**.

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