Calculate the pH of a solution obtained by mixing 50ml of` 0.2`M HCl with `49.9` mL of `0.2`m NaOH solution.

Answer

To calculate the pH of the solution obtained by mixing 50 mL of 0.2 M HCl with 49.9 mL of 0.2 M NaOH, we can follow these steps:

### Step 1: Calculate the number of moles of HCl
To find the number of moles of HCl, we use the formula:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume (in L)} \]
Given:
- Molarity of HCl = 0.2 M
- Volume of HCl = 50 mL = 0.050 L

\[ \text{Moles of HCl} = 0.2 \, \text{mol/L} \times 0.050 \, \text{L} = 0.01 \, \text{mol} \]

### Step 2: Calculate the number of moles of NaOH
Similarly, we calculate the number of moles of NaOH:
\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume (in L)} \]
Given:
- Molarity of NaOH = 0.2 M
- Volume of NaOH = 49.9 mL = 0.0499 L

\[ \text{Moles of NaOH} = 0.2 \, \text{mol/L} \times 0.0499 \, \text{L} = 0.00998 \, \text{mol} \]

### Step 3: Determine the limiting reactant
Since HCl and NaOH react in a 1:1 ratio, we can compare the moles:
- Moles of HCl = 0.01 mol
- Moles of NaOH = 0.00998 mol

NaOH is the limiting reactant, and it will completely react with HCl.

### Step 4: Calculate the remaining moles of HCl
After the reaction, the remaining moles of HCl can be calculated as follows:
\[ \text{Remaining moles of HCl} = \text{Initial moles of HCl} - \text{Moles of NaOH} \]
\[ \text{Remaining moles of HCl} = 0.01 \, \text{mol} - 0.00998 \, \text{mol} = 0.00002 \, \text{mol} \]

### Step 5: Calculate the total volume of the solution
The total volume of the solution after mixing is:
\[ \text{Total Volume} = 50 \, \text{mL} + 49.9 \, \text{mL} = 99.9 \, \text{mL} = 0.0999 \, \text{L} \]

### Step 6: Calculate the concentration of H⁺ ions
The concentration of H⁺ ions from the remaining HCl is given by:
\[ [\text{H}^+] = \frac{\text{Remaining moles of HCl}}{\text{Total Volume (in L)}} \]
\[ [\text{H}^+] = \frac{0.00002 \, \text{mol}}{0.0999 \, \text{L}} \approx 0.0002002 \, \text{M} \]

### Step 7: Calculate the pH
Finally, we can calculate the pH using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
\[ \text{pH} = -\log(0.0002002) \approx 3.699 \]

### Final Answer
The pH of the solution is approximately **3.699**.

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