`100 mL` of `0.15 M HCl` is mixed with `100 mL` of `0.005M HCl` , what is the `pH` of the following solution approxmately

Options

A. 2.5
B. 1.5
C. 2
D. 1

Explanation

To find the pH of the solution formed by mixing 100 mL of 0.15 M HCl with 100 mL of 0.005 M HCl, we can follow these steps:

### Step 1: Calculate the total moles of HCl in each solution.

1. **For the first solution (0.15 M HCl)**:
\[ \text{Moles of HCl}_1 = M_1 \times V_1 = 0.15 \, \text{mol/L} \times 0.1 \, \text{L} = 0.015 \, \text{mol} \]

2. **For the second solution (0.005 M HCl)**:
\[ \text{Moles of HCl}_2 = M_2 \times V_2 = 0.005 \, \text{mol/L} \times 0.1 \, \text{L} = 0.0005 \, \text{mol} \]

### Step 2: Calculate the total moles of HCl after mixing.

\[ \text{Total moles of HCl} = \text{Moles of HCl}_1 + \text{Moles of HCl}_2 = 0.015 \, \text{mol} + 0.0005 \, \text{mol} = 0.0155 \, \text{mol} \]

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

\[ \text{Total Volume} = V_1 + V_2 = 100 \, \text{mL} + 100 \, \text{mL} = 200 \, \text{mL} = 0.2 \, \text{L} \]

### Step 4: Calculate the resultant molarity of HCl in the mixed solution.

\[ \text{Molarity (M)} = \frac{\text{Total moles of HCl}}{\text{Total Volume}} = \frac{0.0155 \, \text{mol}}{0.2 \, \text{L}} = 0.0775 \, \text{M} \]

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

Since HCl is a strong acid, it completely dissociates in solution. Therefore, the concentration of hydrogen ions \([H^+]\) is equal to the molarity of HCl.

\[ \text{pH} = -\log[H^+] = -\log(0.0775) \]

Calculating this gives:

\[ \text{pH} \approx 2.11 \]

### Conclusion

The pH of the mixed solution is approximately **2.11**.

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