Calculate pH solution:
`0.1 M H_(2)SO_(4) (50ml) +0.4 M HCl (50 ml)`

Answer

To calculate the pH of the solution formed by mixing 0.1 M H₂SO₄ (50 mL) and 0.4 M HCl (50 mL), we can follow these steps:

### Step 1: Calculate the concentration of H⁺ ions from H₂SO₄
H₂SO₄ is a strong acid and fully dissociates in solution. The dissociation reaction is:
\[ \text{H}_2\text{SO}_4 \rightarrow 2 \text{H}^+ + \text{SO}_4^{2-} \]

Given:
- Concentration of H₂SO₄ = 0.1 M
- Volume of H₂SO₄ = 50 mL = 0.050 L

The number of moles of H₂SO₄ is:
\[ \text{Moles of H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.050 \, \text{L} = 0.005 \, \text{mol} \]

Since each mole of H₂SO₄ produces 2 moles of H⁺ ions:
\[ \text{Moles of H}^+ \text{ from H}_2\text{SO}_4 = 2 \times 0.005 \, \text{mol} = 0.010 \, \text{mol} \]

### Step 2: Calculate the concentration of H⁺ ions from HCl
HCl is also a strong acid and fully dissociates:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]

Given:
- Concentration of HCl = 0.4 M
- Volume of HCl = 50 mL = 0.050 L

The number of moles of HCl is:
\[ \text{Moles of HCl} = 0.4 \, \text{mol/L} \times 0.050 \, \text{L} = 0.020 \, \text{mol} \]

Thus, the moles of H⁺ ions from HCl is:
\[ \text{Moles of H}^+ \text{ from HCl} = 0.020 \, \text{mol} \]

### Step 3: Calculate the total moles of H⁺ ions
Now, we can find the total moles of H⁺ ions in the mixture:
\[ \text{Total moles of H}^+ = 0.010 \, \text{mol} + 0.020 \, \text{mol} = 0.030 \, \text{mol} \]

### Step 4: Calculate the total volume of the mixture
The total volume of the mixture is:
\[ \text{Total Volume} = 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.100 \, \text{L} \]

### Step 5: Calculate the concentration of H⁺ ions in the mixture
The concentration of H⁺ ions in the mixture is:
\[ \text{Concentration of H}^+ = \frac{\text{Total moles of H}^+}{\text{Total Volume}} = \frac{0.030 \, \text{mol}}{0.100 \, \text{L}} = 0.300 \, \text{M} \]

### Step 6: Calculate the pH of the solution
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the concentration of H⁺ ions:
\[ \text{pH} = -\log(0.300) \]

Using a calculator:
\[ \text{pH} \approx 0.522 \]

### Final Answer
The pH of the solution is approximately **0.522**.

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