If 200 ml of 0.031 M solution of `H _(2) SO_(4)` is added to 84 ml of a 0.150 M KOH solution. What is the pH of the resulting solution? (log 7 = 0.845)

Explanation

To find the pH of the resulting solution when 200 ml of 0.031 M H₂SO₄ is mixed with 84 ml of 0.150 M KOH, we can follow these steps:

### Step 1: Calculate the millimoles of H₂SO₄
- **Formula**: Millimoles = Molarity × Volume (in ml)
- **Calculation**:
\[ \text{Millimoles of H₂SO₄} = 0.031 \, \text{M} \times 200 \, \text{ml} = 6.2 \, \text{mmoles} \]
- Since H₂SO₄ dissociates into 2 H⁺ ions, the total millimoles of H⁺ will be:
\[ \text{Millimoles of H⁺} = 2 \times 6.2 = 12.4 \, \text{mmoles} \]

### Step 2: Calculate the millimoles of KOH
- **Calculation**:
\[ \text{Millimoles of KOH} = 0.150 \, \text{M} \times 84 \, \text{ml} = 12.6 \, \text{mmoles} \]
- KOH dissociates into K⁺ and OH⁻ ions, so the millimoles of OH⁻ will also be 12.6 mmoles.

### Step 3: Determine the limiting reactant
- We have 12.4 mmoles of H⁺ and 12.6 mmoles of OH⁻.
- Since H⁺ is less than OH⁻, H⁺ is the limiting reactant.

### Step 4: Calculate the excess OH⁻
- **Excess OH⁻**:
\[ \text{Excess OH⁻} = 12.6 \, \text{mmoles} - 12.4 \, \text{mmoles} = 0.2 \, \text{mmoles} \]

### Step 5: Calculate the total volume of the solution
- **Total Volume**:
\[ \text{Total Volume} = 200 \, \text{ml} + 84 \, \text{ml} = 284 \, \text{ml} \]

### Step 6: Calculate the concentration of excess OH⁻
- **Concentration of OH⁻**:
\[ \text{Concentration of OH⁻} = \frac{0.2 \, \text{mmoles}}{284 \, \text{ml}} = \frac{0.2}{284} \, \text{M} \approx 7.04 \times 10^{-4} \, \text{M} \]

### Step 7: Calculate pOH
- **Formula**:
\[ \text{pOH} = -\log[\text{OH⁻}] \]
- **Calculation**:
\[ \text{pOH} = -\log(7.04 \times 10^{-4}) = -(\log 7.04 + \log 10^{-4}) = -0.845 - (-4) = 3.155 \]

### Step 8: Calculate pH
- **Formula**:
\[ \text{pH} + \text{pOH} = 14 \]
- **Calculation**:
\[ \text{pH} = 14 - 3.155 = 10.845 \]

### Final Answer
The pH of the resulting solution is approximately **10.85**.

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