When 400 mL of `0.2 M H _(2) SO _(4)` soltution is mixed with 600 mL of `0.1 M` NaOH solution, the increase in temperature of the final solution is _______`xx 10 ^(-2) K.` (Round off to the nearest integer).
[Use`: H ^(+) (aq) + OH ^(-) (aq) to mol ^(-1) ]`
`Delta H =- 57.1 kJ mol ^(-1)]`
Specific heat of `H _(2) O = 4. 18 J K ^(-1) g ^(-1)`
density of `H _(2) O = 1. 0 g cm ^(-3)`
Assume no change in volume of solutio on mixing.

Answer

To solve the problem, we need to find the increase in temperature when 400 mL of 0.2 M H₂SO₄ solution is mixed with 600 mL of 0.1 M NaOH solution. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the moles of H₂SO₄ and NaOH
1. **Calculate moles of H₂SO₄:**
\[ \text{Moles of H₂SO₄} = \text{Molarity} \times \text{Volume} = 0.2 \, \text{mol/L} \times 0.4 \, \text{L} = 0.08 \, \text{mol} \]
Since H₂SO₄ is a diprotic acid, it will release 2 moles of H⁺ for every mole of H₂SO₄. Therefore, the number of moles of H⁺ ions is:
\[ \text{Moles of H⁺} = 0.08 \, \text{mol} \times 2 = 0.16 \, \text{mol} \]

2. **Calculate moles of NaOH:**
\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.6 \, \text{L} = 0.06 \, \text{mol} \]

### Step 2: Determine the limiting reagent
- The reaction is:
\[ \text{H⁺} + \text{OH⁻} \rightarrow \text{H₂O} \]
- Here, we have 0.16 mol of H⁺ and 0.06 mol of OH⁻. Since NaOH (OH⁻) is the limiting reagent, it will completely react.

### Step 3: Calculate the heat released during the reaction
- The enthalpy change (ΔH) for the reaction is given as -57.1 kJ/mol for the neutralization of H⁺ and OH⁻.
- The heat released (Q) can be calculated using the moles of the limiting reagent:
\[ Q = \text{moles of OH⁻} \times \Delta H = 0.06 \, \text{mol} \times (-57.1 \, \text{kJ/mol}) = -3.426 \, \text{kJ} \]

### Step 4: Convert heat released to joules
\[ Q = -3.426 \, \text{kJ} = -3426 \, \text{J} \]

### Step 5: Calculate the temperature change (ΔT)
- Using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \( m \) is the mass of the solution (in grams),
- \( c \) is the specific heat capacity (4.18 J/g·K),
- \( \Delta T \) is the change in temperature.

- The total volume of the solution is:
\[ V_{\text{total}} = 400 \, \text{mL} + 600 \, \text{mL} = 1000 \, \text{mL} \]
Since the density of water is 1 g/mL, the mass of the solution is:
\[ m = 1000 \, \text{g} \]

- Rearranging the formula to find ΔT:
\[ \Delta T = \frac{Q}{m \cdot c} = \frac{-3426 \, \text{J}}{1000 \, \text{g} \cdot 4.18 \, \text{J/g·K}} = -0.8196 \, \text{K} \]

### Step 6: Convert ΔT to the required format
- Since we need the increase in temperature, we take the absolute value:
\[ \Delta T = 0.8196 \, \text{K} = 81.96 \times 10^{-2} \, \text{K} \]
- Rounding off to the nearest integer gives us:
\[ \Delta T \approx 82 \times 10^{-2} \, \text{K} \]

### Final Answer
The increase in temperature of the final solution is **82 x 10^(-2) K**.

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