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Step-by-Step Solution
Step 1: Determine moles of H+ ions from H2SO4
• Molarity of H2SO4 = 0.2 M
• Volume of H2SO4 solution = 400 mL = 0.400 L
• Each mole of H2SO4 provides 2 moles of H+.
Thus, moles of H+ =
\bigl(0.400 \,\text{L} \times 0.2 \,\text{mol L}^{-1}\bigr)\times 2 = 0.16 \,\text{mol}.
Step 2: Determine moles of OH− from NaOH
• Molarity of NaOH = 0.1 M
• Volume of NaOH solution = 600 mL = 0.600 L
Moles of OH− =
0.600 \,\text{L} \times 0.1 \,\text{mol L}^{-1} = 0.06 \,\text{mol}.
Step 3: Identify the limiting reagent
• H+ moles = 0.16
• OH− moles = 0.06
Since OH− is fewer, OH− is the limiting reagent.
Step 4: Calculate heat released during neutralization
The neutralization reaction is:
\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \to \text{H}_2\text{O}.
• Heat of neutralization ( \Delta H_{\gamma} ) = -57.1 \,\text{kJ mol}^{-1} = -57.1 \times 10^3 \,\text{J mol}^{-1}.
• Since 0.06 mol of OH− reacts completely, heat released =
0.06 \,\text{mol} \times 57.1 \times 10^3 \,\text{J mol}^{-1} = 0.06 \times 57.1 \times 10^3 \,\text{J}.
Numerically:
0.06 \times 57.1 \approx 3.426 \quad \text{(in kJ)},
which is
3.426 \times 10^3 = 3426 \,\text{J}.
Step 5: Equate heat released to heat gained by the solution
Assume the total volume of the mixed solution is approximately 1.0 L (since volumes are additive and density is 1.0 g cm−3), so the mass of the solution is about 1000 g.
Let \Delta T = increase in temperature (in K).
Using \text{Heat gained} = (\text{mass}) \times (\text{specific heat}) \times (\Delta T) :
3426 \,\text{J} = (1000 \,\text{g}) \times (4.18 \,\text{J g}^{-1}\text{K}^{-1}) \times \Delta T.
Therefore,
\Delta T = \frac{3426}{1000 \times 4.18} \,\text{K} \approx 0.82 \,\text{K}.
Step 6: Express the final answer in the desired form
The increase in temperature is about 0.82 \,\text{K} , which can be written as
82 \times 10^{-2} \,\text{K}.
Final Answer
82 \times 10^{-2} \,\text{K}.
