Question

200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is \(-57.1\) kJ. The increase in temperature in \(^\circ C\) of the system on mixing is \(x \times 10^{-2}\). The value of \(x\) is _________. (Nearest integer)
[ Given : Specific heat of water \(= 4.18 \, \text{J g}^{-1} \text{K}^{-1}\)
Density of water \(= 1.00 \, \text{g cm}^{-3}\) ]
(Assume no volume change on mixing)

Hint:

Always identify the limiting reagent in neutralization problems. The heat released is proportional to the number of moles of water formed (or the moles of limiting \(H^+\) or \(OH^-\)).

Answer 1

Correct Answer: 82

Step 1: Understanding the Concept:
The heat released during neutralization increases the temperature of the total solution volume. We first find the limiting reagent and the total heat evolved.
Step 2: Detailed Explanation:
Moles of \(HCl = M \times V = 0.2 \times 0.2 = 0.04 \, \text{mol}\)
Moles of \(NaOH = M \times V = 0.1 \times 0.3 = 0.03 \, \text{mol}\)
NaOH is the limiting reagent.
Heat released (\(Q\)) \(= \text{moles of } H_2O \text{ formed} \times \text{Molar heat of neutralization}\)
\[ Q = 0.03 \times 57.1 \, \text{kJ} = 1.713 \, \text{kJ} = 1713 \, \text{J} \]
Total mass of solution (\(m\)) \(\approx (200 + 300) \, \text{mL} \times 1 \, \text{g/mL} = 500 \, \text{g}\)
Using the formula \(Q = m \cdot C \cdot \Delta T\):
\[ 1713 = 500 \times 4.18 \times \Delta T \]
\[ 1713 = 2090 \times \Delta T \]
\[ \Delta T = \frac{1713}{2090} \approx 0.819617 \, ^\circ C \]
The question asks for \(x\) where \(\Delta T = x \times 10^{-2}\):
\[ 0.8196 = 81.96 \times 10^{-2} \]
Step 3: Final Answer:
The nearest integer value of \(x\) is 82.