How much heat is liberated when 100 mL of 0.1 M NaOH are completely neutralised by 100 mL of 0.1 M HCI -

Options

A. `-57` kJ
B. `-0.57 kJ`
C. `-5.7 kJ `
D. `-0.05` kJ

Answer

To determine how much heat is liberated when 100 mL of 0.1 M NaOH is completely neutralized by 100 mL of 0.1 M HCl, we can follow these steps:

### Step 1: Write the Neutralization Reaction
The neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) can be represented as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{H}_2\text{O} + \text{NaCl} \]

### Step 2: Determine the Moles of Reactants
First, we need to calculate the number of moles of NaOH and HCl used in the reaction.

Given:
- Volume of NaOH = 100 mL = 0.1 L
- Concentration of NaOH = 0.1 M

Using the formula for moles:
\[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.01 \, \text{mol} \]

Similarly, for HCl:
- Volume of HCl = 100 mL = 0.1 L
- Concentration of HCl = 0.1 M

\[ \text{Moles of HCl} = 0.1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.01 \, \text{mol} \]

### Step 3: Identify the Heat of Neutralization
The heat of neutralization (\( \Delta H_{\text{neutralization}} \)) for the reaction of a strong acid with a strong base is typically around -57.3 kJ/mol. This negative sign indicates that heat is released during the reaction.

### Step 4: Calculate the Total Heat Released
Since both NaOH and HCl are present in equal moles (0.01 mol), we can calculate the total heat released using the heat of neutralization:
\[ \text{Total Heat Released} = \text{Moles of HCl} \times \Delta H_{\text{neutralization}} \]
\[ \text{Total Heat Released} = 0.01 \, \text{mol} \times (-57.3 \, \text{kJ/mol}) = -0.573 \, \text{kJ} \]

### Step 5: Interpret the Result
The negative sign indicates that this amount of heat is released during the reaction. Therefore, the heat liberated when 100 mL of 0.1 M NaOH is completely neutralized by 100 mL of 0.1 M HCl is:
\[ \text{Heat Liberated} = -0.573 \, \text{kJ} \]

### Final Answer
The heat liberated is **-0.573 kJ**.

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