When 5.0 mL of a 1.0 M HCl solution is mixed with 5.0 mL of a 0.1 M NaOH solution, temperature of solution is increased by `2^@C` predicted accurately from this observation?

Options

A. If 10 mL of same HCl is mixed with 10 mL of same NaOH, temperature rise will be `4^@C`
B. If 10 mL of 0.05 HCl is mixed with 10 mL of 0.05 M NaCl the temperature rise will be `2^@C`
C. If 5 mL of 0.1 M HCl is mixed with 5 mL of 0.1 M `NH_3` solution, the temperature rise will be less than `2^@C`
D. If 5 mL 0.1 M is mixed with 5 mL of 0.1 M NaOH, the temperature rise will be less than `2^@C`

Explanation

To solve the problem, we need to analyze the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) and the implications of the temperature change observed.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between HCl and NaOH can be written as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
This is a neutralization reaction.

2. **Calculate Moles of Reactants**:
- For HCl:
\[ \text{Molarity} = 1.0 \, \text{M}, \quad \text{Volume} = 5.0 \, \text{mL} = 0.005 \, \text{L} \]
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 1.0 \, \text{mol/L} \times 0.005 \, \text{L} = 0.005 \, \text{mol} \]

- For NaOH:
\[ \text{Molarity} = 0.1 \, \text{M}, \quad \text{Volume} = 5.0 \, \text{mL} = 0.005 \, \text{L} \]
\[ \text{Moles of NaOH} = 0.1 \, \text{mol/L} \times 0.005 \, \text{L} = 0.0005 \, \text{mol} \]

3. **Determine Limiting Reactant**:
Since HCl has 0.005 moles and NaOH has 0.0005 moles, NaOH is the limiting reactant. This means that all of the NaOH will react, and there will be excess HCl remaining.

4. **Temperature Change**:
The temperature of the solution increased by 2°C. This indicates that the reaction is exothermic, meaning it releases heat. The amount of heat released is related to the moles of the limiting reactant (NaOH in this case).

5. **Predicting Temperature Change with a Weak Acid**:
If we were to replace HCl (a strong acid) with acetic acid (a weak acid), the reaction would not go to completion as with a strong acid. Therefore, the heat released would be less, leading to a smaller temperature increase.

Since the question states that the temperature rise is less than 2°C when mixing a weak acid with NaOH, we can conclude that the temperature rise would be less than that observed with HCl.

6. **Conclusion**:
Based on the above analysis, the correct prediction from the observation is that if a weak acid (like acetic acid) is used instead of HCl, the temperature rise would be less than 2°C.

### Final Answer:
The predicted outcome is that if a weak acid (like acetic acid) is mixed with NaOH, the temperature rise will be less than 2°C.