Calculate the enthalpy change when `50 mL` of `0.01 M Ca(OH)_(2)` reacts with `25mL` of `0.01 M HCI`. Given that `DeltaH^(Theta)` neutralisaiton of strong acid and string base is `140 kcal mol^(-1)`

Options

A. `14 cal`
B. `35 cal`
C. `10 cal`
D. `7.5 cal`

Explanation

To calculate the enthalpy change when `50 mL` of `0.01 M Ca(OH)₂` reacts with `25 mL` of `0.01 M HCl`, we will follow these steps:

### Step 1: Calculate the number of moles of HCl
To find the number of moles of HCl, we use the formula:

\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in liters)} \]

Given:
- Molarity of HCl = `0.01 M`
- Volume of HCl = `25 mL = 0.025 L`

Calculating the number of moles of HCl:

\[ \text{Number of moles of HCl} = 0.01 \, \text{mol/L} \times 0.025 \, \text{L} = 2.5 \times 10^{-4} \, \text{moles} \]

### Step 2: Calculate the number of moles of Ca(OH)₂
Now, we will calculate the number of moles of Ca(OH)₂ using the same formula:

Given:
- Molarity of Ca(OH)₂ = `0.01 M`
- Volume of Ca(OH)₂ = `50 mL = 0.050 L`

Calculating the number of moles of Ca(OH)₂:

\[ \text{Number of moles of Ca(OH)₂} = 0.01 \, \text{mol/L} \times 0.050 \, \text{L} = 5.0 \times 10^{-4} \, \text{moles} \]

### Step 3: Calculate the number of moles of OH⁻ ions
Each molecule of Ca(OH)₂ produces 2 OH⁻ ions upon dissociation. Therefore, the number of moles of OH⁻ ions is:

\[ \text{Number of moles of OH⁻} = 2 \times \text{Number of moles of Ca(OH)₂} = 2 \times 5.0 \times 10^{-4} = 1.0 \times 10^{-3} \, \text{moles} \]

### Step 4: Identify the limiting reactant
In the neutralization reaction, H⁺ ions from HCl will react with OH⁻ ions from Ca(OH)₂. We have:

- Moles of H⁺ = `2.5 × 10⁻⁴` moles
- Moles of OH⁻ = `1.0 × 10⁻³` moles

Since H⁺ is the limiting reactant, it will determine the extent of the reaction.

### Step 5: Calculate the enthalpy change
The enthalpy change for the neutralization of a strong acid with a strong base is given as `140 kcal/mol`. We will calculate the enthalpy change for the moles of H⁺ that react:

\[ \Delta H = \text{Number of moles of H⁺} \times \Delta H^{\circ}_{\text{neutralization}} \]

Substituting the values:

\[ \Delta H = 2.5 \times 10^{-4} \, \text{moles} \times 140 \, \text{kcal/mol} \]

Calculating:

\[ \Delta H = 0.035 \, \text{kcal} = 35 \, \text{cal} \]

### Final Answer
The enthalpy change when `50 mL` of `0.01 M Ca(OH)₂` reacts with `25 mL` of `0.01 M HCl` is `35 calories`.

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