What is the enthalpy of neutralization of HF against a strong base?
Given : `H^(+)(aq)+OH^(-)(aq)rarrH_(2)O(l),Delta_(r)H^(@)=-56kJ//mol`
`Delta_(s)H^(@)(HF,aq)=-329kJ//mol,Delta_(s)H^(@)(H_(2)O,l)=-285kJ//mol`
`Delta_(s)H^(@)(F^(-),aq)=-320 " kJ"//"mol"`

Options

A. `-17` kJ/mol
B. `-38` kJ/mol
C. `-47` kJ/mol
D. `-43` kJ/mol

Explanation

To calculate the enthalpy of neutralization of HF against a strong base, we can use the given enthalpy values and the reaction involved. Here’s a step-by-step solution:

### Step 1: Write the Neutralization Reaction
The neutralization of hydrofluoric acid (HF) with a strong base (like NaOH) can be represented as:
\[ \text{HF (aq)} + \text{OH}^- (aq) \rightarrow \text{F}^- (aq) + \text{H}_2\text{O} (l) \]

### Step 2: Identify the Enthalpy Change
The enthalpy change for the neutralization reaction can be expressed as:
\[ \Delta H_{\text{neutralization}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \]

### Step 3: Write the Enthalpy of Products
The products of the reaction are F⁻ (aq) and H₂O (l). The enthalpy values provided are:
- \( \Delta H_f^0(\text{F}^-) = -320 \, \text{kJ/mol} \)
- \( \Delta H_f^0(\text{H}_2\text{O}) = -285 \, \text{kJ/mol} \)

Thus, the total enthalpy of the products is:
\[ \Delta H_{\text{products}} = \Delta H_f^0(\text{F}^-) + \Delta H_f^0(\text{H}_2\text{O}) = -320 + (-285) = -605 \, \text{kJ/mol} \]

### Step 4: Write the Enthalpy of Reactants
The reactants are HF and OH⁻. The enthalpy values provided are:
- \( \Delta H_f^0(\text{HF}) = -329 \, \text{kJ/mol} \)
- The enthalpy of OH⁻ can be derived from the known reaction:
\[ \text{H}^+ (aq) + \text{OH}^- (aq) \rightarrow \text{H}_2\text{O} (l) \]
Given \( \Delta H_r^0 = -56 \, \text{kJ/mol} \), we can find the enthalpy of OH⁻:
\[ -56 = 0 + \Delta H_f^0(\text{OH}^-) - 285 \]
\[ \Delta H_f^0(\text{OH}^-) = -56 + 285 = -229 \, \text{kJ/mol} \]

Thus, the total enthalpy of the reactants is:
\[ \Delta H_{\text{reactants}} = \Delta H_f^0(\text{HF}) + \Delta H_f^0(\text{OH}^-) = -329 + (-229) = -558 \, \text{kJ/mol} \]

### Step 5: Calculate the Enthalpy of Neutralization
Now we can substitute the values into the equation for the enthalpy of neutralization:
\[ \Delta H_{\text{neutralization}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \]
\[ \Delta H_{\text{neutralization}} = -605 - (-558) = -605 + 558 = -47 \, \text{kJ/mol} \]

### Final Answer
The enthalpy of neutralization of HF against a strong base is:
\[ \Delta H_{\text{neutralization}} = -47 \, \text{kJ/mol} \]

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