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 follow these steps:

### Step 1: Write the neutralization reaction
The neutralization reaction of 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 changes
We need to use the given standard enthalpy values:
- \(\Delta_s H^\circ (\text{HF}, aq) = -329 \, \text{kJ/mol}\)
- \(\Delta_s H^\circ (\text{H}_2\text{O}, l) = -285 \, \text{kJ/mol}\)
- \(\Delta_s H^\circ (\text{F}^-, aq) = -320 \, \text{kJ/mol}\)
- The enthalpy of neutralization for the reaction of \(H^+\) and \(OH^-\) is given as:
\(\Delta_r H^\circ = -56 \, \text{kJ/mol}\)

### Step 3: Calculate the enthalpy of the products
The products of the reaction are \( \text{F}^-_{(aq)} \) and \( \text{H}_2\text{O}_{(l)} \). The total enthalpy for the products is:
\[ \Delta H_{products} = \Delta_s H^\circ (\text{F}^-, aq) + \Delta_s H^\circ (\text{H}_2\text{O}, l) \]
\[ \Delta H_{products} = (-320 \, \text{kJ/mol}) + (-285 \, \text{kJ/mol}) = -605 \, \text{kJ/mol} \]

### Step 4: Calculate the enthalpy of the reactants
The reactants are \( \text{HF}_{(aq)} \) and \( \text{OH}^-_{(aq)} \). The total enthalpy for the reactants is:
\[ \Delta H_{reactants} = \Delta_s H^\circ (\text{HF}, aq) + \Delta_s H^\circ (\text{OH}^-, aq) \]
We need to find the sublimation enthalpy of \( \text{OH}^- \). The enthalpy of the reaction \( H^+ + OH^- \rightarrow H_2O \) is given as \(-56 \, \text{kJ/mol}\), which means:
\[ \Delta_s H^\circ (\text{OH}^-, aq) = -56 \, \text{kJ/mol} - \Delta_s H^\circ (\text{H}_2\text{O}, l) \]
\[ \Delta_s H^\circ (\text{OH}^-, aq) = -56 \, \text{kJ/mol} + 285 \, \text{kJ/mol} = -229 \, \text{kJ/mol} \]
Now, substituting this value into the reactants' enthalpy:
\[ \Delta H_{reactants} = (-329 \, \text{kJ/mol}) + (-229 \, \text{kJ/mol}) = -558 \, \text{kJ/mol} \]

### Step 5: Calculate the enthalpy of neutralization
Now we can find the enthalpy of neutralization (\(\Delta H_{neutralization}\)):
\[ \Delta H_{neutralization} = \Delta H_{products} - \Delta H_{reactants} \]
\[ \Delta H_{neutralization} = (-605 \, \text{kJ/mol}) - (-558 \, \text{kJ/mol}) = -47 \, \text{kJ/mol} \]

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