If enthalpy of neutralisation of HCl by NaOH is -57 kJ `" mol"^(-1)` and with `NH_(4)OH` is -50 kJ `" mol"^(-1)`. Calculate enthalpy of ionisation of `NH_(4)OH` (aq).

Answer

To calculate the enthalpy of ionization of \( NH_4OH \) (aq), we can use the given enthalpy values of neutralization and apply Hess's Law. Here's the step-by-step solution:

### Step 1: Understand the reactions involved
The enthalpy of neutralization for a strong acid (HCl) reacting with a strong base (NaOH) is given as -57 kJ/mol. This reaction can be represented as:
\[ H^+ (aq) + OH^- (aq) \rightarrow H_2O (l) \quad \Delta H = -57 \text{ kJ/mol} \]

The enthalpy of neutralization for a weak base \( NH_4OH \) with HCl is given as -50 kJ/mol:
\[ H^+ (aq) + NH_4OH (aq) \rightarrow NH_4Cl (aq) + H_2O (l) \quad \Delta H = -50 \text{ kJ/mol} \]

### Step 2: Set up the equation using Hess's Law
According to Hess's Law, the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We can express the reaction involving \( NH_4OH \) in two steps:
1. Ionization of \( NH_4OH \):
\[ NH_4OH (aq) \rightarrow NH_4^+ (aq) + OH^- (aq) \quad \Delta H = ? \text{ (this is what we want to find)} \]
2. Neutralization of \( H^+ \) with \( OH^- \):
\[ H^+ (aq) + OH^- (aq) \rightarrow H_2O (l) \quad \Delta H = -57 \text{ kJ/mol} \]

Combining these two steps gives us:
\[ H^+ (aq) + NH_4OH (aq) \rightarrow NH_4^+ (aq) + H_2O (l) \]

### Step 3: Write the enthalpy change for the overall reaction
The overall enthalpy change for the reaction with \( NH_4OH \) is given as -50 kJ/mol:
\[ \Delta H_{total} = \Delta H_{ionization} + (-57 \text{ kJ/mol}) = -50 \text{ kJ/mol} \]

### Step 4: Solve for the enthalpy of ionization
Now we can rearrange the equation to find the enthalpy of ionization:
\[ \Delta H_{ionization} - 57 = -50 \]
\[ \Delta H_{ionization} = -50 + 57 \]
\[ \Delta H_{ionization} = 7 \text{ kJ/mol} \]

### Final Answer
The enthalpy of ionization of \( NH_4OH \) (aq) is \( 7 \text{ kJ/mol} \).

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