25.0 mL of 1.0M HCl is combined with 35.0 mL of 0.5 M NaOH. The initial temperatures of the solutions is `25^@C`, the density of the solution is 1.0 g/mL, the specific heat capacity of the solution is `4.184 J//g .^(@)C`, the reaction is completed in insulated beaker, and the standard enthalpy of reaction for `H^(+)(aq) + OH^(-)(aq) rarr H_(2)O (l)` is `-56 kJ//mol` . What is the final temperature of the solution?

Options

A. `27^@C`
B. `28.9^@C`
C. `30.1^@C`
D. `32.8^@C`

Explanation

To find the final temperature of the solution after mixing HCl and NaOH, we can follow these steps:

### Step 1: Calculate the millimoles of HCl
We know the formula for millimoles:
\[ \text{Millimoles} = \text{Molarity} \times \text{Volume (in L)} \]
For HCl:
\[ \text{Volume} = 25.0 \, \text{mL} = 0.025 \, \text{L} \]
\[ \text{Molarity} = 1.0 \, \text{M} \]
\[ \text{Millimoles of HCl} = 1.0 \, \text{mol/L} \times 0.025 \, \text{L} = 25.0 \, \text{mmol} \]

### Step 2: Calculate the millimoles of NaOH
Using the same formula:
\[ \text{Volume} = 35.0 \, \text{mL} = 0.035 \, \text{L} \]
\[ \text{Molarity} = 0.5 \, \text{M} \]
\[ \text{Millimoles of NaOH} = 0.5 \, \text{mol/L} \times 0.035 \, \text{L} = 17.5 \, \text{mmol} \]

### Step 3: Determine the limiting reagent
The reaction is:
\[ \text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) \]
Since 17.5 mmol of NaOH will react with 17.5 mmol of HCl, NaOH is the limiting reagent.

### Step 4: Calculate the heat released (Q)
The standard enthalpy of the reaction is given as -56 kJ/mol. For 17.5 mmol:
\[ Q = \left(-56 \, \text{kJ/mol}\right) \times \left(17.5 \, \text{mmol} \times \frac{1 \, \text{mol}}{1000 \, \text{mmol}}\right) = -0.98 \, \text{kJ} = -980 \, \text{J} \]

### Step 5: Calculate the total mass of the solution
The total volume of the solution is:
\[ \text{Total Volume} = 25.0 \, \text{mL} + 35.0 \, \text{mL} = 60.0 \, \text{mL} \]
Given the density of the solution is 1.0 g/mL:
\[ \text{Mass} = \text{Volume} \times \text{Density} = 60.0 \, \text{mL} \times 1.0 \, \text{g/mL} = 60.0 \, \text{g} \]

### Step 6: Calculate the change in temperature (ΔT)
Using the formula:
\[ Q = mc\Delta T \]
Where:
- \( Q = -980 \, \text{J} \)
- \( m = 60.0 \, \text{g} \)
- \( c = 4.184 \, \text{J/g°C} \)

Rearranging gives:
\[ \Delta T = \frac{Q}{mc} = \frac{-980 \, \text{J}}{60.0 \, \text{g} \times 4.184 \, \text{J/g°C}} \approx -3.9 \, \text{°C} \]

### Step 7: Calculate the final temperature (Tf)
The initial temperature (Ti) is 25.0 °C:
\[ T_f = T_i + \Delta T = 25.0 \, \text{°C} + (-3.9 \, \text{°C}) = 25.0 \, \text{°C} - 3.9 \, \text{°C} = 21.1 \, \text{°C} \]

### Final Answer
The final temperature of the solution is approximately **21.1 °C**.

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