The molar heat of solution (ΔHsoln) of sodium hydroxide is -445.1 kJ/mol. In a certain experiment, 5.00 g of solid sodium hydroxide is completely dissolved in 1.000 L of 20.0 °C water in a foam cup calorimeter. Assuming no heat loss, calculate the final temperature of the water.
The Correct Answer and Explanation is:
To calculate the final temperature of the water when sodium hydroxide (NaOH) dissolves, we can use the heat released by the dissolution process. The heat released will raise the temperature of the water. Here’s the step-by-step solution:
Step 1: Calculate the moles of NaOH
The given mass of NaOH is 5.00 g, and its molar mass is approximately 40.0 g/mol. To find the number of moles of NaOH: moles of NaOH=mass of NaOHmolar mass of NaOH=5.00 g40.0 g/mol=0.125 mol\text{moles of NaOH} = \frac{\text{mass of NaOH}}{\text{molar mass of NaOH}} = \frac{5.00 \, \text{g}}{40.0 \, \text{g/mol}} = 0.125 \, \text{mol}moles of NaOH=molar mass of NaOHmass of NaOH=40.0g/mol5.00g=0.125mol
Step 2: Calculate the heat released during the dissolution
The heat of solution (ΔHsoln) for NaOH is -445.1 kJ/mol. This value tells us how much heat is released per mole of NaOH dissolved. Since NaOH dissociates in water, the heat released is: heat released=moles of NaOH×ΔHsoln=0.125 mol×(−445.1 kJ/mol)=−55.64 kJ\text{heat released} = \text{moles of NaOH} \times \Delta H_{\text{soln}} = 0.125 \, \text{mol} \times (-445.1 \, \text{kJ/mol}) = -55.64 \, \text{kJ}heat released=moles of NaOH×ΔHsoln=0.125mol×(−445.1kJ/mol)=−55.64kJ
Thus, 55.64 kJ of heat is released to the water.
Step 3: Calculate the temperature change of the water
The heat released will increase the temperature of the water. The specific heat capacity of water is 4.18 J/g°C4.18 \, \text{J/g°C}4.18J/g°C, and the mass of the water is 1.000 L, which is equivalent to 1000 g of water. We can now use the heat equation: q=m⋅c⋅ΔTq = m \cdot c \cdot \Delta Tq=m⋅c⋅ΔT
Where:
- qqq is the heat absorbed by the water, 55.64 kJ = 55640 J,
- mmm is the mass of the water, 1000 g,
- ccc is the specific heat capacity of water, 4.18 J/g°C4.18 \, \text{J/g°C}4.18J/g°C,
- ΔT\Delta TΔT is the change in temperature.
Rearranging the equation to solve for ΔT\Delta TΔT: ΔT=qm⋅c=55640 J1000 g×4.18 J/g°C=13.3°C\Delta T = \frac{q}{m \cdot c} = \frac{55640 \, \text{J}}{1000 \, \text{g} \times 4.18 \, \text{J/g°C}} = 13.3°CΔT=m⋅cq=1000g×4.18J/g°C55640J=13.3°C
Step 4: Calculate the final temperature
The initial temperature of the water is 20.0°C. Adding the temperature change: final temperature=20.0°C+13.3°C=33.3°C\text{final temperature} = 20.0°C + 13.3°C = 33.3°Cfinal temperature=20.0°C+13.3°C=33.3°C
Final Answer:
The final temperature of the water will be 33.3°C.
This calculation assumes no heat loss to the surroundings and that all the heat released by the dissolving NaOH is absorbed by the water.
