# 4.10: Heat of Solution

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When preparing dilutions of concentrated sulfuric acid, the directions usually call for adding the acid slowly to water with frequent stirring. When this acid is mixed with water, a great deal of heat is released in the dissolution process. If water were added to acid, the water would quickly heat and splatter, causing harm to the person making the solution.

## Heat of Solution

Enthalpy changes also occur when a solute undergoes the physical process of dissolving into a solvent. Hot packs and cold packs (see figure below) use this property. Many hot packs use calcium chloride, which releases heat when it dissolves, according to the equation below.

$\ce{CaCl_2} \left (s \right) \rightarrow \ce{Ca^{2+}} \left( aq \right) + 2 \ce{Cl^-} \left( aq \right) + 82.8 \: \text{kJ}\nonumber$

The molar heat of solution $$\left( \Delta H_\text{soln} \right)$$ of a substance is the heat absorbed or released when one mole of the substance is dissolved in water. For calcium chloride, $$\Delta H_\text{soln} = -82.8 \: \text{kJ/mol}$$. Figure $$\PageIndex{1}$$: Chemical hot packs and cold packs work because of the heats of solution of the chemicals inside them. When the bag is squeezed, an inner pouch bursts, allowing the chemical to dissolve in water. Heat is released in the hot pack and absorbed in the cold pack. (CC BY-NC; CK-12)

Many cold packs use ammonium nitrate, which absorbs heat from the surroundings when it dissolves.

$\ce{NH_4NO_3} \left( s \right) + 25.7 \: \text{kJ} \rightarrow \ce{NH_4^+} \left( aq \right) + \ce{NO_3^-} \left( aq \right)\nonumber$

Cold packs are typically used to treat muscle strains and sore joints. The cold pack is activated and applied to the affected area. As the ammonium nitrate dissolves, it absorbs heat from the body and helps to limit swelling. For ammonium nitrate, $$\Delta H_\text{soln} = 25.7 \: \text{kJ/mol}$$.

##### Example $$\PageIndex{1}$$

The molar heat of solution, $$\Delta H_\text{soln}$$, of $$\ce{NaOH}$$ is $$-44.51 \: \text{kJ/mol}$$. In a certain experiment, $$50.0 \: \text{g}$$ of $$\ce{NaOH}$$ is completely dissolved in $$1.000 \: \text{L}$$ of $$20.0^\text{o} \text{C}$$ water in a foam cup calorimeter. Assuming no heat loss, calculate the final temperature of the water.

###### Known
• Mass $$\ce{NaOH} = 50.0 \: \text{g}$$
• Molar mass $$\ce{NaOH} = 40.00 \: \text{g/mol}$$
• $$\Delta H_\text{soln} \: \left( \ce{NaOH} \right) = -44.51 \: \text{kJ/mol}$$
• Mass $$\ce{H_2O} = 1.000 \: \text{kg} = 1000. \: \text{g}$$ (assumes density $$= 1.00 \: \text{g/mL}$$)
• $$T_\text{initial} \: \left( \ce{H_2O} \right) = 20.0^\text{o} \text{C}$$
• $$c_p \: \left( \ce{H_2O} \right) = 4.18 \: \text{J/g}^\text{o} \text{C}$$
###### Unknown

This is a multiple-step problem:

1) Grams $$\ce{NaOH}$$ is converted to moles.

2) Moles is multiplied by the molar heat of solution.

3) The joules of heat released in the dissolution process is used with the specific heat equation and the total mass of the solution to calculate the $$\Delta T$$.

4) The $$T_\text{final}$$ is determined from $$\Delta T$$.

###### Step 2: Solve.

$50.0 \: \text{g} \: \ce{NaOH} \times \frac{1 \: \text{mol} \: \ce{NaOH}}{40.00 \: \text{g} \: \ce{NaOH}} \times \frac{-44.51 \: \text{kJ}}{1 \: \text{mol} \: \ce{NaOH}} \times \frac{1000 \: \text{J}}{1 \: \text{kJ}} = -5.56 \times 10^4 \: \text{J}\nonumber$

$\Delta T = \frac{\Delta H}{c_p \times m} = \frac{-5.56 \times 10^4 \: \text{J}}{4.18 \: \text{J/g}^\text{o} \text{C} \times 1050 \: \text{g}} = 13.2^\text{o} \text{C}\nonumber$

$T_\text{final} = 20.0^\text{o} \text{C} + 13.2^\text{o} \text{C} = 33.2^\text{o} \text{C}\nonumber$

• The molar heat of solution $$\left( \Delta H_\text{soln} \right)$$ of a substance is the heat absorbed or released when one mole of the substance is dissolved in water.