Enthalpies of Solution and Hydration

A2 · 13 min

When an ionic solid dissolves in water, two things happen at once: the lattice is pulled apart, which costs energy, and the freed ions are surrounded by water molecules, which releases energy. The balance between these two decides whether dissolving is exothermic or endothermic. This note defines the enthalpy changes of solution and hydration, builds the energy cycle that links them to lattice energy, and explains what makes a hydration enthalpy large. The same ideas explain the solubility trends of Group 2 hydroxides and sulfates.

Two new enthalpy changes

Definition

The enthalpy change of solution, $\Delta H^{\ominus}_{\text{sol}}$ , is the enthalpy change when one mole of an ionic substance dissolves in a sufficient volume of water to form an infinitely dilute solution, under standard conditions.

The enthalpy change of hydration, $\Delta H^{\ominus}_{\text{hyd}}$ , is the enthalpy change when one mole of a specified gaseous ion dissolves in a sufficient volume of water to form an infinitely dilute solution, under standard conditions.

"Infinitely dilute" means so much water is used that adding more water causes no further enthalpy change. The ions are then far enough apart that they no longer interact with each other, only with water.

\ce{NaCl(s) + aq -> Na^+(aq) + Cl^-(aq)} \qquad \Delta H^{\ominus}_{\text{sol}}

\ce{Na^+(g) + aq -> Na^+(aq)} \qquad \Delta H^{\ominus}_{\text{hyd}}(\ce{Na+})

\ce{Cl^-(g) + aq -> Cl^-(aq)} \qquad \Delta H^{\ominus}_{\text{hyd}}(\ce{Cl-})

The symbol $\ce{aq}$ on the left stands for "excess water". Hydration enthalpies are for one ion at a time, starting from the gaseous ion. Enthalpies of solution start from the solid.

Why hydration is always exothermic

Water is a polar molecule. The oxygen atom carries a partial negative charge ( $\delta-$ ) and its lone pairs, and each hydrogen carries a partial positive charge ( $\delta+$ ).

Around a cation, water molecules orient with their $\delta-$ oxygen atoms pointing towards the ion.
Around an anion, water molecules orient with their $\delta+$ hydrogen atoms pointing towards the ion.

These ion–dipole attractions form when the gaseous ion enters water. Forming attractions releases energy, so every hydration enthalpy is negative.

Watch out

Do not say water molecules form "bonds" to the ion by sharing electrons, or that the oxygen "bonds to the anion". In simple hydration the attraction is electrostatic: $\delta-$ oxygen to cations, $\delta+$ hydrogen to anions. (Coordinate bonds from water to transition metal ions are a separate idea, covered in complex ions.)

The solution energy cycle

You cannot easily measure hydration enthalpies directly, but you can link them to two quantities you can find: the enthalpy change of solution (measured in a calorimeter) and the lattice energy (from a Born–Haber cycle).

Think of dissolving as two imaginary steps.

Break the lattice into gaseous ions. This is the reverse of lattice energy, so its enthalpy change is $-\Delta H^{\ominus}_{\text{latt}}$ (positive).
Hydrate each gaseous ion. The enthalpy change is the sum of the hydration enthalpies of all the ions in the formula (negative).

Energy cycle for dissolving sodium chloride. Going from NaCl(s) to the solution directly (ΔH sol) equals going up against the lattice energy arrow and then down the hydration arrow.

By Hess's law, the route from solid to solution directly equals the route via the gaseous ions:

Linking solution, hydration and lattice energy

\Delta H^{\ominus}_{\text{sol}} = \sum \Delta H^{\ominus}_{\text{hyd}}(\text{ions}) - \Delta H^{\ominus}_{\text{latt}}

Multiply each hydration enthalpy by the number of that ion in the formula. For $\ce{MgCl2}$ :

\Delta H^{\ominus}_{\text{sol}} = \Delta H^{\ominus}_{\text{hyd}}(\ce{Mg^{2+}}) + 2\Delta H^{\ominus}_{\text{hyd}}(\ce{Cl-}) - \Delta H^{\ominus}_{\text{latt}}(\ce{MgCl2})

Because lattice energy is negative, subtracting it adds a large positive number. So $\Delta H^{\ominus}_{\text{sol}}$ is the small difference between two large quantities:

If the hydration enthalpies release more energy than is needed to break the lattice, $\Delta H^{\ominus}_{\text{sol}}$ is negative and the solution warms up.
If they release less, $\Delta H^{\ominus}_{\text{sol}}$ is positive and the solution cools down (ammonium nitrate in cold packs is the classic example).

Tip

Because $\Delta H^{\ominus}_{\text{sol}}$ is a small difference between large numbers, an uncertainty of a few kJ mol⁻¹ in either term can change its sign. Different data books give slightly different hydration enthalpies, so two textbooks may disagree about a calculated $\Delta H^{\ominus}_{\text{sol}}$ . Always use the data given in the question.

What makes a hydration enthalpy large

The strength of the ion–dipole attraction depends on how concentrated the ion's charge is, which chemists call its charge density.

Factors affecting the magnitude of hydration enthalpy

Ionic charge. A higher charge attracts water molecules more strongly, so $\Delta H^{\ominus}_{\text{hyd}}$ is more exothermic.
Ionic radius. A smaller ion lets water molecules approach its centre more closely, so the attraction is stronger and $\Delta H^{\ominus}_{\text{hyd}}$ is more exothermic.

Typical values (kJ mol⁻¹; values differ by a few percent between data sources):

Cation	radius / nm	$\Delta H^{\ominus}_{\text{hyd}}$	Anion	radius / nm	$\Delta H^{\ominus}_{\text{hyd}}$
$\ce{Li+}$	0.076	–519	$\ce{F-}$	0.133	–506
$\ce{Na+}$	0.102	–406	$\ce{Cl-}$	0.181	–364
$\ce{K+}$	0.138	–322	$\ce{Br-}$	0.196	–335
$\ce{Mg^{2+}}$	0.072	–1920	$\ce{I-}$	0.220	–293
$\ce{Ca^{2+}}$	0.100	–1650	$\ce{OH-}$	—	–460
$\ce{Sr^{2+}}$	0.118	–1480
$\ce{Ba^{2+}}$	0.135	–1360

Read the table in two directions.

Down a group (Li to K, Mg to Ba, F to I) the ions get larger and the hydration enthalpy becomes less exothermic.
Across from 1+ to 2+ at similar radius ( $\ce{Na+}$ 0.102 nm and $\ce{Ca^{2+}}$ 0.100 nm) the hydration enthalpy roughly quadruples, from $-406$ to $-1650$ . Charge has a big effect.

The same two factors control lattice energy, which is why the solubility of a compound is hard to predict without numbers: making ions smaller or more highly charged makes both the lattice energy and the hydration enthalpies more exothermic. Whether a compound becomes more or less soluble depends on which effect grows faster. This is exactly the argument used for Group 2 hydroxides and sulfates.

Solving an energy cycle problem

Write the equation for the enthalpy change you want, with state symbols.
Sketch the triangle: gaseous ions at the top, solid at bottom left, aqueous ions at bottom right.
Write the Hess's law equation: $\Delta H^{\ominus}_{\text{sol}} = \sum \Delta H^{\ominus}_{\text{hyd}} - \Delta H^{\ominus}_{\text{latt}}$ , with multipliers for each ion.
Substitute with brackets around negative numbers and rearrange for the unknown.
Check the sign of the answer makes sense: lattice energy and hydration enthalpies are negative; $\Delta H^{\ominus}_{\text{sol}}$ can be either sign and is usually small.

Worked examples

Enthalpy change of solution of sodium chloride

Calculate $\Delta H^{\ominus}_{\text{sol}}$ for $\ce{NaCl}$ given: $\Delta H^{\ominus}_{\text{latt}}(\ce{NaCl}) = -787$ ; $\Delta H^{\ominus}_{\text{hyd}}(\ce{Na+}) = -406$ ; $\Delta H^{\ominus}_{\text{hyd}}(\ce{Cl-}) = -364\ \text{kJ mol}^{-1}$ .

Solution

\Delta H^{\ominus}_{\text{sol}} = (-406) + (-364) - (-787) = -770 + 787 = +17\ \text{kJ mol}^{-1}

The value is positive: slightly more energy is needed to separate the ions than is released by hydrating them. (The measured value is about $+4\ \text{kJ mol}^{-1}$ ; the difference shows how sensitive a small difference of large numbers is to the data used.)

Lattice energy from enthalpy of solution

For magnesium chloride, $\Delta H^{\ominus}_{\text{sol}} = -155\ \text{kJ mol}^{-1}$ . Using $\Delta H^{\ominus}_{\text{hyd}}(\ce{Mg^{2+}}) = -1920$ and $\Delta H^{\ominus}_{\text{hyd}}(\ce{Cl-}) = -364\ \text{kJ mol}^{-1}$ , calculate the lattice energy of $\ce{MgCl2}$ .

Solution

\Delta H^{\ominus}_{\text{sol}} = \Delta H^{\ominus}_{\text{hyd}}(\ce{Mg^{2+}}) + 2\Delta H^{\ominus}_{\text{hyd}}(\ce{Cl-}) - \Delta H^{\ominus}_{\text{latt}}

-155 = -1920 + 2(-364) - \Delta H^{\ominus}_{\text{latt}}

\Delta H^{\ominus}_{\text{latt}} = -1920 - 728 + 155 = -2493\ \text{kJ mol}^{-1}

This agrees reasonably with the Born–Haber value of about $-2520\ \text{kJ mol}^{-1}$ .

Finding a hydration enthalpy

For calcium chloride, $\Delta H^{\ominus}_{\text{sol}} = -83\ \text{kJ mol}^{-1}$ and $\Delta H^{\ominus}_{\text{latt}} = -2258\ \text{kJ mol}^{-1}$ . Given $\Delta H^{\ominus}_{\text{hyd}}(\ce{Cl-}) = -364\ \text{kJ mol}^{-1}$ , calculate $\Delta H^{\ominus}_{\text{hyd}}(\ce{Ca^{2+}})$ .

Solution

-83 = \Delta H^{\ominus}_{\text{hyd}}(\ce{Ca^{2+}}) + 2(-364) - (-2258)

\Delta H^{\ominus}_{\text{hyd}}(\ce{Ca^{2+}}) = -83 + 728 - 2258 = -1613\ \text{kJ mol}^{-1}

Explaining hydration enthalpies

Explain why the hydration enthalpy of $\ce{Mg^{2+}}$ ( $-1920\ \text{kJ mol}^{-1}$ ) is more exothermic than that of $\ce{Na+}$ ( $-406$ ) and that of $\ce{Ca^{2+}}$ ( $-1650$ ).

Solution

Hydration releases energy because the $\delta-$ oxygen atoms of polar water molecules are attracted to the cation.

$\ce{Mg^{2+}}$ has twice the charge of $\ce{Na+}$ and is also smaller (0.072 nm against 0.102 nm), so it has a much higher charge density and attracts water molecules far more strongly.
$\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ have the same charge, but $\ce{Mg^{2+}}$ is smaller, so water molecules get closer to its centre and are attracted more strongly.

Enthalpy of solution and solubility trend (exam-hard)

Data for two Group 2 sulfates (kJ mol⁻¹):

	$\Delta H^{\ominus}_{\text{latt}}$	$\Delta H^{\ominus}_{\text{hyd}}(\ce{M^{2+}})$
$\ce{MgSO4}$	–2833	–1920
$\ce{BaSO4}$	–2469	–1360

$\Delta H^{\ominus}_{\text{hyd}}(\ce{SO4^{2-}}) = -1059\ \text{kJ mol}^{-1}$ .

(a) Calculate $\Delta H^{\ominus}_{\text{sol}}$ for each sulfate. (b) Magnesium sulfate is soluble in water; barium sulfate is almost insoluble. Explain how your answers are consistent with this, referring to the relative changes in lattice energy and hydration enthalpy.

Solution

(a)

\ce{MgSO4}: \quad -1920 + (-1059) - (-2833) = -146\ \text{kJ mol}^{-1}

\ce{BaSO4}: \quad -1360 + (-1059) - (-2469) = +50\ \text{kJ mol}^{-1}

(b) From $\ce{Mg^{2+}}$ to $\ce{Ba^{2+}}$ the cation hydration enthalpy becomes less exothermic by $560\ \text{kJ mol}^{-1}$ , but the lattice energy becomes less exothermic by only $364\ \text{kJ mol}^{-1}$ . The sulfate ion is large, so increasing the cation radius makes little difference to the inter-ionic distance and hence to the lattice energy, while the hydration of the smaller cation changes a lot. The hydration term falls faster than the lattice term, so $\Delta H^{\ominus}_{\text{sol}}$ becomes more endothermic down the group and the sulfates become less soluble.

Is a negative enthalpy of solution needed to dissolve?

No. Many salts with positive $\Delta H^{\ominus}_{\text{sol}}$ , such as ammonium nitrate and potassium chloride, dissolve readily. Dissolving spreads ions out through the solvent, which increases the entropy of the system. Whether dissolving is feasible depends on the Gibbs free energy change, $\Delta G = \Delta H - T\Delta S$ , which combines both factors (see entropy and Gibbs free energy). The enthalpy change of solution is one useful guide, not the whole story.

Common mistakes

Sign of the lattice term. It is $-\Delta H^{\ominus}_{\text{latt}}$ in the solution cycle, because the lattice is broken. Writing $+\Delta H^{\ominus}_{\text{latt}}$ is the single most common error.
Forgetting multipliers. $\ce{CaCl2}$ has two chloride ions to hydrate.
Hydration of a compound. Hydration enthalpy is for a single gaseous ion, never for "NaCl".
State symbols. Hydration starts from $\text{(g)}$ ions; solution starts from the $\text{(s)}$ compound; both end with $\text{(aq)}$ .
Explaining hydration with "bonds". Describe the attraction between the ion and the oppositely charged end of polar water molecules.

Exam technique

Exam tip

The definitions of $\Delta H^{\ominus}_{\text{sol}}$ and $\Delta H^{\ominus}_{\text{hyd}}$ are 2-mark items. Include "one mole", the correct starting state (solid or gaseous ion), "infinitely dilute solution" (or "so that no further enthalpy change occurs on dilution") and "standard conditions".
When asked to "construct an energy cycle", draw the triangle with all three species boxes, state symbols, and labelled arrows. Arrow directions matter: the lattice energy arrow points from gaseous ions to solid.
When asked to explain a trend in $\Delta H^{\ominus}_{\text{hyd}}$ , refer to ionic charge and ionic radius and to the attraction between the ion and water molecules (charge density). Mention the polar nature of water.
In solubility trend questions, the mark scheme rewards comparing how much the lattice energy changes with how much the hydration enthalpy changes.

Summary

$\Delta H^{\ominus}_{\text{sol}}$ : one mole of ionic solid dissolves to give an infinitely dilute solution. It may be positive or negative.
$\Delta H^{\ominus}_{\text{hyd}}$ : one mole of a gaseous ion dissolves to give an infinitely dilute solution. Always negative (ion–dipole attraction).
$\Delta H^{\ominus}_{\text{sol}} = \sum \Delta H^{\ominus}_{\text{hyd}} - \Delta H^{\ominus}_{\text{latt}}$ .
Hydration enthalpy is more exothermic for ions with higher charge and smaller radius (higher charge density).
$\Delta H^{\ominus}_{\text{sol}}$ is a small difference between two large numbers; solubility also depends on entropy.

Practice questions

Question

Define the term enthalpy change of hydration.
Write equations, with state symbols, for (a) the enthalpy change of solution of magnesium chloride; (b) the enthalpy change of hydration of the calcium ion.
Calculate $\Delta H^{\ominus}_{\text{sol}}$ of $\ce{LiF}$ : $\Delta H^{\ominus}_{\text{latt}} = -1049$ ; $\Delta H^{\ominus}_{\text{hyd}}(\ce{Li+}) = -519$ ; $\Delta H^{\ominus}_{\text{hyd}}(\ce{F-}) = -506\ \text{kJ mol}^{-1}$ .
Calculate $\Delta H^{\ominus}_{\text{sol}}$ of $\ce{KBr}$ : $\Delta H^{\ominus}_{\text{latt}} = -689$ ; $\Delta H^{\ominus}_{\text{hyd}}(\ce{K+}) = -322$ ; $\Delta H^{\ominus}_{\text{hyd}}(\ce{Br-}) = -335\ \text{kJ mol}^{-1}$ .
For sodium bromide, $\Delta H^{\ominus}_{\text{sol}} = -1\ \text{kJ mol}^{-1}$ and $\Delta H^{\ominus}_{\text{latt}} = -747\ \text{kJ mol}^{-1}$ . Given $\Delta H^{\ominus}_{\text{hyd}}(\ce{Na+}) = -406\ \text{kJ mol}^{-1}$ , calculate $\Delta H^{\ominus}_{\text{hyd}}(\ce{Br-})$ .
For strontium chloride, $\Delta H^{\ominus}_{\text{sol}} = -52\ \text{kJ mol}^{-1}$ . Using $\Delta H^{\ominus}_{\text{hyd}}(\ce{Sr^{2+}}) = -1480$ and $\Delta H^{\ominus}_{\text{hyd}}(\ce{Cl-}) = -364\ \text{kJ mol}^{-1}$ , calculate the lattice energy of $\ce{SrCl2}$ .
Place $\ce{K+}$ , $\ce{Na+}$ , $\ce{Mg^{2+}}$ and $\ce{Al^{3+}}$ in order of increasingly exothermic hydration enthalpy and explain your order.
Explain, with reference to the structure of water, why the hydration enthalpy of $\ce{F-}$ is more exothermic than that of $\ce{Cl-}$ .
Ammonium nitrate has $\Delta H^{\ominus}_{\text{sol}} = +26\ \text{kJ mol}^{-1}$ , yet it dissolves readily in water at room temperature. Explain why.
A student uses data-book values to calculate $\Delta H^{\ominus}_{\text{sol}}(\ce{NaCl}) = +17\ \text{kJ mol}^{-1}$ , while a calorimetry experiment gives $+4\ \text{kJ mol}^{-1}$ . Each data-book value has an uncertainty of about $\pm 1\%$ . Explain why the calculated value is unreliable, using the size of the terms involved.

Answers

The enthalpy change when one mole of a specified gaseous ion dissolves in a sufficient volume of water to form an infinitely dilute solution, under standard conditions.
(a) $\ce{MgCl2(s) + aq -> Mg^{2+}(aq) + 2Cl^-(aq)}$ (b) $\ce{Ca^{2+}(g) + aq -> Ca^{2+}(aq)}$
$\Delta H^{\ominus}_{\text{sol}} = -519 - 506 + 1049 = +24\ \text{kJ mol}^{-1}$ .
$\Delta H^{\ominus}_{\text{sol}} = -322 - 335 + 689 = +32\ \text{kJ mol}^{-1}$ .
$-1 = -406 + \Delta H^{\ominus}_{\text{hyd}}(\ce{Br-}) + 747$ , so $\Delta H^{\ominus}_{\text{hyd}}(\ce{Br-}) = -1 + 406 - 747 = -342\ \text{kJ mol}^{-1}$ .
$-52 = -1480 + 2(-364) - \Delta H^{\ominus}_{\text{latt}}$ , so $\Delta H^{\ominus}_{\text{latt}} = -1480 - 728 + 52 = -2156\ \text{kJ mol}^{-1}$ .
$\ce{K+} < \ce{Na+} < \ce{Mg^{2+}} < \ce{Al^{3+}}$ . Charge increases from 1+ to 3+ and radius decreases along this order, so charge density increases. Water molecules (via their $\delta-$ oxygen atoms) are attracted more strongly, so more energy is released.
Water is polar; the $\delta+$ hydrogen atoms are attracted to anions. $\ce{F-}$ and $\ce{Cl-}$ have the same charge, but $\ce{F-}$ is much smaller, so the hydrogen atoms of water can get closer to its centre and the ion–dipole attraction is stronger. More energy is released on hydration.
Although dissolving is endothermic, it greatly increases the entropy of the system: the ordered lattice breaks up into ions spread through the solution. At room temperature $T\Delta S$ is larger than $\Delta H$ , so $\Delta G = \Delta H - T\Delta S$ is negative and dissolving is feasible.
$\Delta H^{\ominus}_{\text{sol}}$ is found from $-770 + 787$ , a small difference between two large numbers. A 1% uncertainty in each term is about $\pm 8\ \text{kJ mol}^{-1}$ , so the combined uncertainty (up to about $\pm 16\ \text{kJ mol}^{-1}$ ) is as large as the answer itself. The calculated value can only be trusted to show that $\Delta H^{\ominus}_{\text{sol}}$ is small; it cannot reliably fix its value or even its sign.

Two new enthalpy changes

Why hydration is always exothermic

The solution energy cycle

What makes a hydration enthalpy large

Worked examples

Is a negative enthalpy of solution needed to dissolve?

Exam technique

Summary

Practice questions

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