Solute meet solvent
“One of the most striking properties of water is its ability to dissolve many substances, forming aqueous solutions. Solutions are very important kinds of matter— important for industry and for life. The ocean is an aqueous solution that contains thousands of components: ions of the metals and nonmetals, complex inorganic ions, many different organic substances. The properties of solutions have been extensively studied, and it has been found that they can be correlated in large part by some simple laws.” — Linus Pauling (Pauling 1970).
Key concepts:
Solutions are created when a substance or substances are dissolved in a liquid. The dissolved substances are called solutes while the liquid doing the dissolving is called the solvent and is present in the largest amount.
Here, the ionic compound, potassium dichromate, is dissolved in water. Notice how the resulting solution changed color from that of the pure liquid and the dissolved substance is homogeneously distributed throughout.
Molecular Equation
\[\mathrm{K_2Cr_2O_7}(s) \longrightarrow \mathrm{K_2Cr_2O_7}(aq)\]
Ionic Equation
\[\mathrm{K_2Cr_2O_7}(s) \longrightarrow \mathrm{2K^+}(aq) + \mathrm{Cr_2O_7^{2-}}(aq)\]
Some substances readily dissolve in a liquid while others do not. One rule of thumb is “like dissolves like” and refers to the intermolecular forces that the solute and solvent exhibit. For example, water readily dissolves sucrose (it is miscible) as both can form hydrogen bonds.
Water and oil do not readily mix (it is immiscible) as water forms hydrogen bonds whereas hydrocarbons exhibit dispersion.
Intermolecular forces can help us predict if one substance will be soluble in another.
Solutions are usually liquid but can be a solid or gas.
Solution | Solute | Solvent |
---|---|---|
air | O2(g) |
N2(g) |
soft drinks | CO2(g) |
H2O(l) |
palladium hydride | H2(g) |
Pd(s) |
rubbing alcohol | H2O(l) |
C3H8O(l) |
saltwater | NaCl(s) |
H2O(l) |
brass | Zn(s) |
Cu(s) |
Many ionic compounds dissociate when dissolved in water, resulting in dissolved ions in solution. These are called electrolytes.
Those that do not form ions when dissolved in solution are called non-electrolytes. It is important to track the number of resulting solute particles that end up in solution. For example, when one particle of NaCl dissolves in water, the resulting number of dissolved solute particles is two and can be seen with the ionic equation.
\[\mathrm{NaCl}(s) \longrightarrow \mathrm{Na^+}(aq) + \mathrm{Cl^-}(aq)\]
The ratio of dissolved particles for NaCl is 2:1 (two resulting particles for every one particle dissolved).
Non-electrolytes, however, do not dissociate and the resulting number of dissolved particles equals the number of particles introduced into the solution. A sugar molecule such as sucrose (C12H22O11) is a non-electrolyte. If one molecule of sucrose is dissolved in water, the resulting number of solute particles in solution is one.
\[\mathrm{C_{12}H_{22}O_{11}}(s) \longrightarrow \mathrm{C_{12}H_{22}O_{11}}(aq)\]
The ratio of dissolved particles for non-electrolytes is 1:1 (one resulting particle for every one particle dissolved).
Three moles of NaCl are dissolved in water. How many solute particles (in mol) are there?
500.0 g of MgSO4 (m.m. = 120.366 g mol–1) is dissolved in water. How many solute particles (in mol) are there?
Problem 1
\[\mathrm{NaCl}(s) \longrightarrow \mathrm{Na^+}(aq) + \mathrm{Cl^-}(aq) \qquad i =2\] \[3~\mathrm{mol~NaCl} \times 2 = 6~\mathrm{mol}\]
Problem 2
\[\mathrm{MgSO_4}(s) \longrightarrow \mathrm{Mg^{2+}}(aq) + \mathrm{SO_4^{2-}}(aq) \qquad i = 2\] \[500.0~\mathrm{g~MgSO_4} \left(\dfrac{\mathrm{mol}}{120.366~\mathrm{g}} \right ) 2 = 8.31~\mathrm{mol}\]
The Van’t Hoff Factor (i) is simply the ratio for different substances.
Substance | Solute | iideal | ireal |
---|---|---|---|
non-electrolyte |
no dissociation |
1 |
1 |
NaCl |
Na+ + Cl– |
2 |
1.9 |
MgSO4 |
Mg2+ + SO42– |
2 |
1.3 |
MgCl2 |
Mg+ + 2Cl– |
3 |
2.7 |
K2SO4 |
2K+ + SO42– |
3 |
2.6 |
FeCl3 |
Fe + 3Cl– |
4 |
3.4 |
Ion Pairing
Ionic compounds dissociate when dissolved in solution. At very low concentrations, these ions are not very likely to “find” each other in a vast ocean of solvent particles. However, at higher concentration, the probability of the ions finding each other increases. Ions of opposite charge will be attracted to each other and form ion pairs, thereby decreasing the number of dissolved particles slightly.
iideal assumes no ion pairing whereas ireal is the actual ratio of dissolved solute particles at some concentration.
Joe: For the last time, I’m pretty sure what’s killing the crops is this Brawndo stuff.
Secretary of State: But Brawndo’s got what plants crave. It’s got electrolytes.
Attorney General: So wait a minute. What you’re saying is that you want us to put water on the crops.
Joe: Yes.
Attorney General: Water. Like out the toilet?
Joe: Well, I mean, it doesn’t have to be out of the toilet, but, yeah, that’s the idea.
Secretary of State: But Brawndo’s got what plants crave.
Attorney General: It’s got electrolytes.
As noted before, electrolytes (substances that result in ions when dissolved), form electrically conductive solutions. The more ions that are dissolved, the more conductive the solution is. Therefore, conductivity can be used to determine
Strong electrolytes include:
Weak electrolytes include:
Below is a table of substances. Predict which ones are a strong, weak, or non-electrolyte.
Number | Substance | Formula |
---|---|---|
1 |
Distilled water |
H2O |
2 |
Tap water |
H2O |
3 |
Sodium chloride |
NaCl |
4 |
Sugar |
|
5 |
Vinegar |
CH3COOH |
6 |
Hydrochloric acid |
HCl |
7 |
Sodium hydroxide |
NaOH |
8 |
Ethanol |
C2H5OH |
9 |
Barium sulfate |
BaSO4 |
The mixing of substances to form a solution can be endothermic or exothermic. If the final state (i.e. solution) has a lower enthalpy than the initial (non-mixed) state, the mixing process is exothermic.
Sulfuric acid + Water
However, if the final state has a higher enthalpy than the initial state, the mixing process is endothermic. This is precisely how cold-packs work. A salt such as ammonium nitrate (NH4NO3) is dissolved in water and heat is absorbed making it feel cold.
Ammonium chloride + Water
The solubility of a substance is the maximum amount of solute that can be dissolved in an amount of solvent at a certain temperature. Therefore, it describes the concentration of a saturated solution. Solubility changes with temperature.
Substance | 10 °C | 25 °C | 50 °C |
---|---|---|---|
LiCl |
73.8 |
84.5 |
94.1 |
NaCl |
32.7 |
36 |
36.7 |
KCl |
30.9 |
35.5 |
43 |
NaOH |
63.9 |
100 |
170.3 |
Temperature affects solubility. If solubility increases with increasing temperature, the dissolution process is endothermic. If solubility decreases with increasing temperature, the dissolution process is exothermic.
Concentration is a measure of solute to solvent ratios and can be expressed in many ways. The units chosen generally depend on the state of matter of the solute and/or solvent.
\[r = \dfrac{\mathrm{g~solute}}{100~\mathrm{g~solvent}}\]
\[\rho = \dfrac{\mathrm{g~solute}}{\mathrm{L~solution}}\] Mass concentration is the density of a component in a mixture (mass over volume), hence the use of the Greek letter \(\rho\) to represent mass concentration as it is used to represent density,
\[M = \dfrac{\mathrm{mol~solute}}{\mathrm{L~solution}}\] Molarity is derived from mass concentration by converting grams of solute into moles of solute via molar mass.
\[m = \dfrac{\mathrm{mol~solute}}{\mathrm{kg~solvent}}\]
\[\chi_{\mathrm{solute}} = \dfrac{n_{\mathrm{solute}}}{n_{\mathrm{solution}}}\] \[\chi_{\mathrm{solvent}} = \dfrac{n_{\mathrm{solvent}}}{n_{\mathrm{solution}}}\]
\[\mathrm{mol~\%~solute} = \chi_{\mathrm{solute}} \times 100\%\] \[\mathrm{mol~\%~solvent} = \chi_{\mathrm{solvent}} \times 100\%\]
\[\omega_{\mathrm{solute}} = \dfrac{m_{\mathrm{solute}}}{m_{\mathrm{solution}}}\] \[\omega_{\mathrm{solvent}} = \dfrac{m_{\mathrm{solvent}}}{m_{\mathrm{solution}}}\]
\[\mathrm{mass~\%~solute} = \omega_{\mathrm{solute}} \times 100\%\] \[\mathrm{mass~\%~solvent} = \omega_{\mathrm{solvent}} \times 100\%\]
\[m/v~\% = \dfrac{m_{\mathrm{solute~(g)}}}{V_{\mathrm{solution~(mL)}}} \times 100\%\]
\[\dfrac{m_\mathrm{{solute}}}{m_\mathrm{{solution}}} \times \mathrm{~multiplication~factor}\]
Unit and multiplication factor
\[\dfrac{V_\mathrm{{solute}}}{V_\mathrm{{solution}}} \times \mathrm{~multiplication~factor}\]
Unit and multiplication factor
This tutorial outlines how to take a solubility (in g/100g) and convert it to other concentration units.
We will be using the solubility of NaCl in water at 25 °C of 36g/100g as reported by Wikipedia.
We need to know the volume of the saturated solution which requires us to look up the density of the solution from other sources (remember, densities are not additive).
Below is a table of aqueous sodium chloride solution densities (in kg L–1) at various concentrations and temperatures (borrowed from handymath.com).
Table: Densities (in kg/L) of NaCl in water at various concentrations and temperatures.
Conc. (wt%) | 0 °C | 10 °C | 25 °C | 40 °C | 60 °C | 80 °C | 100 °C |
---|---|---|---|---|---|---|---|
1 | 1.0075 | 1.0071 | 1.0041 | 0.9991 | 0.9900 | 0.9785 | 0.9651 |
2 | 1.0151 | 1.0144 | 1.0111 | 1.0059 | 0.9967 | 0.9852 | 0.9719 |
4 | 1.0304 | 1.0292 | 1.0253 | 1.0198 | 1.0103 | 0.9988 | 0.9855 |
8 | 1.0612 | 1.0591 | 1.0541 | 1.0480 | 1.0381 | 1.0264 | 1.0134 |
12 | 1.0924 | 1.0895 | 1.0837 | 1.0770 | 1.0667 | 1.0549 | 1.0420 |
16 | 1.1242 | 1.1206 | 1.1140 | 1.1069 | 1.0962 | 1.0842 | 1.0713 |
20 | 1.1566 | 1.1525 | 1.1453 | 1.1377 | 1.1268 | 1.1146 | 1.1017 |
24 | 1.1900 | 1.1856 | 1.1778 | 1.1697 | 1.1584 | 1.1463 | 1.1331 |
26 | 1.2071 | 1.2025 | 1.1944 | 1.1861 | 1.1747 | 1.1626 | 1.1492 |
To use this data, we need to know the concentration (in wt%) of a 36 g NaCl / 100 g water solution to choose the correct density at 25 °C. We will calculate it below.
1. Find the mass of a saturated solution
Given that the solubility of NaCl in water (at 25 °C) is 36 g of NaCl in 100 g of water, the mass of the saturated solution is
\[\begin{align*} m_{\mathrm{soln.}} &= m_{\mathrm{solute}} + m_{\mathrm{solvent}} \\[1.5ex] &= 36.0~\mathrm{g~NaCl} + 100~\mathrm{g~H_2O}\\[1.5ex] &= 136.0~\mathrm{g} \end{align*}\]
2. Find mass fraction
We now calculate the mass fraction of NaCl.
\[\begin{align*} \omega_{\mathrm{solu.}} &= \dfrac{m_{\mathrm{solu.}}}{m_{\mathrm{soln.}}}\\[1.5ex] &= \dfrac{36.0~\mathrm{g}}{136.0~\mathrm{g}}\\[1.5ex] &= 0.26471 \end{align*}\]
3. Find mass %
It is important to note that “% by mass” has a variety of names. “wt%” is “% by mass.”
\[\begin{align*} \mathrm{mass}~\% &= \omega_{\mathrm{solu.}} \times 100\%\\[1.5ex] &= 0.26471 \times 100\% \\[1.5ex] &= 26.471\% \end{align*}\]
4. Find the density of solution
Cross-referencing the solution density table, we see that a saturated aqueous solution of NaCl with a wt% of 26 (from step 3) at 25 °C has a density of 1.1944 kg L–1.
We convert this density to g mL–1 as follows
\[\begin{align*} \dfrac{1.1944~\mathrm{kg}}{\mathrm{L}} \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) = 1.1944~\mathrm{g~mL^{-1}} \end{align*}\]
5. Find the volume of solution
Using density and mass, we can find volume.
\[\begin{align*} d_{\mathrm{soln.}} = \dfrac{m_{\mathrm{soln.}}}{V_{\mathrm{soln.}}} \longrightarrow V_{\mathrm{soln.}} &= \dfrac{m_{\mathrm{soln.}}}{d_{\mathrm{soln.}}} \\[1.5ex] &= \dfrac{136.0~\mathrm{g}}{\mathrm{1.1944~g~mL^{-1}}}\\[1.5ex] &= 113.865~\mathrm{mL} \\ &= 0.113865~\mathrm{L} \end{align*}\]
We can now determine solubility in other concentration units.
6. Organize data
Property | Value |
---|---|
msolu |
36 g |
msolv |
100 g |
msoln |
136 g |
mass fraction |
0.26471 |
mass % |
26.471 |
dsoln |
1.1944 g mL–1 |
Vsoln |
0.113865 L |
\[\begin{align*} \rho &= \dfrac{\mathrm{g~solute}}{\mathrm{L~solution}}\\[1.5ex] &= \dfrac{36.0~\mathrm{g}}{\mathrm{0.113865~\mathrm{L}}}\\[1.5ex] &= 316.16~\mathrm{g~L^{-1}} \end{align*}\]
We can now determine the molar solubility (M ; mol L–1) of NaCl (at 25 °C). First convert 36.0 g NaCl to moles.
\[\begin{align*} n_{\mathrm{solu.}} &= 36.0~\mathrm{g~NaCl} \left (\dfrac{\mathrm{mol~NaCl}}{58.44~\mathrm{g~NaCl}} \right )\\[1.5ex] &= 0.616~\mathrm{mol~NaCl} \end{align*}\]
We can now calculate the molar solubility (M).
\[\begin{align*} M &= \dfrac{\mathrm{mol~solute}}{\mathrm{L~solution}}\\[1.5ex] &= \dfrac{0.616~\mathrm{mol~NaCl}}{0.113865~\mathrm{L}}\\[1.5ex] &= 5.41~\mathrm{mol~L^{-1}} \end{align*}\]
We use dimensional analysis to convert 100 g of water into 1 kg of water.
\[\begin{align*} m &= \dfrac{\mathrm{mol~solute}}{\mathrm{kg~solvent}}\\[1.5ex] &= \dfrac{0.616~\mathrm{mol~NaCl}}{100~\mathrm{g~water}} \left ( \dfrac{1000~\mathrm{g}}{1~\mathrm{kg}} \right ) \\[1.5ex] &= 6.16~\mathrm{mol~kg^{-1}} \end{align*}\]
To find mole fraction, we must obtain the solute and solvent amount in moles. We already know how much NaCl (in mol) we have. Let’s determine moles of water.
\[100~\mathrm{g~H_2O} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) = 5.549~\mathrm{mol~H_2O}\]
Now find the mole fraction.
\[\begin{align*} \chi_{\mathrm{solu}} &= \dfrac{n_{\mathrm{solu.}}}{n_{\mathrm{soln.}}}\\[1.5ex] &= \dfrac{0.616~\mathrm{mol~NaCl}}{0.616~\mathrm{mol~NaCl} + 5.549~\mathrm{mol~water}}\\[1.5ex] &= 0.09992 \end{align*}\]
Multiply the mole fraction by 100% to obtain the mole %.
\[\begin{align*} \mathrm{mol~\%} &= \chi_{\mathrm{solu.}}\times 100\% \\[1.5ex] &= 0.09992 \times 100\%\\[1.5ex] &= 9.992\% \end{align*}\]
The m/v% of the solution (in g mL–1) can be found as follows.
\[\begin{align*} m/v~\% &= \dfrac{m_{\mathrm{solute~(g)}}}{V_{\mathrm{solution~(mL)}}} \times 100\% \\[1.5ex] &= \dfrac{36~\mathrm{g~NaCl}}{1138.65~\mathrm{mL}} \\[1.5ex] &= 3.16\% \end{align*}\]
The parts per million (ppm) concentration unit is reserved for very dilute solutions. However, we can still calculate the concentration (in ppm by mass) of a saturated aqueous NaCl solution.
\[\begin{align*} \mathrm{ppm} &= \dfrac{m_{\mathrm{solu.}}}{m_{\mathrm{soln.}}} \times 10^6 \\[1.5ex] &= \dfrac{36.0~\mathrm{g~NaCl}}{136~\mathrm{g~soln.}} \times 10^6 \\[1.5ex] &= 2.65\times 10^5~\mathrm{ppm~(by~mass)} \end{align*}\]
The parts per million (ppm) concentration unit is reserved for very dilute solutions. However, we can still calculate the concentration (in ppm by volume) of a saturated aqueous NaCl solution.
First, determine the volume of 36.0 g NaCl by using the density of NaCl (dNaCl = 2.165 g cm–3).
\[\begin{align*} d_{\mathrm{NaCl}} = \dfrac{m_{\mathrm{NaCl}}}{V_{\mathrm{NaCl}}} \longrightarrow V_{\mathrm{NaCl}} &= \dfrac{m_{\mathrm{NaCl}}}{d_{\mathrm{NaCl}}} \\[1.5ex] &= \dfrac{36.0~\mathrm{g~NaCl}}{2.165~\mathrm{g~cm^{-3}}}\\[1.5ex] &= 16.628~\mathrm{cm^3} \end{align*}\]
Recall that cm3 is a mL.
\[\begin{align*} \mathrm{ppm} &= \dfrac{V_{\mathrm{solu.}}}{V_{\mathrm{soln.}}} \times 10^6 \\[1.5ex] &= \dfrac{16.628~\mathrm{mL~NaCl}}{113.865~\mathrm{mL~soln.}} \times 10^6 \\[1.5ex] &= 1.46\times 10^5~\mathrm{ppm~(by~volume)} \end{align*}\]
Below are the tabulated concentrations of a saturated aqueous solution of NaCl at 25 °C.
Concentration | Unit | Value |
---|---|---|
g/100g | g/100g | 36.0 |
mass concentration | g L–1 | 316.2 |
mass fraction | unitless | 0.2647 |
mass % | % | 26.47 |
mole fraction | unitless | 0.0999 |
mole % | % | 9.99 |
molarity | mol L–1 | 5.41 |
molality | mol kg–1 | 6.16 |
m/v% | unitless | 3.16 |
ppm (by mass) | ppm | 2.65 × 105 |
ppm (by volume) | ppm | 1.46× 105 |
Let us now examine the aqueous solubility of gases. The presented mass % data is from the CRC Handbook of Chemistry and Physics 97th ed. (p. 5-314) (William M. Haynes 2016).
The solubility of many gases are quite small and decrease with increasing temperature.
Notice the large solubility of carbon dioxide in water relative to other gases. Earth’s oceans absorb about 30% of the CO2 that is released into the atmosphere. Carbon dioxide can react with water to produce carbonic acid via the following process
\[\mathrm{CO_2}(aq) + \mathrm{H_2O}(l) \longrightarrow \mathrm{H_2CO_3}(aq)\] This leads to ocean acidification. Ocean water is slightly basic but has become more acidic (ΔpH = –0.11) since the industrial revolution.
Though increasing temperature lowers the solubility of a gas, increasing pressure will raise the solubility of a gas. Henry’s Law states that the concentration of a gas is proportional to the partial pressure of the gas above the liquid.
The ratio of gas concentration (Cg) vs pressure (Pg) is constant (called Henry’s constant; kH) and is unique to the identity of the gas. Across modest concentrations, the relationship is linear and can be described by the following equation
\[C_{\mathrm{g}} = k_{\mathrm{H}} P_{\mathrm{p}}\]
For example, Henry’s constant for carbon dioxide is reported to be 3 × 10–4 mol m–3 Pa–1 at room temperature (Sander 2015). Let us convert the concentration (mol m–3) to molarity using the following known relationships:
\[\begin{align*} k_{\mathrm{H}} &= \dfrac{3\times 10^{-4}~\mathrm{mol}}{\mathrm{m^3~Pa}} \left ( \dfrac{1~\mathrm{m^3}}{10^3~\mathrm{L}} \right ) \left ( \dfrac{101,325~\mathrm{Pa}}{1~\mathrm{atm}} \right ) \\[1.5ex] &= 0.03204~\mathrm{mol~L^{-1}~atm^{-1}} \end{align*}\]
Therefore, for every 1 atm of pressure, the concentration of carbon dioxide in water increases by 0.03204 mol L–1. We can plot the predicted solubility of CO2 vs a small range of pressures.
Let us now examine the aqueous solubility of inorganic compounds (mass % data from (William M. Haynes 2016)). Compare the data to the solubility rules
Ionic compounds may or may not be soluble in water. Solubility rules tell us which compounds are soluble in water and to what extent they are soluble.
Soluble Ions |
Exception |
---|---|
Alkali metals (Group I) Li+, Na+, K+, etc. |
none |
Ammonium Ions NH4+ |
none |
Nitrates, acetates, chlorates, and perchlorate NO3–, C2H3O2–, ClO3–, ClO4– |
none |
Binary compounds of halogens (X) with metals (M) MCl, MBr, MI, etc. |
F–, Ag+, Pb2+, Hg2+ |
Sulfates SO42– |
Ba2+, Sr2+, Ca2+, Pb2+, Ag+, and Hg2+ |
Slightly Soluble Ions | Exception |
Sulfates of lead, silver, and mercury SO42– with Pb2+, Ag+, and Hg2+ |
none |
Hydroxides of alkaline earth metals (Group II) OH– with Ca2+, Sr2+, etc. |
Ba2+ |
Insoluble Ions | Exception |
Sulfides S2– |
Ca2+, Ba2+, Sr2+, Mg2+, Na+, K+, and NH4+ |
Hydroxides OH– |
Alkali metals (Group I), transition metals, Al3+, and NH4+ |
Carbonates, oxalates, chromates, and phosphates CO32–, C2O42–, CrO42–, and PO43– |
Alkali metals (Group I) and NH4+ |
Alkali metals (Group I) are soluble without exception. Notice that hydroxides (OH–) are soluble when paired with alkali metals.
Ammonium (NH4+) ions are soluble without exception.
Binary compounds of halogens (X) with metals (M) are soluble except F–, Ag+, Pb2+ Ag+, and Hg2+. Note that AgCl2 is water soluble.
Sulfates (SO42–) are soluble except when paired with Ba2+, Sr2+, Ca2+, Pb2+, Ag+, and Hg2+.
Sodium Sulfate
The solubility of sodium sulfate increases with increasing temperature until 32.384 °C (with a mass % of around 30) after which the solubility begins to decrease. This is due to the salt changing phase.
At low temperature (< 32.384 °C), sodium sulfate is a salt hydrate meaning it is an ionic compound with a number of water molecules that are enclosed within its crystal lattice. The chemical formula for this is Na2SO4 · 10H2O and is called mirabilite. Interestingly, the cations are [Na(OH2)6]+ in this structure.
At higher temperatures, the waters are released (“melt away”) from the crystal lattice to give anhydrous thenardite (Na2SO4). This process is given below.
\[
\begin{align*}
\mathrm{Na_2SO_4\cdot10H_2O}(aq) &\longrightarrow \mathrm{Na_2SO_4}(aq) + 10\mathrm{H_2O}(l) \\[1.25ex]
\mathrm{low~}T ~~~~~~~~~~~~~~~ &\phantom{\longrightarrow} ~~~~~~~~~~~~~~~~~~~ \mathrm{high~}T
\end{align*}
\]
Sulfates of lead, silver, and mercury are slightly soluble.
Hydroxides of alkaline earth metals (Group II) are slightly soluble with the exception of Ba2+ which is highly soluble.
Hydrocarbons do not mix well with water. The longer the hydrocarbon, the more dispersion and the lower the solubility.
Structures
Alcohols contain an –OH functional group and can increase the solubility of a compound.
Structures
Aromatic compounds contain one or more rings with delocalized pi electrons and are not very soluble in water.
Structures
Modifying aromatic compounds by adding other atoms or functional groups to the aromatic ring can increase the solubility of the molecule depending on the functional group.
Properties of pure liquids change when a solution is formed. The greater the concentration of the solute(s) in the solution, the larger the change in these properties. These properties are known as colligative properties meaning the change in property of the solution does not depend upon the identity of the solute, it only depends upon the number of solute particles.
The vapor pressure of a solvent is lower than that of the pure liquid. The higher the concentration of the solution, the more the vapor pressure of the solvent is lowered. This relationship can be given as
\[P_{\mathrm{A}} = \chi_{\mathrm{A}} P^{\circ}_{\mathrm{A}}\]
where χA is the mole fraction of the solute, P°A is the vapor pressure of the pure liquid, and PA is the vapor pressure of the solvent.
Raoult’s Law
Raoult’s Law relates the vapor pressure, P, of a solution, the solute concentration (mole fraction: χ) and the vapor pressure of the pure solvent P°. The lowered vapor pressure of the solution is directly proportional to the concentration of the solution. The more solute, the lower the vapor pressure of the solution. Raoult’s Law works well for solutions that behave ideally (i.e. dilute solutions). For solutions containing non-volatile solutes, the vapor pressure of the solution (Psolution) is directly proportional to the mole fraction of the solvent times the vapor pressure of the pure solvent (P°solvent). The more concentrated the solution, the smaller the mole fraction of the solvent resulting in a lower vapor pressure.
\[P_{\mathrm{solution}} = \chi_i P_{\mathrm{solvent}}^{\circ}\]
For solutions containing volatile solutes, the partial pressure of each, individual component in the solution will be lower than the vapor pressure of each pure solute. The vapor pressure of each component in the solution (i) is proportional to its concentration in the solution such that
\[P_{i} = \chi_{i} P^{\circ}_{i}\] Each component (i) has a lower vapor pressure than its pure substance form. The lower the concentration of the component, the lower its vapor pressure in the solution. Note that non-volatile solutes have a partial pressure of zero.
Adding the vapor pressures together would give the total vapor pressure of the solution such that
\[P_{\text {soln }}=\sum_{i=1}^{n}\left(P_{\text {solute}}\right)_{i}+P_{\text {solvent}}\]
For example, an aqueous solution containing dissolved benzene is a two-component solution where the solvent (water) is volatile and the solute (benzene) is a volatile non-electrolyte. Therefore, i = water and benzene and the vapor pressure of both would be determined (using the equation above).
122 g of sugar, sucrose (C12H22O11; m.w. = 342.3 g mol–1) is added to 350 g of water at 25 °C. What is the vapor pressure (in atm) of the solution? P°water = 0.0313 atm
Determine moles of solute.
\[n_{\mathrm{solute}} = 122~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{342.30~\mathrm{g}} \right ) = 0.356\]
Determine moles of solvent.
\[n_{\mathrm{solvent}} = 350~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) = 19.42\] Determine the mole fraction of solvent.
\[\chi_{\mathrm{water}} = \dfrac{n_{\mathrm{solvent}}}{n_{\mathrm{solution}}} = \dfrac{19.42}{0.356+19.42}= 0.982\]
Determine vapor pressure of solution.
\[P_{\mathrm{solution}} = \chi_i P_{\mathrm{solvent}}^{\circ} = 0.982 \times 0.0313~\mathrm{atm} = 0.0307~\mathrm{atm}\]
The freezing point of a solution is lower than that of the pure liquid. The higher the concentration of the solution, the more the freezing point is lowered. This relationship can be given as
\[\Delta T_{\mathrm{f}} = i m K_{\mathrm{f}}\]
where i is the Van’t Hoff factor of the solute, m is the concentration of the solution, Kf is the molal freezing point depression constant for the solvent (or cryoscopic constant), and ΔTf is the change in the freezing point from the pure solvent.
The molal freezing (Kf) and boiling (Kb) point constants for a solvent can be determined experimentally and are reported as positive values.
We will analyze how Kf is determined for benzene, a common solvent. The normal freezing point of benzene is 5.5 °C.
Below is table (click to expand) a of solutes dissolved in benzene at 1g/100g solvent concentration as well as the resulting change in freezing point (Raoult 1882). All listed solutes are neutral and do not dissociate in solution (i.e. they are non-electrolytes; i = 1).
Solute | Formula | m.w. (g/mol) | ΔT (°C) | m | Kf |
---|---|---|---|---|---|
methyl iodide | CH3I | 142 | –0.335 | 0.070 | 5.04 |
chloroform | CHCl3 | 119.5 | –0.428 | 0.084 | 5.11 |
carbon tetrachloride | CCl4 | 154 | –0.333 | 0.065 | 5.12 |
carbon disulfide | CS2 | 76 | –0.654 | 0.132 | 4.97 |
ethyl iodide | C2H5I | 156 | –0.331 | 0.064 | 5.16 |
ethyl bromide | C2H5Br | 109 | –0.461 | 0.092 | 5.02 |
hexane | C6H14 | 86 | –0.597 | 0.116 | 5.13 |
ethylene chloride | C2H5Cl | 99 | –0.491 | 0.101 | 4.86 |
turpentine (a-pinene) | C10H16 | 136 | –0.366 | 0.074 | 4.98 |
nitrobenzene | C6H5NO2 | 123 | –0.390 | 0.081 | 4.80 |
naphthalene | C10H8 | 128 | –0.391 | 0.078 | 5.00 |
anthracene | C14H10 | 178 | –0.287 | 0.056 | 5.12 |
methyl nitrate | CH3NO3 | 77 | –0.640 | 0.130 | 4.93 |
dimethyl oxalate | C4H6O4 | 118 | –0.417 | 0.085 | 4.92 |
methyl salicylate | C8H8O3 | 152 | –0.339 | 0.066 | 5.15 |
diethyl ether | (C2H5)2O | 74 | –0.671 | 0.135 | 4.97 |
diethyl sulfide | C4H10S | 90 | –0.576 | 0.111 | 5.18 |
ethyl cyanide | C3H5N | 55 | –0.938 | 0.182 | 5.16 |
ethyl formate | C3H6O2 | 74 | –0.666 | 0.135 | 4.93 |
ethyl valerate | C7H14O2 | 130 | –0.384 | 0.077 | 5.00 |
allyl thiocyanate | C4H5NS | 99 | –0.519 | 0.101 | 5.14 |
nitroglycerine | C3H5N3O9 | 227 | –0.220 | 0.044 | 4.99 |
tributyrin | C15H26O6 | 302 | –0.161 | 0.033 | 4.87 |
triolein | C57H104O6 | 884 | –0.056 | 0.011 | 4.98 |
acetaldehyde | C2H4O | 44 | –1.107 | 0.227 | 4.87 |
chloral | C2HCl3O | 147.5 | –0.342 | 0.068 | 5.03 |
benzaldehyde | C7H6O | 106 | –0.473 | 0.094 | 5.01 |
camphor | C10H16O | 152 | –0.338 | 0.066 | 5.14 |
acetone | C3H6O | 58 | –0.850 | 0.172 | 4.93 |
5-Nonanone | C9H18O | 142 | –0.359 | 0.070 | 5.10 |
Mean | 5.02 | ||||
Min | 4.80 | ||||
Max | 5.18 | ||||
Std. Dev | 0.11 |
Freezing points of each solution were measured. It was found that the lowered freezing point of each solution was directly proportional to the concentration of the solution (here given as molality) regardless of the identity of the solute. Here we plot the freezing point of benzene vs concentration.
A linear regression of the data gives a very good fit (R2 = 0.9973). Therefore, the freezing point of benzene solutions is strongly correlated to the concentration of the solution (i.e. the number of dissolved solute particles) and is not affected by the type of solute being dissolved.
The Kf values are simply the ratio of the change in freezing point vs. the number of particles in solution. These values for benzene are remarkably similar (the average value of Raoult’s findings is 5.02 °C m–1). This is very close the currently accepted value of 5.12 °C m–1.
Given that
\[\Delta T = imK_{\mathrm{f}}\] then
\[K_{\mathrm{f}} = \dfrac{\Delta T}{im}\]
For all species listed in the table above, i = 1. The change in a freezing point (ΔT ) can be measured. Therefore, to obtain Kf, simply punch in the concentration (m) of the solution and its change in freezing point.
Example: Determine the Kf for methyl iodide using the experimental data given above
\[\begin{align*} K_{\mathrm{f}} &= \dfrac{\Delta T}{im}\\[1.5ex] &= \dfrac{0.335~\mathrm{^{\circ}C}}{1\times 0.070~m}\\[1.5ex] &= 4.76~\mathrm{^{\circ}C~}m^{-1} \end{align*}\]
Note: In the table above, Kf for methyl iodide is listed as 5.04 as this is what is reported in Raoult’s original paper. However, we surmise that this is likely a typographical error as we see the value should be 4.76 here.
122 g of sugar, sucrose (C12H22O11; m.w. = 342.3 g mol–1) is added to 350 g of water at 25 °C. What is the freezing point (in °C) of the solution?
Kf,water = 1.86 °C m–1
Determine moles of solute.
\[n_{\mathrm{solute}} = 122~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{342.30~\mathrm{g}} \right ) = 0.356\]
Determine molality of solution.
\[m = \dfrac{n_{\mathrm{solute}}}{\mathrm{kg~solvent}} = \dfrac{0.356~\mathrm{mol}}{0.350~\mathrm{kg}} = 1.017\]
Determine change in freezing point.
\[\begin{align*} \Delta T_{\mathrm{f}} &= iK_{\mathrm{f}}m = \left ( 1 \right ) \left ( 1.86~^{\circ}\mathrm{C}~m^{-1} \right ) \left ( 1.017~m \right ) = 1.89~^{\circ}\mathrm{C} \\[1.5ex] \end{align*}\]
Determine final freezing point of the solution.
\[\begin{align*} T_{\mathrm{f}} &= 0~^{\circ}\mathrm{C} - 1.89~^{\circ}\mathrm{C} = -1.89~^{\circ}\mathrm{C} \\ \end{align*}\]
The boiling point of a solution is higher than that of the pure liquid. The higher the concentration of the solution, the more the boiling point is raised This relationship can be given as
\[\Delta T_{\mathrm{b}} = i m K_{\mathrm{b}}\]
where i is the van’t Hoff factor of the solute, m is the concentration of the solution, Kb is the molal boiling point elevation constant (or ebullioscopic constant) for the solvent, and ΔTb is the change in the boiling point from the pure solvent.
122 g of sugar, sucrose (C12H22O11; m.w. = 342.3 g mol–1) is added to 350 g of water at 25 °C. What is the boiling point (in °C) of the solution?
Kb,water = 0.513 °C m–1
Determine moles of solute.
\[n_{\mathrm{solute}} = 122~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{342.30~\mathrm{g}} \right ) = 0.356\]
Determine molality of solution.
\[m = \dfrac{n_{\mathrm{solute}}}{\mathrm{kg~solvent}} = \dfrac{0.356~\mathrm{mol}}{0.350~\mathrm{kg}} = 1.017\]
Determine change in boiling point.
\[\begin{align*} \Delta T_{\mathrm{b}} &= iK_{\mathrm{b}}m = \left ( 1 \right ) \left ( 0.513~^{\circ}\mathrm{C}~m^{-1} \right ) \left ( 1.017~m \right ) = 0.522~^{\circ}\mathrm{C} \end{align*}\]
Determine final boiling point of the solution.
\[\begin{align*} T_{\mathrm{b}} &= 100~^{\circ}\mathrm{C} + 0.522~^{\circ}\mathrm{C} = 100.522~^{\circ}\mathrm{C} \end{align*}\]
A solution and pure solvent are initially separated by an osmotic membrane. Solvent molecules move to higher concentration (osmosis). The osmotic pressure is defined as the external pressure required to give zero net movement of solvent across the membrane.
This can be seen with carrots submersed in salt water (left) and pure water (right) overnight.
Osmotic pressure is important in the medical field. Consider blood, a vital solution for our bodies. If the solution is too concentrated, blood cells will become hypertonic and shrivel. If the solution is too dilute, blood cells can become hypotonic.
The osmotic pressure of the solution is higher than that of the pure liquid. The higher the concentration of the solution, the higher the osmotic pressure it exhibits. The relationship can be given as
\[\Pi = iMRT \]
where i is the Van’t Hoff factor of the solute, M is the concentration of the solution, R is the gas constant, T is the temperature of the solution, and Π is the osmotic pressure of the solution.
122 g of sugar, sucrose (C12H22O11; m.w. = 342.3 g mol–1) is added to enough of water to have a 1 L solution (at 25 °C). What is the osmotic pressure (in °C) of the solution?
Determine moles of solute.
\[n_{\mathrm{solute}} = 122~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{342.30~\mathrm{g}} \right ) = 0.356\]
Determine molarity of solution.
\[M = \dfrac{n_{\mathrm{solu}}}{\mathrm{L~soln}} = \dfrac{0.356~\mathrm{mol}}{1~\mathrm{L}} = 0.356\]
Determine the osmotic pressure.
\[\begin{align*} \Pi &= iMRT \\[1.5ex] &= (1)(0.356~\mathrm{mol~L^{-1}})(0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}})(298.15~\mathrm{K}) \\[1.5ex] &= 8.71~\mathrm{atm} \end{align*}\]
For attribution, please cite this work as
Dornshuld (2022, April 3). Solutions. Retrieved from https://dornshuld.chemistry.msstate.edu/notes/ch11/solutions/index.html
BibTeX citation
@misc{dornshuld2022solutions, author = {Dornshuld, Eric Van}, title = {Solutions}, url = {https://dornshuld.chemistry.msstate.edu/notes/ch11/solutions/index.html}, year = {2022} }