# Solubility

Solubility indicates the amount of solute that can be dissolved in an amount of solvent under certain conditions. This is generally given as the weight ratio percentage and has the form

$\dfrac{\mathrm{g~solute}}{100~\mathrm{g~solvent}}$

This is to be read as “parts by weight per 100 parts by weight of solvent”. However, solubilities can be given in any appropriate unit of concentration with careful conversions.

Example:

Wikipedia lists the solubility of sodium chloride (NaCl) in water at 25 °C as

• 360 g L–1

Careful examination further down in the article reveals that this solubility is actually

• 360 g NaCl per 1 kg of water

(The author of article simply assumed 1 L of water being approximately 1 kg of water.)

Therefore, 360 g NaCl per 1 kg of water is

• 36 g NaCl per 100 g of water

giving us a weight ratio percentage

$\mathrm{weight~ratio~percentage} = \dfrac{36.0~\mathrm{g~NaCl}}{100~\mathrm{g~H_2O}}\times 100\% = 36.0$

How does one convert solubility in weight ratio percentage into other concentration units?

## Converting solubility to other concentration units

Solubility can be converted from weight ratio percentage into other forms of concentration units.

$\mathrm{g~solute}/\mathrm{100~g~solvent} \longrightarrow \mathrm{other~concentration~units}$ However, 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 (wt%) and temperatures (°C)

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

Clearly, 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.

### Determining the volume of the saturated solution

1. Determine the mass of a saturated solution of aqueous NaCl from solubility.

We know (from above) the solubility of NaCl in water (at 25 °C) is 36 g of NaCl in 100 g of water. Therefore, 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. Determine the mass fraction of the solute.

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. Determine the % by mass concentration of the solute in the solution.

It is important to note that “% by mass” has a variety of names.

\begin{align*} \mathrm{mass}~\% &= \omega_{\mathrm{solu.}} \times 100\%\\[1.5ex] &= 0.26471 \times 100\% \\[1.5ex] &= 26.471\% \end{align*}

4. Determine the volume of the saturated solution.

Cross-referencing the solution density table, we see that a saturated aqueous solution of NaCl with a concentration (in wt%) of 26 (from step 3) at 25 °C is 1.1944 kg L–1 or 1.1944 g mL–1.

\begin{align*} d_{\mathrm{soln.}} &= \dfrac{m_{\mathrm{soln.}}}{V_{\mathrm{soln.}}} \longrightarrow \\[1.5ex] 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.

### Find mass concentration

We can determine the mass concentration.

\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*}

### Find molarity

We can now determine the molar solubility (M; mol L–1) of NaCl (at 25 °C). We first determine the number of moles of 36.0 g NaCl in the saturated solution.

\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{g~L^{-1}} \end{align*}

### Find molality

We can find the molality of the saturated solution by first determining the number of moles of NaCl in the saturated solution.

\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*}

Now determine the molality (m). 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*}

### Find mole fraction and mole percent

From the reported solubility of NaCl in water (at 25 °C)

• 36 g NaCl / 100 g of water

In moles

• 0.616 mol NaCl / 5.549 mol water

the mole fraction is

\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*}

The mole percent is therefore

\begin{align*} \mathrm{mol~\%} &= \chi_{\mathrm{solu.}}\times 100\% \\[1.5ex] &= 0.09992 \times 100\%\\[1.5ex] &= 9.992\% \end{align*}

### Find parts per million (by mass)

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*}

### Find parts per million (by volume)

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 \\[1.5ex] 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*}

### Summary

Below are the tabulated concentrations of a saturated aqueous solution of NaCl at 25 °C.

Concentration Unit Value
weight ratio % % 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
ppm (by mass) ppm 2.65 × 105
ppm (by volume) ppm 1.46× 105

## Practice Problems

### Problem 1

According to the following solubility curve plot for some solids1, how many more moles of sugar (C12H22O11) can be dissolved in water when the temperature is increased from 0 $$^{\circ}$$C to 40 $$^{\circ}$$C in 100 g of H2O? Solution

180 g sugar in 100 g H$$_2$$O at 0 $$^{\circ}$$C

240 g sugar in 100 g H$$_2$$O at 40 $$^{\circ}$$C

\begin{align*} \mathrm{mol~sugar~at~0 ^{\circ}C} = 180~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{342.3~\mathrm{g}} \right ) = 0.526~\mathrm{mol}\\ \mathrm{mol~sugar~at~40 ^{\circ}C} = 240~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{342.3~\mathrm{g}} \right ) = 0.701~\mathrm{mol}\\ 0.701~\mathrm{mol} - 0.526~\mathrm{mol} = 0.175~\mathrm{mol} \end{align*}

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