# 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 g/100g and has the form

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

This can 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

How does one convert solubility g/100g into other concentration units?

## Tutorial: Solubility Conversions

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.

### Volume of solution

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

### 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*}

### Molarity

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*}

### Molality

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*}

### Mole fraction

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*}

### Mole %

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*}

### Mass Volume %

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*}

### 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*}

### 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 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
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

## 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 °C to 40 °C in 100 g of H2O? 180 g sugar in 100 g H2O at 0 °C

240 g sugar in 100 g H2O at 40 °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*}

### Problem 2

An aqueous NaCl (58.44 g mol–1) solution has a concentration of 2.5% by mass. What is the g/100g, molar, molal, and concentration of the solution? The density of the solution is 1.0183 g mL–1.

Find the mass of a 1 L solution.

\begin{align*} m_{\mathrm{soln}} &= V_{\mathrm{soln}} \times d_{\mathrm{soln}} \\[1.5ex] &= 1~\mathrm{L} \left ( \dfrac{10^3~\mathrm{mL}}{1~\mathrm{L}} \right ) \left ( \dfrac{1.0183~\mathrm{g}}{\mathrm{mL}} \right ) \\[1.5ex] &= 1018.3~\mathrm{g} \end{align*}

Assume 100g of solution and get the mass fraction of the solute. The percent becomes g of solute.

\begin{align*} 2.5\% \longrightarrow \dfrac{2.5~\mathrm{g~solute}}{100~\mathrm{g~solution}} \end{align*}

Get mass of solute, NaCl

\begin{align*} m_{\mathrm{solute}} &= m_{\mathrm{soln}} \times \mathrm{\omega_{\mathrm{NaCl}}} \\[1.5ex] &= 1018.3~\mathrm{g~soln} \times \left ( \dfrac{2.5~\mathrm{g~solute}}{100~\mathrm{g~solution}} \right ) \\[1.5ex] &= 25.458~\mathrm{g} \end{align*}

Get moles of solute, NaCl

\begin{align*} n_{\mathrm{solute}} &= m_{\mathrm{solute}} \left( \dfrac{\mathrm{g}}{\mathrm{mol}} \right )\\[1.5ex] &= 25.458~\mathrm{g~NaCl} \left ( \dfrac{\mathrm{mol~NaCl}}{58.44~\mathrm{g~NaCl}} \right ) \\[1.5ex] &= 0.4356~\mathrm{mol} \end{align*}

Get molar concentration (M)

\begin{align*} M &= \dfrac{\mathrm{mol~solute}}{\mathrm{L~solution}} \\[1.5ex] &= \dfrac{0.4356~\mathrm{mol~NaCl}}{1~\mathrm{L}} \\[1.5ex] &= 0.436~\mathrm{mol~L^{-1}} \end{align*}

Get mass of solvent, H2O

\begin{align*} m_{\mathrm{solvent}} &= m_{\mathrm{solution}} - m_{\mathrm{solute}}\\[1.5ex] &= 1018.3~\mathrm{g} - 25.458~\mathrm{g~solute}\\[1.5ex] &= 992.842~\mathrm{g} \end{align*}

Get molal concentration (m)

\begin{align*} m &= \dfrac{\mathrm{mol~solute}}{\mathrm{kg~solvent}} \\[1.5ex] &= \dfrac{0.4356~\mathrm{mol~NaCl}}{992.842~\mathrm{g~solvent}} \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \\[1.5ex] &= 0.439~\mathrm{mol~kg^{-1}} \end{align*}

Get g/100g.

\begin{align*} \mathrm{g/100g} &= \dfrac{m_{solute}}{m_{solvent}} \times 100 \\[1.5ex] &= \dfrac{25.458~\mathrm{g~NaCl}}{992.42~\mathrm{g~H_2O}} \times 100~\mathrm{g~H_2O} \\[1.5ex] &= 2.56~\mathrm{g/100g} \end{align*}

Previous
Next