Solubility

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 (d_NaCl = 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.

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?

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.