Solubility

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

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 (d_NaCl = 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.

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.