## Solubility Table (Table of Solubilities)

Water is a commonly used solvent, so it is very useful to construct a table of solubilities based on the mass of a solute that will dissolve in a given volume of water.

- Water is the solvent.
- The solute is the substance being dissolved in the water.

Solubility tables (tables of solubility) usually tabulate the maximum *mass in grams* of the solute that can be dissolved in *100 mL* of water.

This data in the table is relevant only to the temperature given for the table.
Commonly, solubility data is given for 25°C.

At 25°C and 101.3 kPa (1 atm) the density of water is 1.00 g mL^{-1}

That is, 1 g of water has a volume of 1 mL at 25°C

Therefore, 100 g of water will have a volume of 100 × 1 mL = 100 mL at 25°C

So at 25°C and 101.3 kPa, the solubility of a solute in water given as mass in grams per 100 g water is the same as the solubility of the solute given as mass in grams per 100 mL of water.

That is, at 25°C: solubility in g/100 g water = solubility in g/100 mL water

Example: the solubility of sodium chloride (NaCl) in water at 25°C is 36 g/100 mL , or, 36 g/100 g

A solute is usually considered to be soluble in water if more than 1.0 g can be dissolved in 100 mL of water.

A solute is usually considered to be insoluble in water if less than 0.1 g can be dissolved in 100 mL of water.

It is important to note that "insoluble" does NOT necessarily mean that the solute cannot be dissolved, it only means that an extremely small amount of the solute can be dissolved in the solvent.

If the solubility of a substance is 0 g/100 mL, then, and only then, can we say that none of the solute dissolves in the solvent.

A solute is usually considered to be slightly soluble, or sparingly soluble, in water if between 0.1 and 1.0 g can be dissolved in 100 mL of water.

The solubility table below gives the maximum mass of solute in grams that can be dissolved in 100 mL of water at 25°C.

Solubilities of some common substances in water at 25^{o}C |

Solute Name |
Solute Formula |
Solubility g/100 mL water |
Description of Solubility |

silver nitrate |
AgNO_{3}(s) |
245 |
soluble |

sodium hydroxide |
NaOH(s) |
80 |

sucrose |
C_{12}H_{22}O_{11}(s) |
70 |

lead(II) nitrate |
Pb(NO_{3})_{2}(s) |
60 |

ammonia |
NH_{3}(g) |
48 |

glucose |
C_{6}H_{12}O_{6}(l) |
45 |

ammonium chloride |
NH_{4}Cl(s) |
39 |

sodium chloride |
NaCl(s) |
36 |

potassium chloride |
KCl(s) |
35 |

magnesium sulfate |
MgSO_{4}(s) |
18 |

copper(II) sulfate |
CuSO_{4}(s) |
14 |

bromine |
Br_{2}(g) |
3.5 |

chlorine |
Cl_{2}(g) |
0.6 |
slightly soluble |

carbon dioxide |
CO_{2}(g) |
0.15 |

calcium hydroxide |
Ca(OH)_{2}(s) |
0.12 |

iodine |
I_{2}(s) |
0.03 |
insoluble |

barium sulfate |
BaSO_{4}(s) |
0.001 |

silver chloride |
AgCl(s) |
0.0002 |

sulfur |
S(s) |
0 |

Sodium chloride, NaCl, is a soluble salt. From the solubility table above we see that the solubility of sodium chloride is 36 g/100 mL water at 25°C.

- 1 g of NaCl
_{(s)} will dissolve in 100 mL of water at 25°C.

This solution will be an *unsaturated* aqueous solution of sodium chloride.
- 25 g of NaCl
_{(s)} will dissolve in 100 mL of water at 25°C.

This solution will be an *unsaturated* aqueous solution of sodium chloride.
- 36g of NaCl
_{(s)} will dissolve in 100 mL of water at 25°C.

This solution will be a *saturated* aqueous solution of sodium chloride, but there will be no visible solid NaCl in the vessel.
- If 40 g of NaCl
_{(s)} is added to 100 mL of water at 25°C, only 36g of NaCl_{(s)} will dissolve, 4 g of the NaCl_{(s)} will not dissolve.

This solution will be a *saturated* aqueous solution of sodium chloride and there will be visible solid NaCl in the vessel.

We can say that this excess NaCl_{(s)} precipitates out of the solution.

## Worked Example 1 (using the StoPGoPS approach to problem solving)

*Question 1:* A student has been given 250 mL of water at 25°C and needs to add enough calcium hydroxide to make a saturated solution.

The solubility of calcium hydroxide at 25°C is 0.12 g/100 mL water.

What is the minimum mass, in grams, of calcium hydroxide that the student must add to the water?

*Response:*

*What is the question asking you to do?*

Calculate the mass of calcium hydroxide in grams.

m(calcium hydroxide) = ? g

*What information have you been given in the question?*

volume of water = 250 mL (25°C)

solubility of calcium hydroxide = 0.12 g/100 mL water at 25°C

*What is the relationship between what you have been given and what you need to find out?*

m(calcium hydroxide,g)/250 mL : 0.12 g/100 mL

*Perform the calculation*

- Write the equation:
*m(calcium hydroxide)* 250 mL |
= |
*0.12 g* 100 mL |

- Rearrange the equation by multiplying both sides of the equation by 250 mL:
*m(calcium hydroxide) × *~~250 mL~~
~~250 mL~~ |
= |
*0.12 g × 250 *~~mL~~ 100 ~~mL~~ |

m(calcium hydroxide) |
= |
*0.12 g × 250 * 100 |

- Solve the equation:
m(calcium hydroxide) |
= |
*0.12 g × 250 * 100 |

m(calcium hydroxide) |
= |
0.30 g |

*Is your answer plausible?*

250 mL is 2.5 times more than 100 mL, so if 0.12 g calcium hydroxide dissolved in 100 mL water produces a saturated solution, then 0.12 × 2.5 = 3.0 g calcium hydroxide dissolved in 250 mL water will produce a saturated solutution.

*State your solution to the problem.*

The minimum mass of calcium hydroxide required is 0.30 g

## Worked Example 2 (using the StoPGoPS approach to problem solving)

*Question 2:* A student adds 0.01 g of solid iodine to 20 mL of water at 25°C.

The solubility of iodine is 0.03 g/100 mL water at 25°C.

How much iodine, in grams, should precipitate out of the solution?

*Response:*

*What is the question asking you to do?*

Calculate the mass of iodine precipitated in grams.

m(undissolved iodine) = ? g

*What information have you been given in the question?*

m(total iodine) = 0.01 g

volume of water = 20 mL (25°C)

solubility of iodine = 0.03 g/100 mL water at 25°C

*What is the relationship between what you have been given and what you need to find out?*

m(dissolved iodine)/20 mL : 0.03 g/100 mL

m(undissolved iodine) = m(total iodine) - m(dissolved iodine)

*Step 1: Calculate mass of iodine that dissolves*

- Write the equation:
*m(dissolved iodine)* 20 mL |
= |
*0.03 g* 100 mL |

- Rearrange the equation by multiplying both sides of the equation by 20 mL:
*m(dissolved iodine) × *~~20 mL~~
~~20 mL~~ |
= |
*0.03 g × 20 *~~mL~~ 100 ~~mL~~ |

m(dissolved iodine) |
= |
*0.03 g × 20 * 100 |

- Solve the equation:
m(dissolved iodine) |
= |
0.006 g |

*Step 2: Calculate mass of iodine that does not dissolve*

m(undissolved iodine) = m(total iodine) - m(dissolved iodine)

m(undissolved iodine) = 0.01 g - 0.006 g = 0.004 g

*Is your answer plausible?*

Since we have been asked to calculate the mass of iodine that precipitates out of solution, we know that the mass of iodine added must have exceeded 0.03 g/100 mL water.

0.03 g per 100 mL = (0.03 g / 100 = 0.0003 g) per 1 mL

0.0003 × 20 = 0.006 g per 20 mL

0.01 g is greater than 0.006 g so our answer is reasonable.

Since only 0.006 g can dissolve in 20 mL, the rest, 0.01 - 0.006 = 0.004 g must not have dissolved.

*State your solution to the problem.*

The mass of iodine that precipitates out of solution is 0.004 g