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Solubility Rules

Key Concepts

  1. List the anions and cations in the solution
  2. List the possible ionic compounds that could be produced
  3. Use the solubility rules (list, table or chart) to decide if either of the ionic compounds are insoluble and will therefore form a precipitate.

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Solubility Rules as a List

The solubility of ionic compounds in water at 25°C, in general:

Worked Example Using the List of Solubility Rules

Question: A student adds an aqueous solution of silver nitrate, AgNO3(aq), to dilute hydrochloric acid, HCl(aq), at 25°C.

Will a precipitate form?

Response:

  1. List the anions and cations in the solution

        ions from AgNO3(aq)
      silver ion
    Ag+(aq)
    nitrate ion
    NO3-(aq)
    ions from HCl(aq) hydrogen ion
    H+(aq)
       
    chloride ion
    Cl-(aq)
       

  2. List the possible ionic compounds that could be produced (remember that an ionic compound is composed of anions and cations so there will be 2 possibilities)

        ions from AgNO3(aq)
      silver ion
    Ag+(aq)
    nitrate ion
    NO3-(aq)
    ions from HCl(aq) hydrogen ion
    H+(aq)
    X hydrogen nitrate
    (nitric acid)
    HNO3
    chloride ion
    Cl-(aq)
    silver chloride
    AgCl
    X

  3. Use the solubility rules listed to decide if either of the ionic compounds are insoluble and will therefore form a precipitate.

    (i) All nitrates are soluble, so hydrogen nitrate (nitric acid) is soluble and will not form a precipitate, HNO3(aq).

    (ii) All chlorides are soluble EXCEPT those of silver, lead and mercury(I), so silver chloride is insoluble and will form a precipitate, AgCl(s)

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Solubility Rules as a Table

If you need to memorise the solubility rules for ionic compounds in water at 25°C, then the list above is useful.
However, all that information can be placed in a table as shown below, which makes it easier to locate solubility information for a particular ionic compound made up of a particular cation and a particular anion.
This table is designed to answer the question,

"If I have a compound made up of this cation and this anion, will it dissolve in water at 25°C to form an aqueous solution, or will it form a solid (precipitate)?"

Solubility Rules for Aqueous Solutions at 25°C
Negative Ions
(Anions)
+ Positive Ions
(Cations)
= Solubility of compounds in water Example
any anion + alkali ions
(Li+,Na+,K+,Rb+,Cs+,Fr+)
= soluble sodium fluoride is soluble, NaF(aq)
any anion + hydrogen ion
[H+(aq)]
= soluble hydrogen chloride is soluble, HCl(aq)
any anion + ammonium ion
(NH4+)
= soluble ammonium chloride is soluble, NH4Cl(aq)
nitrate (NO3-) + any cation = soluble potassium nitrate is soluble, KNO3(aq)
acetate (CH3COO-) + any cation = soluble sodium acetate is soluble, CH3COONa(aq)
chloride (Cl-)

bromide (Br-)

iodide (I-)

+ silver (Ag+)
lead (Pb2+)
mercury (Hg22+)
copper (Cu+)
thallium (Tl+)
= low solubility (insoluble) silver chloride forms a white precipitate (a white solid), AgCl(s)
+ any other cation = soluble potassium bromide is soluble, KBr(aq)
sulfate (SO42-) + calcium (Ca2+)
strontium (Sr2+)
barium (Ba2+)
silver (Ag+)
lead (Pb2+)
radium (Ra2+)
= low solubility (insoluble) barium sulfate forms a white precipitate (a white solid), BaSO4(s)

+ any other cation = soluble copper(II) sulfate is soluble, CuSO4(aq)
sulfide (S2-) + alkali metal ions (Li+,Na+,K+,Rb+,Cs+,Fr+)
alkali earth metal ions (Be2+,Mg2+,Ca2+,Sr2+,Ba2+,Ra2+)
and hydrogen ion (H+(aq))
and ammonium ion (NH4+)
= soluble magnesium sulfide is soluble, MgS(aq)
+ any other cation = low solubility (insoluble) zinc sulfide is insoluble, ZnS(s)
Hydroxide (OH-) + alkali metal ions (Li+,Na+,K+,Rb+,Cs+,Fr+)
hydrogen ion (H+(aq))
ammonium ion (NH4+)
strontium ion (Sr2+)
barium ion (Ba2+)
radium ion (Ra2+)
thallium ion (Tl+)
= soluble strontium hydroxide is soluble, Sr(OH)2(aq)

+ any other cation = low solubility (insoluble) silver hydroxide is insoluble (forms a precipitate), AgOH(s)
phosphate (PO43-)

carbonate (CO32-)

sulfite (SO32-)

+ alkali metal ions (Li+,Na+,K+,Rb+,Cs+,Fr+)
hydrogen ions (H+(aq))
ammonium ions (NH4+)
= soluble ammonium phosphate is soluble, (NH4)3PO4(aq)
+ any other cation = low solubility (insoluble) magnesium carbonate is insoluble, MgCO3(s)

Worked Example Using the Solubility Rules Table

Question: A student adds an aqueous solution of barium nitrate, Ba(NO3)2(aq), to dilute sulfuric acid, H2SO4(aq), at 25°C.

Will a precipitate form?

Response:

  1. List the anions and cations in the solution

        ions from Ba(NO3)2(aq)
      barium ion
    Ba2+(aq)
    nitrate ion
    NO3-(aq)
    ions from H2SO4(aq) hydrogen ion
    H+(aq)
       
    sulfate ion
    SO42-(aq)
       

  2. List the possible ionic compounds that could be produced (remember that an ionic compound is composed of anions and cations so there will be 2 possibilities)

        ions from Ba(NO3)2(aq)
      barium ion
    Ba2+(aq)
    nitrate ion
    NO3-(aq)
    ions from H2SO4(aq) hydrogen ion
    H+(aq)
    X hydrogen nitrate
    (nitric acid)
    HNO3
    sulfate ion
    SO42-(aq)
    barium sulfate
    BaSO4
    X

  3. Use the solubility rules in the table to decide if either of the ionic compounds are insoluble and will therefore form a precipitate.

    (i) Go to the row labelled "nitrates (NO3-)" under the heading "Negative Ions (anions)", fourth from the top of the table).
    Go across the row to the column for "Positive Ions (Cations)".
    Read, "nitrate (NO3-) + any cation = soluble compound"
    So hydrogen nitrate (nitric acid) is soluble and will not form a precipitate, HNO3(aq).

    (ii) Go to the row labelled "sulfate (SO42-)" under the heading "Negative Ions (anions)", seventh from the top of the table).
    Go across the row to the column for "Positive Ions (Cations)".
    Read, "sulfate (SO42-) + barium (Ba2+) = insoluble compound"
    So barium sulfate is insoluble and will form a precipitate, BaSO4(s) .

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Solubility Rules as a Chart

The solubility rules for aqueous solutions at 25°C can also be used to construct a different sort of table which we will call a solubility chart.

The solubility chart below is designed to answer the question,

"Will an ionic compound made up of this cation and this anion dissolve in water at 25°C?"

The following key is used:

yes = The compound is soluble, it will all dissolve in water at 25°C

no = The compound is insoluble, there is no appreciable concentration of ions in the water at 25°C, the solid compound will precipitate out of solution.

slightly = A very small amount of the compound will dissolve in water at 25°C, that is, there is a very, very, small concentration of ions in solution.

... = no data available

reacts = The compound does not dissolve in water (which is a physical change), instead it reacts with water (which is a chemical change)

anions →

cations ↓

chloride
Cl-
bromide
Br-
iodide
I-
oxide
O2-
sulfide
S2-
hydroxide
OH-
carbonate
CO32-
chromate
CrO42-
sulfate
SO42-
acetate
CH3COO-
nitrate
NO3-
ammonium
NH4+
yes yes yes ... yes yes yes yes yes yes yes
sodium ion
Na+
yes yes yes reacts yes yes yes yes yes yes yes
potassium ion
K+
yes yes yes reacts yes yes yes yes yes yes yes
magnesium ion
Mg2+
yes yes yes no reacts no slightly ... yes yes yes
calcium ion
Ca2+
yes yes yes reacts reacts slightly no slightly slightly yes yes
barium ion
Ba2+
yes yes yes yes yes yes no no no yes yes
manganese(II) ion
Mn2+
yes yes yes no no no no ... yes yes yes
iron(II) ion
Fe2+
yes yes yes no no no no ... yes yes yes
iron(III) ion
Fe3+
yes yes ... no ... no ... no yes no yes
copper(II) ion
Cu2+
yes yes ... no no no no no yes yes yes
nickel(II) ion
Ni2+
yes yes yes no no no no no yes yes yes
cadmium(II) ion
Cd2+
yes yes yes no no no no no yes yes yes
zinc ion
Zn2+
yes yes yes no no no no yes yes yes yes
tin(II) ion
Sn2+
yes yes slightly no no no ... ... yes yes ...
mercury(I) ion
Hg22+
yes slightly no no no no no no reacts yes yes
lead(II) ion
Pb2+
slightly slightly no no no no no no no yes yes
silver ion
Ag+
no no no no no ... no no slightly slightly yes
cations ↑

anions →

chloride
Cl-
bromide
Br-
iodide
I-
oxide
O2-
sulfide
S2-
hydroxide
OH-
carbonate
CO32-
chromate
CrO42-
sulfate
SO42-
acetate
CH3COO-
nitrate
NO3-

Worked Example Using the Solubility Rules Chart

Question: A student adds an aqueous solution of lead(II) nitrate, Pb(NO3)2(aq), to an aqueous solution of sodium iodide, NaI(aq), at 25°C.

Will a precipitate form?

Response:

  1. List the anions and cations in the solution

        ions from Pb(NO3)2(aq)
      lead(II) ion
    Pb2+(aq)
    nitrate ion
    NO3-(aq)
    ions from NaI(aq) sodium ion
    Na+(aq)
       
    iodide ion
    I-(aq)
       

  2. List the possible ionic compounds that could be produced (remember that an ionic compound is composed of anions and cations so there will be 2 possibilities)

        ions from Pb(NO3)2(aq)
      lead(II) ion
    Pb2+(aq)
    nitrate ion
    NO3-(aq)
    ions from NaI(aq) sodium ion
    Na+(aq)
    X sodium nitrate
    NaNO3
    iodide ion
    I-(aq)
    lead(II) iodide
    PbI2
    X

  3. Use the solubility rules in the chart to decide if either of the ionic compounds are insoluble and will therefore form a precipitate.

    (i) Go to the row labelled "sodium ion Na+" under the heading "cations", (second from the top of the table).
    Go across the row of "anions" and look at the last column labelled "nitrate NO3-".
    Read, "yes"
    So sodium nitrate is soluble and will not form a precipitate, NaNO3(aq).

    (ii) Go to the row labelled "lead(II) ion Pb2+" under the heading "cations", (second from the bottom of the table).
    Go across the row of "anions" to the third column labelled "iodide I-".
    Read, "no"
    So lead(II) iodide is insoluble and will form a precipitate, PbI2(s) .

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1. Solubility Tables (tables of solubility) give a quantitative measure of how much solute dissolves in water at 25°C.

Solubility Curves (graphs) give a quantitative measure of how much solute will dissolve in water at different temperatures.

2. A solute is said to be soluble if its solubility is greater than 1 g/100 g of water (1 g/100 mL water at 25°C)

3. A solute is said to be insoluble if its solubility is less than 0.1 g/100 g of water (0.1 g/100 mL water at 25°C)

A solute is said to be slightly soluble, or sparingly soluble, if its solubility is between 0.1 g/100 g of water and 1.0 g/100 g (between 0.1 g/100 mL water and 1.0 g/100 mL at 25°C)