Inorganic Salt Dissociation Equations

salt inorganic is every ionic substance (formed by a cation other than hydronium / H⁺ and anion other than hydroxide / OH^-) which, when dissolved in water, undergoes the phenomenon of dissociation. In this process, the cation and anion present are released, as shown in dissociation equation to follow:

XY_(here) → X⁺ + Y^-

The dissociation equations for inorganic salts always feature:

• The abbreviation (aq): Indicates that salt is mixed with water;

• An arrow: Indicates the release of ions;

• X⁺: Cation referring to the first group or symbol of the salt formula;

• Y^-: Anion referring to the group or symbol after the X of the salt formula.

To assemble a dissociation equation of a salt, it is necessary to know well the formula of the inorganic salt that will be dissociated. As a rule, the formula of an inorganic salt can have the following characteristics:

a) Dissociation equation of a salt with an X cation and a Y anion without any index written on either of them.

The charge of both has different signs and equal values, and the value is always defined by the anion. For example:

Example 1: NaCl - Sodium Chloride

As the Cl anion has a charge value of -1, the Na cation has a charge of +1. Thus, the ions are represented by Na⁺¹ and Cl^-1, and the dissociation equation for this salt is:

NaCl_(here) → In⁺¹ + Cl^-1

Example 2: MgS - Magnesium Sulfide

As the S anion has a charge value of -2, the Mg cation has a charge of +2. Thus, the ions are represented by Mg⁺² and S^-2, and the dissociation equation for this salt is:

MgS_(here) → Mg⁺² + S^-2

b) Dissociation equation for a salt that has a cation with an index written right after it and no index written after the anion.

In this case, the anion charge is the index written in front of the cation, and the cation charge has a value of 1, since there is no number as an anion index. For example:

Example 1: K₂S - Potassium Sulfide

As the K cation has the index 2, anion charge is -2. Already the cation will have a +1 charge because there is no index written on the anion. Thus, the ions are represented by K⁺¹ and S^-2, and the dissociation equation for this salt is:

K₂s_(here) → 2K⁺¹ + S^-2

It is necessary to place the coefficient 2 to the left of the K, since in the formula for salt there is 2 K.

c) Dissociation equation of a salt with an X cation with no index written after it and an anion Y presenting the element oxygen with an index written right after it.

In this case, the index written after the oxygen must be disregarded, and the charge of the cation and the anion will have different signs and equal values, the value always being defined by the anion. For example:

Example 1: NaClO₄- Sodium perchlorate

Like the ClO anion₄ presents load -1, the Na cation has a +1 charge. So the ions are represented by Na⁺¹ and ClO₄^-1. The dissociation equation for this salt is:

NaClO_4(aq) → In⁺¹ + ClO₄^-1

Example 2: MgCO₃- Magnesium carbonate

Like the anion CO₃ presents load -2, the Mg cation has a +2 charge. So the ions are represented by Mg⁺² and ClO₃^-2, and the dissociation equation is:

MgCO_3(aq) → Mg⁺² + ClO₃^-2

Example 3: AlPO₄- Aluminum phosphate

Like the anion po₄ presents load -3, the Al cation has a +3 charge. So the ions are represented by Al⁺³ and PO₄^-3, and the dissociation equation for this salt is:

AlPO_4(aq) → Al⁺³ + PO₄^-3

d) Dissociation equation of a salt that has a cation X with an index written right after it and the anion Y showing the element oxygen and an index written right after it too.

In this case, the cation index is the anion charge, and the cation charge is equal to 1, as there is only the index immediately after oxygen. For example:

Example 1: K₂ONLY₃- Potassium sulphite

As the K cation has the index 2, anion charge is -2. Already cation K has +1 charge because there is no index written on the anion after the 3, which belongs to oxygen. Thus, the ions are represented by K⁺¹ and SO₃^-2, and the dissociation equation for this salt is:

K₂ONLY_3(aq) → 2K⁺¹ + OS₃^-2

It is necessary to place the coefficient 2 to the left of the K, since in the formula for salt there is 2 K.

Example 2: Au₃BO₃- Gold borate I

As the Au cation has an index 3, the charge of the anion BO₃ é -3. Already the cation has a +1 charge because there is no index written on the anion after the 3, which belongs to oxygen. Thus, the ions are represented by Au⁺¹ and Bo₃^-3, and the dissociation equation for this salt is:

Au₃BO_3(aq) → 3 Au⁺¹ + BO₃^-3

It is necessary to place the coefficient 3 to the left of Au, since in the formula for salt there is 3 Au.

Example 3: Ass₄P₂O₆- Copper hypophosphate I

As the copper cation (Cu) has the index 4, the anion charge is -4. already the cation has +1 charge because there is no index written on the anion after the 6, which belongs to oxygen. Thus, the ions are represented by Cu⁺¹ and P₂O₆^-4, and the dissociation equation for this salt is:

Ass₄P₂O_6(aq) → 4 Cu⁺¹ + P₂O₆^-4

It is necessary to place the coefficient 4 to the left of the copper cation (Cu), since in the salt formula there are 4 copper cations.

e) Dissociation equation of a salt with an X cation with no index written after it and the anion Y inside parentheses with a written index.

In this case, the index after the anion parentheses is the charge of the cation, and the charge of the anion is 1, since there is no index written after the cation. For example:

Example 1: Mg (ClO₂)₂- Magnesium Chlorite

As the ClO anion₂ presents the index 2 after the parentheses, Mg cation charge is +2. already the anion has a charge of -1 because there is no index written after the cation. Thus, the ions are represented by Mg⁺² and ClO₂^-1, and the dissociation equation for this salt is:

Mg (ClO₂)_2(aq) → Mg⁺² + 2 ClO₂^-1

It is necessary to place the coefficient 2 to the left of the ClO₂, since in the salt formula there is 2 ClO₂.

Example 2: Al (NC)₃- Aluminum isocyanide

As the NC anion has the index 3 after the parentheses, Al cation charge is +3. already the anion has a charge of -1 because there is no index written on the cation. Thus, the ions are represented by Al⁺³ and NC^-1, and the dissociation equation for this salt is:

Al (NC)_3(aq) → Al⁺³ + 3 NC^-1

It is necessary to place the coefficient 3 to the left of the NC, since in the formula for salt there are 3 NC.

Example 3: You (MnO₄)₄- Titanium permanganate IV

As the MnO anion₄ presents the index 4 after the parentheses, the charge of Ti cation is +4. already the anion has a charge of -1 because there is no index written on the cation. Thus, the ions are represented by Ti⁺⁴ in the₄^-1, and the dissociation equation for this salt is:

You (MnO₄)_4(aq) → You⁺⁴ + 4 MnO₄^-1

It is necessary to place coefficient 4 to the left of MnO₄, since in the salt formula there is 4 MnO₄.

f) Dissociation equation of a salt that has a cation X with an index written right after it and the anion Y inside parentheses with an index written.

In this case, the index after the anion parentheses is the charge of the cation, and the index after the cation is the charge of the anion. For example:

Example 1: Al₂(ONLY₄)₃- Aluminum sulfate

As the anion SO₄ presents the index 3 after the parentheses, the Al cation charge is +3. already the anion has -2 charge because index 2 is written after the cation. Thus, the ions are represented by Al⁺³ and SO₄^-2, and the dissociation equation for this salt is:

Al₂(ONLY₄)_3(aq) → 2 Al⁺³ + 3 SO₄^-2

It is necessary to place the coefficient 3 to the left of the SO₄ and the coefficient 2 to the left of Al, since in the formula for salt we have 2 Al and 3 SO₄.

Example 2: You₂(Ç₂O₄)₄- Titanium oxalate IV

As the C anion₂O₄ presents the index 4 after the parentheses, Ti cation charge is +4. already the anion has -2 charge because index 2 is written after the cation. Thus, the ions are represented by Ti⁺⁴ and C₂O₄^-2, and the dissociation equation for this salt is:

You₂(Ç₂O₄)_4(aq) → 2 Ti⁺⁴ + 4 C₂O₄^-2

It is necessary to place the coefficient 4 to the left of the C₂O₄ and the coefficient 2 to the left of Ti, since in the salt formula we have 2 Ti and 4 C₂O₄.

Example 3: Faith₄(P₂O₆)₃- Iron III hypophosphate

Like the anion P₂O₆ presents the index 3 after the parentheses, the Fe cation charge is +3. already the anion has -4 charge because index 4 is written right after the cation. Thus, the ions are represented by Fe⁺³ and P₂O₆^-4, and the dissociation equation for this salt is:

Faith₄(P₂O₆)_3(aq) → 4 Fe⁺³ + 3P₂O₆^-2

It is necessary to place coefficient 3 to the left of P₂O₆ and the coefficient 4 to the left of Fe, since in the formula for salt we have 4 Fe and 3 P₂O₆.

By Me. Diogo Lopes Dias

Source: Brazil School - https://brasilescola.uol.com.br/quimica/equacoes-dissociacao-sais-inorganicos.htm