Partial Neutralization Equations

when we react a acid (HX) and a base (MeOH), there is a chemical reaction called neutralization, which originates a salt inorganic and a water molecule. In this reaction, the ionizable hydrogen (present in the acid) reacts with the hydroxyl (present in the base) to form water:

H⁺ + OH^- → H₂O

The equation that represents the phenomenon of neutralization can be expressed as follows:

HX + MeOH → MeX + H₂O

When only a portion of the ionizable hydrogens in the acid reacts with the hydroxyls in the base, or vice versa, the reaction is called partial neutralization.

In the equations that represent partial neutralization, there is the presence of salts with hydrogen (H) and hydroxyl (OH). See below for the general format of a partial neutralization equation:

• Partial Neutralization Equation with excess of hydroxyls:

HX + Me (OH)₂ → MeOHX + H₂O

• Partial Neutralization Equation with excess of ionizable hydrogens:

H₂X + MeOH → MeHX + H₂O

Analyzing the general equations of partial neutralization, we can see that, whenever this type of reaction happens, we have the formation of hydrogenated salt (MeHX) or hydroxylated salt (MeOHX). The formation of either salt depends on the relationship between the amount of hydroxyls in the base and ionizable hydrogens in the acid.

see some examples of partial neutralization equations:

Example 1: Partial neutralization equation between hydrochloric acid (HCl) and magnesium hydroxide [Mg (OH)₂]:

1 HCl + 1 Mg (OH)₂ → MgOHCl + 1 H₂O

Analyzing the partial neutralization equation between the acid and the base in question, we have to:

1. The acid has only one ionizable hydrogen;

2. The base has two hydroxyls;

3. just one hydroxyl it is used in the formation of water because there is only one ionizable hydrogen;

4. The hydroxyl that is not used in the formation of water it is part of the salt formed and is written in the formula of the salt after the metal and before the Cl anion.

Example 2: Partial neutralization equation between phosphoric acid (H₃DUST₄) and potassium hydroxide (KOH).

1 hour₃DUST₄ + 1 KOH → KH₂DUST₄ + 1 hour₂O

Analyzing the partial neutralization equation between the acid and the base in question, we have to:

1. The acid has three ionizable hydrogens;

2. The base has a hydroxyl;

3. Only one ionizable hydrogen it is used in the formation of water because there is only one hydroxyl in the base;

4. You two ionizable hydrogens that are not used in water formation are part of the formed salt and will be written in the salt formula after the metal and before the PO anion₄.

Example 3: Partial neutralization equation between sulfuric acid (H₂ONLY₄) and titanium hydroxide IV [Ti(OH)₄].

1 hour₂ONLY₄ + 1 Ti (OH)₄ → Ti(OH)₂ONLY₄ + 2 H₂O

Analyzing the partial neutralization equation between the acid and the base in question, we have to:

1. The acid has only one ionizable hydrogen;

2. The base has two hydroxyls;

3. Only two hydroxyls they are used in the formation of water because there are only two ionizable hydrogens;

4. The hydroxyls that are not used in the formation of water they are part of the salt formed and are written in the formula of the salt after the metal and before the anion SO₄.

Example 4: Partial neutralization equation between pyrophosphoric acid (H₄P₂O₇) and silver hydroxide (AgOH).

1 hour₄P₂O₇ + 1 AgOH → AgH₃P₂O₇ + 1 hour₂O

1. The acid has four ionizable hydrogens;

2. The base has a hydroxyl;

3. Only one ionizable hydrogen it is used in the formation of water because there is only one hydroxyl in the base;

4. You three ionizable hydrogens that are not used in water formation are part of the formed salt and are written in the salt formula after the metal and before the P anion₂O₇.

By Me. Diogo Lopes Dias

Source: Brazil School - https://brasilescola.uol.com.br/quimica/equacoes-neutralizacao-parcial.htm