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# Net Ionic Equation

A chemical reaction involves the conversion of reactant atoms or molecules to product. During any chemical reaction, reactant molecules are joined together to form a product which has different chemical and physical properties than the reactants. A reaction involves an energy changes and change in electronic configuration of reactant molecules.

The symbolic representation of chemical reaction includes the symbols for all of the products and reactant is known as chemical equation. A chemical equation shows the reactant molecules which involve in reaction and products which are formed during the reaction as well as the amount of each substance involve in reaction. Each chemical compound represents by a chemical formula.

A chemical equation represents by using these chemical formula of reactants and products molecules. For writing the chemical equation, first write down the formulas of substance involves in reaction, than place a plus sign between chemical formulas of substance with an arrow after last reactant molecule.

The physical state of reactant and product molecules is written in parenthesis as (s) for solid, (l) for liquid, (g) for gas and (aq) for aqueous for solution.

The conversion of reactant to product represents by a single headed arrow ‘?' directed from reactant to product. The reversible reaction can move in both directions and represents by double headed arrow '$\Leftrightarrow$'. For balancing the chemical equation, we have to place coefficients in front of formulas to insure the same number of atoms on both sides of arrows.

Some common examples of chemical equations are as follows.

H2CO3(aq) $\to$ H2O(l) + CO2(g)
H2SO3(aq)
$\to$H2O(l) + SO2(g)
NH4OH(aq)$\to$ NH$_{3(g)}$+ H$_2$O(l)

 Related Calculators Buffer Ionic Strength Calculator Ionic Compound Formula Calculator Calculator Equation 2 Step Equation Calculator

## Ionic and Net Ionic Equations

### Net Ionic Equation Definition

Net ionic equation is defined as "the equation that will give the information about the ions that exists when the medium is aqueous." In polar solvents like water many salts get dissolved and in a dissolved state they will be present as cations and anions. In a net ionic equation these ions are represented. Some Point about Ionic and Net Ionic Equation
1. A compound which can conduct electricity in their aqueous solution is called as electrolyte. An electrolyte dissociates into respective ions in solution which are able to conduct electricity.
2. Strong Electrolytes dissociate completely and exist as ions in solutions. For example, strong acids such as HCl, HBr, HI, HNO3, H2SO4 and HClO4 and strong bases like NaOH, KOH etc.
3. All soluble salts are also act as strong electrolyte and dissociate 100 % in solution. While weak electrolyte yields few to no ions in solution as they are partially dissociate in solution and should be represented by the molecular formula.
4. A chemical equation which shows dissociation of electrolyte and written as dissociated ions of electrolyte is known as ionic equation.
5. Generally single and double displacement reactions are used to represent ionic equations. Precipitation reaction, neutralization reaction are best examples of ionic equations.

Some common examples of ionic equations are as follow.

NaCl (aq) + AgNO3 (aq) $\to$ NaNO3(aq) + AgCl(s)
BaCl2 (aq) + Na2SO4 (aq)
$\to$ 2NaCl(aq) + BaSO4(s)
HCl (aq) + FeS(s)
$\to$ FeCl2(aq) + H2S(g)
HCl (aq) + NaOH (aq)
$\to$ NaCl (aq) + H2O(l)
CaCO3(s) + HCl (aq)
$\to$ CaCl2(aq) + CO2(g) + H2O(l)

Let’s write one balanced ionic equation for one of the neutralization reaction between sodium hydroxide and nitric acid in their aqueous solution. First write the balance chemical equation;
NaOH(aq) + HNO3(aq) $\to$ H2O(l) + NaNO3(aq)
Since it’s an acid-base reaction, therefore the product must be water and salt sodium nitrate. Then write the total ionic equation by determining the dissociation of reactants and products in solution. Remember, a strong electrolyte is written as ions while a weak electrolyte is written as molecule because it cannot be dissociate completely in solution. In given reaction, nitric acid(acid), sodium hydroxide(base) and sodium nitrate(salt) are strong electrolyte and water is a non-electrolyte, hence should be written as molecule only. Hence in ionic equation NaOH, HNO3 and NaNO3 are written as ions and water as a molecule.
Na+(aq) + OH-(aq) + H+(aq) + NO3-(aq) ? H2O(l) + Na+(aq) + NO3-(aq)
In total ionic equation, those species which are not undergoing any change during reaction are known as spectator ions. In balanced net equation these spectator ions are not included in equation as they exist with the same oxidation state on both sides of chemical equation. Here in given example, Na+ ion and NO3- ion are spectator ions, therefore must be excluded in net ionic equation. Hence net ionic equation should be written as follow.
OH-(aq) +H+(aq) $\to$ H2O(l)

Use below widget to balance the net ionic equations.

## Writing Net Ionic Equations

A chemical reaction in solution is batter describe by writing their ionic and net ionic equations which indicates the ions that exist in solution. Let’s take an example of double displacement reaction of sodium sulphate and barium chloride in aqueous solution to form barium sulphate as precipitate with sodium chloride in solution.

2NaSO4 (aq) + BaCl2(aq) 2NaCl(aq) + BaSO4(s)

The balanced ionic reaction indicates all soluble ionic materials as ions, followed by (aq) and insoluble ionic solids are written with (s) following their formula. In precipitation reaction all soluble unionized species are written with their molecular formula with (aq). Hence ionic equation of given chemical reaction will be

2Na+(aq) + SO42-(aq) + Ba2+(aq) + 2Cl1-(aq) 2Na+(aq) + 2Cl1-(aq) + BaSO4(s)

For writing the net ionic equation, we have to identify spectator ions which remain unchanged on both side of chemical equation. For example in given reaction sodium and chloride ions remain unchanged on both side and act as spectator ions. Therefore net ionic equation for the above reaction is

Ba2+(aq) + SO42-(aq) BaSO4(s)

Let’s write another example of net ionic equation for the reaction between silver nitrate and lithium bromide. It’s an example of double displacement reaction results the formation of lithium nitrate in solution and silver bromide as precipitate. The balanced chemical equation and ionic equation can be written as follow.

AgNO3(aq) + LiBr(aq) LiNO3(aq) + AgBr(s)
Ag+(aq) + NO3-(aq) + Li+(aq) + Br-(aq)
Li+(aq) + NO3-(aq) + AgBr(s)

Since both reactants of given reaction are soluble, therefore easily ionize in solution and form insoluble silver bromide and soluble lithium nitrate in solution. Since silver bromide is insoluble in nature, so should be written as molecule in ionic equation while rest of all species is written as ions.

In ionic equation, lithium and nitrate ions are common on both sides of chemical equation hence they act as spectator ions in equation and must be removed in net ionic equation. Hence the net ionic equation should look like

Ag+(aq) + Br-(aq) AgBr(s)

## Balancing Net Ionic Equations

For balancing a net ionic equation, first write a balanced molecular chemical equation and balanced complete ionic equation. Than identify spectator ions which remain unchanged during reaction. Cross out these spectator ions from ionic equation and write the leftover as chemical equation. This equation is known as net ionic equation.

Let’s discuss an example of reaction between sodium phosphate and calcium chloride in aqueous solution to form an insoluble white precipitate of calcium phosphate and soluble sodium chloride as product. It is a precipitation reaction and can be described by given balanced chemical equation.

2Na3PO4 (aq) + 3CaCl2 (aq) 6NaCl (aq) + Ca3(PO4)2(s)

Since a complete ionic equation indicates reactants and product molecules which exist as ions in their aqueous solution. Therefore for writing the ionic equation for this chemical equation, start from balanced chemical equation and break all soluble strong electrolyte species into their ions. Remember you have to indicate correct formula and charge of each ion as well as their number in equation. All insoluble compounds remain unchanged in equation and written as molecule only.

Let’s consider all reactant and product of given reaction.
• One mole of Na3PO4 contains 3 moles of sodium ion (Na+) and one mole of phosphate ion (PO43-). Therefore two moles of sodium phosphate will form total of six moles (2 x 3) of Na+ and two moles (2 x 1) of PO43- ions.
2Na3PO4 (aq) 6 Na+(aq) + 2PO43-(aq)
• Another reactant is calcium chloride whose one mole contains one mole of Ca2+ and two moles of Cl-. In balanced reaction, there are three moles of calcium chloride which gives total of 3 moles (3 x 1) of Ca2+ and 6 moles (3 x 2) of Cl- ions.
3CaCl2 (aq) 3 Ca2+(aq) + 6Cl- (aq)
• On product side one mole of sodium chloride contains one mole of Na+ and one mole of Cl-. There are six moles of sodium chloride in balanced reaction which will give total six moles of sodium ion and chloride ions.
6NaCl(aq) 6Na+(aq) + 6Cl-(aq)
• Another product is calcium phosphate which is an insoluble solid, hence it remains unchanged into the complete ionic equation.
Hence for the given reaction, the complete ionic equation is

6Na+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Cl-(aq) 6Na+(aq) + 6Cl-(aq) + Ca3(PO4)2(s)

By using this balance ionic equation we can easily write net ionic equation. For writing net ionic equation, we must identify spectator ions which remain same on both side of ionic equation. In previous ionic equation six sodium ions and six chloride ions are present on both sides of the equation. Hence these are spectator ions and should be excluded in net ionic equation.

Remember, spectator will always have the same formula, charge, number and physical state on both side of balanced ionic equation. Now cross out these two ions from ionic equation to get balance net ionic equation of given reaction.

6 Na+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Cl-(aq) 6Na+(aq) + 6Cl-(aq) +Ca3(PO4)2(s)

Balance net ionic equation: 2PO43-(aq) + 3Ca2+(aq) Ca3(PO4)2(s)

## Examples of Net Ionic Equations

Some examples of net ionic equations with their ionic equations are as follow.

1. 3(NH4)2CO3 (aq) + 2 Al(NO3)3 (aq) $\to$ 6NH4NO3 (aq) + Al2(CO3)3 (s)

• Balance ionic equation:
6NH4+(aq) + 3CO32-(aq) + 2Al3+(aq) + 6NO3-(aq) $\to$ 6NH4+(aq) + 6NO3-(aq) + Al2(CO3)3(s)
• Spectator ions: 6NH4+(aq), 6NO3-(aq)
• Balance net ionic equation: 2Al3+(aq) + 3CO32-(aq) ? Al2(CO3)3 (s)

2. 2NaOH(aq) + H2SO4 (aq) $\to$ Na2SO4 (aq) + 2H2O (l)
• Balance ionic equation:
2Na+(aq) + 2OH-(aq) + 2H+(aq) + SO42-(aq) $\to$ 2Na+(aq) + SO42-(aq) + 2H2O(l)
• Spectator ions: 2Na+(aq), SO42-(aq)
• Balance net ionic equation: 2OH-(aq) + 2H+(aq) ? 2H2O(l)
Or can written as OH-(aq) + H+(aq) ? H2O(l)

3. Mg (s) + 2 HCl(aq) $\to$ MgCl2 (aq) + H2 (g)
• Balance ionic equation:
Mg(s) + 2H+(aq) + 2Cl-(aq) $\to$ Mg2+(aq) + 2Cl- (aq) + H2(g)
• Spectator ions: 2Cl-(aq)
• Balance net ionic equation: Mg(s) + 2H+(aq) ? Mg2+(aq) + H2(g)

4. Zn (s) + CuSO4 (aq)
$\to$ ZnSO4 (aq) + Cu (s)
• Balance ionic equation:
Zn (s) + Cu2+(aq) + SO42-(aq) $\to$ Zn2+(aq) + SO42-(aq) + Cu(s)
• Spectator ions: SO42- (aq)
• Balance net ionic equation: Zn (s) + Cu2+(aq) ? Zn2+(aq) + Cu(s)

5. Na2CO3 (aq) + 2 HNO3 (aq)  $\to$ 2 NaNO3 (aq) + H2O (l) + CO2 (g)
• Balance ionic equation:
2Na+(aq) + CO32-(aq) + 2H+(aq) + 2NO3-(aq) $\to$ 2Na+(aq) + 2NO3-(aq) + H2O(l) + CO2(g)
• Spectator ions: 2Na+(aq), 2NO3-(aq)
• Balance net ionic equation: CO32-(aq) + 2H+(aq) ? H2O (l) + CO2 (g)

6. CaCO3 (s) + 2HCl(aq) $\to$ CaCl2(aq) + H2O (l) + CO2(g)
• Balance ionic equation:
CaCO3 (s) + 2H+(aq) + 2Cl- (aq) $\to$ Ca2+(aq) + 2Cl-(aq) + H2O (l) + CO2(g)
• Spectator ions: 2Cl- (aq)
• Balance net ionic equation: CaCO3(s) + 2H+(aq) ? Ca2+(aq) + H2O (l) + CO2(g)

7. 2HBr(aq) + Ca(OH)2(aq) $\to$ 2H2O (l) + CaBr2(aq)
• Balance ionic equation:
H+(aq) + 2Br-(aq) + Ca2+(aq) + 2OH-(aq) $\to$ 2H2O (l) + Ca2+(aq) + 2Br-(aq)
• Spectator ions: 2 Br-(aq), Ca2+(aq)
• Balance net ionic equation: H+(aq) + 2OH-(aq) ? 2H2O (l)

8. AgNO3(aq) + NaCl(aq)
$\to$ AgCl(s) + NaNO3(aq)
• Balance ionic equation:
Ag+(aq)+ NO3-(aq) + Na+(aq) + Cl-(aq) $\to$ AgCl(s) + Na+(aq) + NO3-(aq)
• Spectator ions: Na+(aq) , NO3-(aq)
• Balance net ionic equation: Ag+(aq) + Cl-(aq) ? AgCl(s)

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