What Are Double Displacement Reactions?

A double displacement reaction — also called a double replacement reaction, exchange reaction, or metathesis reaction — is a type of aqueous reaction in which two ionic compounds swap cation-anion partners. You will encounter all of these names, though "double displacement" and "double replacement" appear most frequently in general chemistry coursework.

The essential concept is simple: cations and anions trade partners. In an ionic compound, the cation is always written first and the anion second. In a double displacement reaction, the cation from one compound pairs with the anion from the other, and vice versa.

The character of a double displacement reaction is defined by what forms in the products. There are 3 major types of double displacement reactions:

• Precipitation reactions — one product is an insoluble solid (a precipitate)
• Acid-base neutralization reactions — the reactants are an acid and a hydroxide base
• Gas-forming reactions — one product decomposes to release a gas (typically CO₂)

Precipitation Reactions

The defining feature of a precipitation reaction is the formation of an insoluble solid product called a precipitate. To work through one, you need to predict the products, write and balance the molecular equation, and then apply solubility rules to assign phases.

Predicting Products and Assigning Charges

Consider lead(II) nitrate reacting with sodium iodide. To predict the products, combine the cation from one compound with the anion from the other. The two new pairings are lead with iodide, and sodium with nitrate.

To write correct formulas, you must know the charges. Lead is a variable-charge metal, but you can determine its charge from context: in lead(II) nitrate, each nitrate carries a −1 charge, and there are two of them, so lead must be +2. Iodide as a monatomic ion is −1. To form a neutral compound, you need two iodide ions for every lead ion, giving PbI₂.

For the second product: sodium (from group 1) is always +1, and nitrate is −1, so they combine in a 1:1 ratio to give NaNO₃. Notice that the formula for sodium nitrate does not have to match the formula of the original lead nitrate — it all comes down to forming electrically neutral compounds.

Balancing and Applying Solubility Rules

Balancing the molecular equation for this reaction:

One lead balances on each side. Two nitrates on the left require a coefficient of 2 in front of sodium nitrate on the right, which in turn requires a coefficient of 2 in front of sodium iodide on the left — giving two sodiums and two iodides on each side. The equation is balanced.

Now assign phases to the products using solubility rules:

• PbI₂: Most iodides are soluble, except those of silver, lead, and mercury. Lead iodide is therefore insoluble — it is a solid precipitate.
• NaNO₃: All nitrates are soluble with no exceptions. Sodium nitrate is aqueous.

3 Types of Strong Electrolytes and the Ionic Equation

The molecular equation is a useful starting point, but it does not fully represent what is happening in solution. To get a more accurate picture, we need to identify the strong electrolytes and write the total ionic equation.

Strong electrolytes dissociate completely into ions in solution. There are exactly 3 types of strong electrolytes:

1. Soluble ionic compounds (ionic and aqueous)
2. Strong acids
3. Strong bases

For the lead nitrate + sodium iodide reaction, the strong electrolyte analysis looks like this:

• Pb(NO₃)₂(aq) — ionic and aqueous → strong electrolyte
• NaI(aq) — ionic and aqueous → strong electrolyte
• PbI₂(s) — ionic but not aqueous (it is insoluble) → not a strong electrolyte
• NaNO₃(aq) — ionic and aqueous → strong electrolyte

In the total ionic equation, every strong electrolyte is written as its constituent ions (because that is how they actually exist in solution). Compounds that are not strong electrolytes remain intact.

Spectator Ions and the Net Ionic Equation

Looking at the total ionic equation, something becomes apparent: some ions appear on both sides of the arrow completely unchanged. These are the spectator ions — they were present in solution throughout the reaction but took no part in it. The analogy is apt: they came to the party alone and left the party alone, just watching.

In the lead iodide example, the Na⁺ and NO₃⁻ ions are the spectators. Two sodium ions arrive in solution and two sodium ions depart in solution, unchanged. The same is true of the two nitrate ions.

The net ionic equation is obtained by canceling the spectator ions from both sides of the total ionic equation. What remains shows the actual chemical transformation:

A Systematic Method for Identifying Spectator Ions

Once you have identified all the strong electrolytes, you can spot spectator ions quickly with one rule: any ion that is part of a strong electrolyte on both the reactant side and the product side is a spectator.

• NO₃⁻ — part of a strong electrolyte on both sides → spectator ion
• Na⁺ — part of a strong electrolyte on both sides → spectator ion
• Pb²⁺ — part of a strong electrolyte on the left, but not on the right → reacts
• I⁻ — part of a strong electrolyte on the left, but not on the right → reacts

With a little practice, you can derive the net ionic equation directly from the molecular equation without writing out the full ionic equation as an intermediate step.

Acid-Base Neutralization Reactions

The second major type of double displacement reaction is acid-base neutralization. In this lesson, we focus specifically on reactions between an acid and a hydroxide base (a base containing OH⁻). Second-semester chemistry will introduce other base types, but hydroxide-based neutralizations follow a consistent, learnable pattern.

The double displacement logic still applies: H⁺ pairs with OH⁻ to form water, and the remaining cation pairs with the remaining anion to form a salt. Remember that "salt" in chemistry means any ionic compound — not just table salt.

Example: Strong Acid + Strong Base (HCl + NaOH)

Trading partners: H goes with OH to form water, and Na goes with Cl to form sodium chloride.

Identifying spectator ions: Cl⁻ is part of a strong electrolyte on both sides (HCl and NaCl), and Na⁺ is part of a strong electrolyte on both sides (NaOH and NaCl). They are both spectators. That leaves only H⁺ and OH⁻ as the active participants.

Net Ionic Equation: Weak Acid + Strong Base

Things change when the acid is weak. A weak acid does not fully dissociate in solution — typically less than 5% of its molecules ionize — so it is a weak electrolyte and must be written as an intact molecule rather than as separate ions in the net ionic equation.

Consider hydrofluoric acid reacting with sodium hydroxide. The products are water and sodium fluoride. (Fluoride salts of group 1 metals are soluble, so NaF is aqueous.) The equation is balanced as-is with a 1:1:1:1 ratio.

Now identify the spectator ions. Only Na⁺ appears as part of a strong electrolyte on both sides — in NaOH on the left and in NaF on the right. It is the only spectator ion. This means HF, OH⁻, and F⁻ all remain in the net ionic equation.

Balancing a More Complex Neutralization: Phosphoric Acid + Sodium Hydroxide

Phosphoric acid (H₃PO₄) reacting with sodium hydroxide is a useful example because balancing it can trip students up. The key is a strategic trick: write water as HOH initially, rather than H₂O.

When you write water as HOH, the H from the acid and the OH from the base are visually distinct, making it much easier to track the atoms during balancing.

The products are water and sodium phosphate. Sodium is +1, phosphate is −3, so you need three sodium ions per phosphate: Na₃PO₄.

Net Ionic Equation: H₃PO₄ + NaOH

Because H₃PO₄ is a weak acid, it is not a strong electrolyte and is written intact in the ionic equation. NaOH (strong base) and Na₃PO₄ (soluble ionic compound) are strong electrolytes and are split into ions. H₂O is a molecular compound and is not split.

Gas-Forming Double Displacement Reactions

The third type of double displacement reaction produces a gas as a product. Technically, this is still an acid-base reaction — just not one involving a hydroxide base. Because it looks and behaves distinctly from a standard neutralization, it is worth treating as its own category. The gas-forming double displacement reaction you are most likely to encounter in general chemistry is the reaction of a strong acid with a carbonate salt.

Example: HCl + Sodium Carbonate

Apply the partner-swapping logic: H pairs with CO₃ (giving carbonic acid, H₂CO₃), and Na pairs with Cl (giving sodium chloride).

Here is the critical point: carbonic acid is unstable and spontaneously decomposes as soon as it forms:

Because the carbonic acid immediately breaks down, it is not considered a final product. You substitute H₂O and CO₂ in its place when writing the molecular equation.

Net Ionic Equation: HCl + Na₂CO₃

To write the net ionic equation, identify the strong electrolytes. HCl is a strong acid, Na₂CO₃ and NaCl are soluble ionic compounds — all three are strong electrolytes and are split into ions. CO₂(g) and H₂O(l) are molecular compounds and are not strong electrolytes and will not be split.