Spectator Ions Net Ionic Equations

Because spectator ions don t actually participate in the chemistry of a reaction, you don t need to include them in a chemical equation. Doing so leads to a needlessly complicated reaction equation, so chemists prefer to write net ionic equations, which omit the spectator ions. A net ionic equation doesn t include every component that may be present in a given beaker. Rather, it includes only those components that actually react. 

Here s a simple recipe for making net ionic equations of your own  

Examine the starting equation to determine which ionic compounds are dissolved, as indicated by the aq) symbol following the compound name. 

Rewrite the equation, explicitly separating dissolved ionic compounds into their component ions. 

Polyatomic ions don t break apcirt in solution, so be sure to familiarize yourself with the common ones (flip to Chapter 6 for details). 

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Moiecuiar equation Compiete ionic equation Spectator ions Net ionic equation... 

Section 4.6 formula equation complete ionic equation spectator ions net ionic equation Section 4.8 acid base... 

Conventional equation Net ionic equation Spectator spectator ion Total ionic equation Section 9.5 Activity series... 

Here, as always, spectator ions such as Na+ are not included in the net ionic equation. 

Canceling the spectator ions leaves the net ionic equation for the reaction, the chemical equation that displays the net change taking place in the reaction ... 

A complete ionic equation expresses a reaction in terms of the ions that are present in solution a net ionic equation is the chemical equation that remains after the cancellation of the spectator ions. 

Because the spectator ions are not shown, the net ionic equation will be the same when any soluble mercury(I) compound is mixed with any soluble iodide. 

The Na+ and NO, ions are spectators, and so they are omitted from the net ionic, equation. 

When the solution in Beaker 1 is mixed with the solution in Beaker 2, a precipitate forms. Using the following table, write the net ionic equation describing the formation of the precipitate, and then identify the spectator ions. 

The contents of Beaker 1 are mixed with those of Beaker 2. If a reaction takes place, write the net ionic equation and indicate the spectator ions. ... 

Each of the following five procedures results in the formation of a precipitate. For each reaction, write the chemical equations describing the formation of the precipitate the overall equation, the complete ionic equation, and the net ionic equation. Identify the spectator ions. 

A net ionic equation contains only those species that participate in a chemical reaction. Notice that neither K nor Cr appears in the equation for the precipitation of Fe (OH). Similarly, neither K nor NO3 appears in the equation for the precipitation of Pbl2. Although these other ions are present in the solution, they undergo no change during the precipitation reaction. Ions that are not involved in the chemical change are referred to as spectator ions. Spectator ions are omitted from net ionic equations. 

C04-0086. Write the baianced net ionic equation for each of these precipitation reactions. Also, identify the spectator ions. 

Written in the latter manner, the equation shows that in effect the sodium ions and the chloride ions have not changed. They began as ions in solution and wound up as those same ions in solution. They are called spectator ions. Since they have not reacted, it is really not necessary to include them in the equation. If they are left out, a net ionic equation results ... 

Net ionic equations must always have the same net charge on each side of the equation. (The same number of each type of spectator ion must be omitted from both sides of the equation.) For example, the equation... 

Note This exercise uses KMn04 and FeS04. These are both soluble salts which dissociate into their ions. The K+ and S042- ions are spectator ions and are omitted from the balanced net ionic equation. 

Balance the two half-equations for charge with electrons and then combine them to produce the net ionic equation. Finally, add in the nitrate spectator ions. 

Words that can be used as topics in essays 5% rale buffer common ion effect equilibrium expression equivalence point Henderson-Hasselbalch equation heterogeneous equilibria homogeneous equilibria indicator ion product, P Ka Kb Kc Keq KP Ksp Kw law of mass action Le Chatelier s principle limiting reactant method of successive approximation net ionic equation percent dissociation pH P Ka P Kb pOH reaction quotient, Q reciprocal rule rule of multiple equilibria solubility spectator ions strong acid strong base van t Hoff equation weak acid weak base... 

Write a balanced net ionic equation. Because H2C03 is a weak acid, it would tend to remain as carbonic acid, especially in the presence of a strong acid like HBr. KBr is an ionic solid, soluble in water therefore, it would exist as separate ions known as spectator ions, which are not written in the net ionic equation. 

These problems involve mixing two solutions. Each solution is a water solution of an ionic compound. From the mixture of the two solutions, at least one insoluble precipitate will form. The other ions present are probably soluble and are called spectator ions they are not included in the net ionic equation. You must know your solubility rules to do these problems. 

We write the net ionic equation by dropping out the spectator ions and showing only those chemical species that are actually involved in the chemical reaction ... 

To get the net ionic equations, we must remove the spectator ions from the total ionic equations. We eliminate anything that appears identical on both sides of the reaction arrow. This cancellation gives ... 

Finally, we eliminate the spectator ions to obtain the net ionic equations ... 

This net ionic equation focuses only on the substances that are actually involved in the reaction. It indicates that an aqueous solution containing Pb2+ (any solution, not just Pb(N03)2(aq)) will react with any solution containing the sulfate ion to form insoluble lead(II) sulfate. If this equation form is used, the spectator ions involved will not be known, but in most cases this is not a particular problem since the focus is really the general reaction, and not the specific one. You will be expected to write the balanced net ionic equation for many of the reactions on the test. 

A—Acetic acid is a weak acid as such it should appear as HC2H302. Potassium hydroxide is a strong base so it will separate into K+ and OH ions. The potassium ion is a spectator ion, and is left out of the net ionic equation. 

In the molecular equation, the reactants and products are shown in their undissociated/unionized form the ionic equation shows the strong electrolytes in the form of ions the net ionic equation drops out all spectator ions and shows only those species that are undergoing chemical change. 

For example, a mixture of iron(III) nitrate and potassium thiocyanate, in aqueous solution, react to form the iron(III) thiocyanate ion, Fe(SCN) (aq). The reactant solutions are nearly colourless. The product solution ranges in colour from orange to blood-red, depending on its concentration. The nitrate and potassium ions are spectators. Therefore, the net ionic equation is... 

H (aq) Cl (aq) hJa (aq) + OH (aq) Fla (aq) "t" Cl (aq) "t" H20(r) AH = —56 kJ Subtracting spectator ions from both sides, the net ionic equation is H+(aq) -I- OH (aq) H20(f) AH = -56 kJ Different combinations of strong Arrhenius acids and bases react with the same exothermic result. Measurements always show the release of 56 kJ of energy per mole of water formed. This makes sense, because the net ionic equation is the same regardless of the specific neutralization reaction that occurs. 

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