Net Ionic Equations

Net ionic equations


Net ionic equations are useful in that they show that the chemicals directly involved in a chemical reaction. They are simpler than the overall equation, and help us focus on the "heart" of chemical changes in a particular reaction. The key to being able to write net ionic equations is the ability to recognize the monoatomic and polyatomic solubility rules, and rules of behavior electrolyte. If you are weak in these areas, a re ¬ view of these concepts is useful before attempting to write net ionic equations. If you feel you should review the polyatomic ions, see Table 3.3 on page 88 of your book ¬ text. Solubility rules for ionic compounds Mon ¬ in the water can be found in Table 4.1, page 130 (References are to Denniston, Topping, Caret, 5 / e.)

Behavior of electrolyte

Once dissolved in water, many materials are divided into ions, or dissociated. This is because water is a polar solvent. The main experimental evidence for this is that the solutions of these materials carry an electrical current. Since an electric current is a stream of charged particles and water molecules are neutral, the only explanation for the current flow is that the dissociation of solute ions to produce Ted ¬ current flow necessary. Any material that the aqueous solution will conduct an electric current (ie, containing ions) is called an electrolyte.

Some of the materials are 100% dissociated into ions in aqueous solution, these materials are called strong electrolytes. Materials that are separated only in part of their ions are called weak electrolytes and materials that do not distinguish at all is called nonelectro ¬ lytes. (Note that the definitions of "strong" and "weak" electrolytes are not related to the overall solution, but part of the material to dissolve the existing format.) In general, weak electrolyte ions separated only minimally (usually less than 5%).

When writing chemical equations for reactions in aqueous solution, it is often use to write them ¬ FUL, which shows the real species in solution (such as ions or molecules that are relevant ¬ applicable), rather than using the full "molecular" formula for all reactants and products. How do you know whether a material is strong, weak or nonelectrolyte when dissolved in water? The following list summarizes the "rules" of the behavior of electrolytes.

First All salts (ionic compounds) are strong electrolytes.

2. Most acids are weak electrolytes (ie, are "weak acids"). This generalization ¬ cluds inorganic and organic acids (ie, containing carbon, which is usually contain C, H and O). The only exceptions to this generalization are common HCl, HBr, HI, HNO3, H2SO4 and HClO4. There are six strong acids.

3. Among the bases, metal hydroxides are strong electrolytes (ie, are "solid basis"). Ammonia (NH3) and organic bases (usually containing formulas as C, H and N, and can be considered derived from ammonia by substitution of one or more atoms of hydrogen and carbon containing groups) are weak.

4. All other materials are non-electrolytes.

The appropriate use of these rules is crucial to write chemical formulas net ionic form. Examples of this process are listed below.

Precipitation reactions

Let's first start and complete chemical equation for how the net ionic equation is de-rived. For example, consider the reaction of lead (II) nitrate with hydrochloric acid, in the form of lead (II) chloride and nitric acid below:

Pb (NO3) 2 (aq) + 2 HCl (aq) PbCl 2 (s) + 2 HNO3 (aq)

This equation can be rewritten completely in ionic form using the solubility rules and rules of behavior of electrolytes. All salts are strong electrolytes, so the lead (II) nitrate are separated. As hydrochloric acid and nitric acid are strong acids (found in the list of exceptions) and therefore, say ¬ partner. Lead (II) is insoluble, but remember that all halides are soluble except silver, lead and mercury halides (I). The above equation is differentiated in their writings ("ion") is the form:

Pb2 + (aq) + 2 NO3-(aq) + 2 H + (aq) + 2 Cl (aq) PbCl 2 (s) + 2 H + (aq) + 2 NO3-(aq)

Note that the stoichiometry of the balanced equation must be maintained. Pb (NO3) 2 is the DIS-associated with the Pb2 + and NO3 ~ 2 and 2 molecules of HCl dissociation reagents indicated in PDD-2 and H 2 + Cl. At this point, you can neutralize the ions that did not participate in Delta of the reaction. Note how the nitrate ions and hydrogen ions remain unchanged on both sides of the equation. Ions that do not change the chemical reaction called spectator ions and can be removed from the equation without destroying the right (as they are removed in exactly the same number on both sides!).

Pb2 + (aq) + 2 NO3-(aq) + 2 H + (aq) + 2 Cl (aq) PbCl 2 (s) + 2 H + (aq) + 2 NO3-(aq)

What remains is the net ionic equation showing that the chemical species involved in chemical processes:

PB2 + (aq) + 2 Cl ~ (aq) PbCl2 (s)

It is also possible to predict the effect of the net ionic equation is given only reagents. For example, suppose you had to determine the net ionic equation is obtained by mixing solutions of barium bromide and sodium:

BaBr2 (aq) + Na2SO4 (aq)?

One way to address this problem is to determine the ions in solution. Two representatives of the reactions of salts, which are completely dissociated in Ba2 +, Br-, Na + and SO42-. We know that the soluble barium bromide, but sodium ions or sulfate ions combine with barium ions to form an insoluble compound? Barium ions and sodium ions, both positively charged, they repel each other, so there is no provision made for training. Similarly, as the bromide and sulfate both have a negative charge, do not expect this combination to form composite. On the other hand, sulfate ions and barium ions could easily form barium sulfate. Now it's just a matter of point of view of solubility rules to see if the barium sul ¬ fate is soluble or not. Rule of the solubility of the sulfates are soluble except those of strontium, barium and lead, silver sulfate and calcium sulfate are partially soluble.

As you can see in these rules, barium sulfate is insoluble. The sodium ions combine with bromide ions to form com ¬ sodium bromide. Under the rules of solubility, sodium bromide are soluble in sodium salts are soluble, while, like most of the halides. Now we can write a complete balanced equation:

BaBr2 (aq) + Na2SO4 (aq) BaSO4 (s) + 2 NaBr (aq)

As before, the above equation be rewritten in the ionized form, showing the soluble species such as ions in solution:

Ba2 + (aq) + 2 Br ~ (aq) + 2 Na + (aq) + SO42-(aq) BaSO4 (s) + 2 Na + (aq) + 2 Br ~ (aq)

Then a cross in the spectator ions:

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

What remains is a balance, the net ionic equation:

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

We next consider the case when you mix a solution of magnesium chloride with copper (II) nitrate:

Cl2 (aq) + Cu (NO3) 2 (aq)?

As before, the reaction most likely ions to "change partners":

MgCl 2 (aq) + Cu (NO3) 2 (aq), Mg (NO3) 2 (?) + CuCl2 (?)

Under the rules of solubility in both Mg (NO3) 2 and CuCl 2 are soluble in water (all nitrates are soluble, most of the halides are soluble)

Cl2 (aq) + Cu (NO3) 2 (aq) Mg (NO3) 2 (aq) + CuCl2 (aq)

Today it is recognized that all four compounds are salts, and completely detached from the water:

Mg 2 + (aq) + 2 Cl ~ (aq) + Cu 2 + (aq) + 2 NO3 ~ (aq)

Mg 2 + (aq) + 2 NO3-(aq) + Cu2 + (aq) + 2 Cl (aq)

When the spectator ions are removed, we observe that all the reagent ions are former side exactly the same as in the product. This means you should forget all the spectator ions ions, leaving nothing behind! In fact, this is exactly what is happening in this "reaction." Because no precipitate or a weak electrolyte, our (good) provided for in this case would be "no reaction".

Acid-base reactions

Net ionic equations are often used on acid-base reactions as well. The key to writing successful on the net ionic equation for the acid-base reactions must be able to distinguish between strong and weak acid or base. The degree of dissociation of an acid is determined by the strength of the acid. For a strong acid dissociation is complete (100%), so we wanted to show acid dissociated into ions in the net ionic equation. In solution, weak acids dissociate to a lesser degree (usually <5%). Since weak acids exist in solution mainly undissoci-tion of molecules, they appear as neutral molecules (ie, not proven to be separated into ions) in the net ionic equation. For the same reason, weak bases also appear as undisso associate in solution. Strong bases are shown separated by metal cations and OH ~.

Organic acids generally contain C, H and O in its molecular structure and in particular a carboxylic acid (COOH or CO 2 H). All organic acids are low. An example would be benzoic acid or HC6H5CO2 C6H5COOH. Ammonia (NH3) and organic bases (usually derived from ammonia) are weak bases. For example methylamine, CH3NH2 is a weak base.

The products are generally produced by the acid-base reaction is a salt and water. You can write the formula of the salt, it helps to know that the salt cation is always going to be the bottom and the anion and the salt is always acid. (One way to remember this is considered one of the consonants and vowels in a single-tion of a platform and an anion of acid.)

Let's apply some rules to write net ionic equations for some of the acid-base reactions. What is the full equation, complete ionic and net ionic reaction of aqueous nitric acid (HNO3) with a potassium hydroxide (KOH)?

HNO3 (aq) + KOH (aq)?

Cations of the base is a potassium ion (K +) and anion of an acid is a nitrate ion (NO3 ~), so that the salt will form KNO3 in water and other products. Since all nitrate solu-tion, potassium nitrate is Aque ¬ organizational units, and water is the liquid phase. Com-plete and balanced equation is:

HNO3 (aq) + KOH (aq) KNO3 (aq) + H2O (l)

Under the rules of electrolyte behavior given above, HNO3, KOH and KNO3 all strong electrolytes and therefore dissociate completely. Water is in the "all other materials" category, it will be a nonelectrolyte and is not represented separated. The total ionic equation is:

H + (aq) + NO 3 ° (aq) + K + (aq) + OH ¯ (aq) K + (aq) + NO 3 ° (aq) + H2O (l)

We then remove the spectator ions (potassium nitrate):

H + (aq) + NO3 ° (aq) + K + (aq) + OH ° (aq) K + (aq) + NO3 ° (aq) + H2O (l)

What remains is the net ionic equation:

H + (aq) + OH ° (aq) H2O (l)

The same net ionic equation is the result of the reaction of a strong acid with strong base. (Try other examples make clear.)

Let's see what happens in response to weak acid strong base. Which is perfect, and ionic net ionic equations for the reaction of benzoic acid (HC6H5CO2) of sodium hydroxide?

The salt will be formed NaC6H5CO2 (Na + and acid substrate C6H5CO2 °). This salt is dissolved because of all the groups IA compounds are soluble. So, the equation is complete:

HC6H5CO2 (aq) + NaOH (aq) NaC6H5CO2 (aq) + H2O (l)

Since benzoic acid is a weak acid (it contains carbon in its molecular structure, and is not one of the six common strong acids), it should not be shown separately in the overall equation Net ionic or ionic . Both NaOH (metal hydroxide) and NaC6H5CO2 (salt) are strong electrolytes. The total ionic equation is:

HC6H5CO2 (aq) + Na + (aq) + OH ° (aq) Na + (aq) + C6H5CO2 ° (aq) + H2O (l)

In this case, sodium ions, the only spectator ions:

HC6H5CO2 (aq) + Na + (aq) + OH ° (aq) Na + (aq) + C6H5CO2 ° (aq) + H2O (l)

There remains Net ionic equation:

HC6H5CO2 (aq) + OH ° (aq) C6H5CO2 ° (aq) + H2O (l)

Finally, a look at the acid-base reaction involving a weak base and a strong acid. What is the full equation, complete ionic and net ionic reaction of HCl with methyl ¬ amine (CH3NH2), a weak organic base? Since hydroxide is not a part of the formula of the weak base, water will not be a product, in this case. The overall reaction is:

HCl (aq) + CH3NH2 (aq) CH3NH3Cl (aq)

HCl must be separated from each other, because it is a strong acid. Since methylamine is a weak base, it will be dissociated into the equation. Salts of organic bases are soluble (like NH4 +) and separated. Total ionic equation, the results are as follows:

H + (aq) + Cl ¯ (aq) + CH3NH2 (aq) + CH3NH3 (aq) + Cl ¯ (aq)

Only a spectator, at the time of the chloride ion:

H + (aq) + Cl (aq) + CH3NH2 (aq) + CH3NH3 (aq) + Cl (aq)

There remains Net ionic equation:

H + (aq) + CH3NH2 (aq) + CH3NH3 (aq)

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