Solubility Rules and Net Ionic Equations

Solubility Rules and Net Ionic Equations
PO43– Na+ 2Ca2+ 3Cl– Al3+ Solubility Rules and Net Ionic Equations

Dissociation of Ionic Compounds in Water
Ionic compounds dissociate into positive ions (cations) and negative ions (anions) when they form a solution. Dissociation equations show the number of each cation and anion produced by a compound when it dissolves in an aqueous solution. NaCl (s) → Pb(NO3)2 (s) → FeCl3 (s) → Note: dissociation equations must be balanced.

Factors Affecting Solubility in Water
1) The Charge on an Ion as the ionic charge increases, attraction between ions increases therefore as charge increases, solubility ________ e.g. chloride salts are __________ soluble than phosphate salts 2) Ion Size smaller ions contain shorter bonds therefore these bonds are stronger than larger ions with the same charge as ion size increases, solubility ________________

Solution Chemistry When mixing 2 aqueous solutions there are 2 possible outcomes: 1. 2.

Double Displacement Reactions Can Be Recognized in 3 Ways:
The formation of a precipitate The formation of a gas The formation of water

Not all ionic compounds dissolve!
Instead of doing experiments all the time to see which ones will dissolve, we use The solubility rules.

Solubility Rules Soluble Insoluble NO3- NH4+, Li+, Na+, K+, Rb+, Cs+
Cl-, Br-, I- except with Ag+, Pb2+, Hg+. SO42- except with Ca2+, Sr2+, Ba2+, Ra2+, Pb2+. CO32-, PO43-, OH- except with NH4+, Li+, Na+, K+, Rb+, Cs+

Which of the following are soluble in water? SrSO4 NaNO3 PbCl2
Not soluble Soluble Not soluble

Precipitation Reactions
When a solid doesn’t dissolve it is called insoluble. A solid that forms when two solutions are mixed is called a precipitate. By definition: -soluble = >1g / 100mL of water - slightly soluble = 0.1g-1g/100mL H2O - insoluble = <0.1g / 100mL H2O low high “insoluble” partial “very soluble” (< 0.1 g/100mL) solubility (> 1 g/100mL)

A couple definitions.. A Total Ionic Equation is a reaction equation which shows all of the ions involved in the reaction. Spectator Ions are ions, which are not important to the net result of the reaction and are therefore passive onlookers. A Net Ionic Equation is a total ionic equation without the spectator ions. It shows only the ions, which are involved in the reaction.

Predict the products of the following reaction: (if no solid precipitate is formed, there is no reaction) Pb(NO3)2(aq) + KI (aq) → Pb(NO3)2(aq) + 2KI (aq) → PbI2(s) + 2KNO3 (aq) We know it is a solid precipitate because it is insoluble according to the solubility rules. Total Ionic Equation (shows all ions involved) Pb2+(aq)+ 2NO3-(aq) + 2K+(aq) + 2I-(aq) → PbI2(s) + 2K+(aq) +2NO3-(aq)

A picture of the reaction:
Because K+ and NO3- remain dissolved, they are called spectator ions and are not included in the net ionic equation. NO3- NO3- Pb2+ K+ I- K+ NO3- PbI2(s) Net Ionic Equation Pb2+ (aq) + 2I-(aq) → PbI2(s)

Predicting Ionic Solubility
Use a solubility chart to predict the solubility of the following compounds in water. State whether they are soluble or insoluble in water. Ca(NO3)2 CuCl2 KBr (NH4)2CO3 Fe(NO3)3 Ca(OH)2 NH4CH3COO Pb(NO3)4 Ca3(PO4)2 CuCl KCH3COO PbCO3 BaSO4 AgOH Ag2SO4 Cr(OH)3

Use a solubility chart to predict the solubility of the following compounds in water. State whether they are soluble or insoluble in water. Ca(NO3)2 soluble CuCl2 KBr (NH4)2CO3 Fe(NO3)3 Ca(OH)2 NH4CH3COO Pb(NO3)4 Ca3(PO4)2 CuCl KCH3COO PbCO3 BaSO4 AgOH Ag2SO4 Cr(OH)3

Use a solubility chart to predict the solubility of the following compounds in water. State whether they are soluble or insoluble in water. Ca(NO3)2 soluble CuCl2 KBr (NH4)2CO3 Fe(NO3)3 Ca(OH)2 NH4CH3COO Pb(NO3)4 Ca3(PO4)2 insoluble CuCl KCH3COO PbCO3 BaSO4 AgOH Ag2SO4 Cr(OH)3

Use a solubility chart to predict the solubility of the following compounds in water. State whether they are soluble or insoluble in water. Ca(NO3)2 soluble CuCl2 KBr (NH4)2CO3 Fe(NO3)3 Ca(OH)2 insoluble NH4CH3COO Pb(NO3)4 Ca3(PO4)2 CuCl KCH3COO PbCO3 BaSO4 AgOH Ag2SO4 Cr(OH)3

General Solubility Rules (Time savers)
Nitrates (NO31-) soluble with all cations. Ammonium (NH41+) soluble with all anions. Alkali Metals (Li1+, Na1+, K1+,…)

Predicting the Precipitate that Forms in a Chemical Reaction
In a chemical reaction between two Ionic solutions, a precipitate may form. If a precipitate forms, one of the products of the reaction is the precipitate, in other words, it is insoluble and it is a solid. We can predict which product forms the precipitate if we have a solubility chart handy.

For example, When a solution of Potassium iodide [KI (aq)] reacts with a solution of Lead (II) nitrate [Pb(NO3)2 (aq)], a yellow precipitate forms. This reaction can be described in words: Potassium iodide solution reacts with Lead (II) nitrate solution to form Potassium nitrate with Lead (II) iodide. One of the products is insoluble (the yellow precipitate)

First Step: Write out the Balanced Reaction Equation: 2KI (aq) + 1Pb(NO3)2(aq)  2KNO3(?) + 1PbI2(?) Second step: Determine which product is insoluble: Name of product Formula Solubility Potassium nitrate Lead (II) iodide

First Step: Write out the Balanced Reaction Equation: 2KI (aq) + 1Pb(NO3)2(aq)  2KNO3(?) + 1PbI2(?) Second step: Determine which product is insoluble: Name of product Formula Solubility Potassium nitrate KNO3 Lead (II) iodide PbI2

First Step: Write out the Balanced Reaction Equation: 2KI (aq) + 1Pb(NO3)2(aq)  2KNO3(?) + 1PbI2(?) Second step: Determine which product is insoluble: Name of product Formula Solubility Potassium nitrate KNO3 Soluble - ie (aq) Lead (II) iodide PbI2

First Step: Write out the Balanced Reaction Equation: 2KI (aq) + 1Pb(NO3)2(aq)  2KNO3(?) + 1PbI2(?) Second step: Determine which product is insoluble: Name of product Formula Solubility Potassium nitrate KNO3 Soluble - ie (aq) Lead (II) iodide PbI2 Insoluble - ie (s)

The Total Ionic Equation shows all of the ions present in the two solutions, and which ones form the precipitate. When writing this equation, you must balance the atoms and the charges, you must also indicate which species are (aq) and which, if any forms the precipitate (s). 2K1+(aq) + 2I1-(aq)

The Total Ionic Equation shows all of the ions present in the two solutions, and which ones form the precipitate. When writing this equation, you must balance the atoms and the charges, you must also indicate which species are (aq) and which, if any forms the precipitate (s). 2K1+(aq) + 2I1-(aq) + 1Pb2+(aq) + 2NO31- (aq) 

The Total Ionic Equation shows all of the ions present in the two solutions, and which ones form the precipitate. When writing this equation, you must balance the atoms and the charges, you must also indicate which species are (aq) and which, if any forms the precipitate (s). 2K1+(aq) + 2I1-(aq) + 1Pb2+(aq) + 2NO31- (aq)  2K1+(aq) + 2NO31-(aq)

The Total Ionic Equation shows all of the ions present in the two solutions, and which ones form the precipitate. When writing this equation, you must balance the atoms and the charges, you must also indicate which species are (aq) and which, if any forms the precipitate (s). 2K1+(aq) + 2I1-(aq) + 1Pb2+(aq) + 2NO31- (aq)  2K1+(aq) + 2NO31-(aq) + 1PbI2(s)

The Net Ionic Equation demonstrates only the ions that form the precipitate. To show the Net Ionic Equation, you get rid of Spectator Ions – ions that do not take part in the reaction (ie., they appear the same on the reactant side as they do on the product side.) 2K1+(aq) + 2I1-(aq) + 1Pb2+(aq) + 2NO31- (aq)  2K1+(aq) + 2NO31-(aq) + 1PbI2(s)

The Net Ionic Equation demonstrates only the ions that form the precipitate. To show the Net Ionic Equation, you get rid of Spectator Ions – ions that do not take part in the reaction (ie., they appear the same on the reactant side as they do on the product side.) Simplified and rearranged: 1Pb2+(aq) + 2I1-(aq)  1PbI2(s)

Summary - Big Ideas Some double displacement reactions form precipitates. Precipitates form in double displacement reactions between two ionic solutions. Precipitates are insoluble ionic compounds, and do not dissolve. A net ionic equation demonstrates only the ions that react.

Net Ionic Equation Write the balanced chemical and net ionic equation for: Na2CO3(aq) + CaCl2(aq) → Step 1: Write the balanced chemical equation Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2NaCl (aq) Insoluble precipitate Step 2: Remove the spectator ions (those that are soluble). Ca2+(aq) + CO32-(aq) → CaCO3(s)

Write the total and net ionic equation for when sodium chloride reacts with silver nitrate
Step 1: Write the chemical equation and indicate whether or not a precipitate is formed. NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s) Step 2: Write the chemical equation with the soluble ionic compounds in their dissociated ions. (Make sure all charges are indicated and the reaction is balanced.) Na+(aq)+ Cl-(aq)+ Ag+(aq) +NO3- (aq)→AgCl(s) + Na+(aq) +NO3- (aq)

Cl-(aq)+ Ag+(aq) →AgCl(s)
Step 3: Identify any spectator ions (ions which are the same in the reactants and products) and eliminate them. Na+(aq)+ Cl-(aq)+ Ag+(aq) +NO3- (aq)→AgCl(s) + Na+(aq) +NO3- (aq) Step 4: Rewrite the total ionic equation without the spectator ions to get the net ionic equation. Ensure the reaction is balanced. Cl-(aq)+ Ag+(aq) →AgCl(s)

Solubility Rules and Net Ionic Equations

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