Unit+4+-+Aqueous+Solutions

> > ex) Given 6.0 mol of HCl per liter of solution, a) how many moles of HCl are in 75 mL of the solution and b) what volume of acid is required to contain one mole of HCl? > To find this, just multiply the moles of ions by the molarity of solution. media type="youtube" key="g8jdCWC10vQ" height="349" width="425"
 * __Molarity: __ ** the concentration of a solute in a solution
 *  Formula: M = moles of solute/liters of solution[[image:acid-base_titration_experiment_a500455.jpg width="225" height="350" align="right"]]
 *  Symbol: [ ] (represents the molarity of a species in a solution)
 * ex) [HCl] means the concentration of ammonia in moles/liter
 * ex) The molarity of a solution containing 1.20 mol of substance in 2.50 L of solution
 * M = 1.20 mol/2.50 L
 * M = 0.480
 * answer: 0.480 M
 *  Molarity can be used to calculate moles of solute given volume of solution, or volume of solution give moles of solute
 * a) 75 x 1 L/1000 mL x 6.0 mol/1 L = .45 moles HCl
 * b) 1 mol HCl x 1 L/6.0 mol HCl = .17 L HCl
 *  Molarity can also be used to find the concentrations of ions in solutions.

**__Net Ionic Equations: __**  equations that show the reaction, excluding the spectator ions, which are in the solution before, during and after the reaction, but do not partake in the reaction itself. This type of equation is a chemical equation for a reaction involving ions in which only those species that actually react are included. > ex) <span style="font-family: 'Cambria','serif'; font-size: 14px;"> Fe3+ <span style="font-family: 'Cambria','serif';">(aq) + <span style="font-family: 'Cambria','serif'; font-size: 14px;">3Cl- <span style="font-family: 'Cambria','serif';">(aq) -> <span style="font-family: 'Cambria','serif'; font-size: 14px;">FeCl3 <span style="font-family: 'Cambria','serif';">(s)
 * <span style="font-family: 'Cambria','serif';">Sometimes when solutions are mixed, **a precipitate ** is formed, which is included in the net ionic equation

<span style="font-family: 'Cambria','serif';"> > ex) In a reaction mixing Ba(NO3)2 and Na2CO3, BaCO3 would form a precipitate and NaNO3 would just dissolve. >> <span style="font-family: 'Cambria','serif'; font-size: 14px;">Ba2+ <span style="font-family: 'Cambria','serif';">(aq) + Co32-(aq) -> <span style="font-family: 'Cambria','serif'; font-size: 14px;">BaCO3(s)
 * <span style="font-family: 'Cambria','serif';">A chart can be used to determine whether or not a precipitate will be formed, based on the solubility of the substances (another chart is on text pg 80)
 * <span style="font-family: 'Cambria','serif';">So the net ionic equation would be:

media type="youtube" key="8RmVwz2fNGc" height="349" width="425"

<span style="color: black; font-family: Cambria,serif; font-size: 10pt;"> __**Acid**__: forms the H+ ion in water

<span style="font-family: Cambria,serif;">//Strong Acids//: completely ionize in water, forming the H+ ion; HCl, HBr, HI, HNO3, HClO4, H2SO4 <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">//Weak Acids:// only partially ionize <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Ex) Chemical reaction of a strong acid:

<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">HCl (aq) -> H+ (aq) + Cl- (aq)

<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">*The reaction goes to completion and there is no HCl left in the solution <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">*With a weak acid, it would be partially ionized, so the solution would contain both the ions and the acid in the solution

<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">**__ Base __ :** forms the OH- ion in water

<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">//Strong Bases:// completely ionize in water, forming the OH- ion; LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2 <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">//Weak Bases//: only partially ionize <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Ex) Chemical reaction of a strong base:

<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Ca(OH)2 (s) -> Ca2+ (aq) + 2OH- (aq)

<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">*The reaction goes to completion and there is no Ca(OH)2 left in the solution <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">*With a weak base, it would be partially ionized, so the solution would contain both the ions and the acid in the solution

__**<span style="font-family: Cambria,serif;">Acid-Base Reactions: **__

<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">1) //Strong acid & strong base:// They completely ionize and become neutral, forming water and salt. <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Ex) A reaction of HCl + Ca(OH)2 <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Net Ionic Equation: H+ (aq) + (OH)- (aq) -> H20 (l) <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">*The Ca2+ and Cl- combine and dissolve

<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">2) //<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Weak acid & strong base: // <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Ex) A reaction of Ca(OH)2 + HF <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Net Ionic Equation: OH- (aq) + HF (aq) -> F- (aq) + H20

<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">3) //<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Strong acid & weak base: // <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Ex) A reaction of HCl + NH3 <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Net Ionic Equation: H+ (aq) + NH3 (aq) -> NH4+ (aq)

<span style="font-family: Cambria,serif;"> __**Titration**__ : The method used to determine the point at which an equation is complete; determines strength


 * <span style="font-family: Cambria,serif;">Formula: MAVA = MBVB
 * <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Indications are used, such as phenolphthalein
 * <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">The equivalence point is when the solution is neutral, so the OH- and H+ concentrations are equal
 * <span style="font-family: Cambria,serif;">ALWAYS write net ionic equations out first

<span style="color: black; font-family: Cambria,serif; font-size: 90%; vertical-align: baseline;">Ex) Given: 25.00 mL of 0.0800 M Ca(OH)2 is required to react with 10.00 mL of HCl. Calculate the molarity of HCl. <span style="color: black; font-family: Cambria,serif; font-size: 90%; line-height: 80%; vertical-align: baseline;">OH- (aq) + H+ (aq) -> H2O <span style="color: black; font-family: Cambria,serif; font-size: 90%; line-height: 80%; margin-bottom: 0pt; margin-left: 0in; margin-right: 0in; margin-top: 0in; vertical-align: baseline;">0.0800 M = X __<span style="color: black; font-family: Cambria,serif; font-size: 90%; line-height: 80%; margin-bottom: 0pt; margin-left: 34.55pt; margin-right: 0in; margin-top: 0in; text-indent: -0.35in; vertical-align: baseline;"> 0.025 L __ __<span style="color: black; font-family: Cambria,serif; font-size: 90%; line-height: 80%; margin-bottom: 0pt; margin-left: 0in; margin-right: 0in; margin-top: 0in; vertical-align: baseline;">X = 0.002 x 2 0.004 mol OH- __ __<span style="color: black; font-family: Cambria,serif; font-size: 90%; line-height: 80%; margin-bottom: 0pt; margin-left: 0in; margin-right: 0in; margin-top: 0in; vertical-align: baseline;">H+ = 0.004 mol H+ __ __<span style="color: black; font-family: Cambria,serif; font-size: 90%; line-height: 80%; margin-bottom: 0pt; margin-left: 0in; margin-right: 0in; margin-top: 0in; vertical-align: baseline;">X = __0.004 __<span style="color: black; font-family: Cambria,serif; font-size: 90%; line-height: 80%; margin-bottom: 0pt; margin-left: 0in; margin-right: 0in; margin-top: 0in; vertical-align: baseline;"> 0.01 __ __<span style="font-size: 90%; line-height: 80%; margin-bottom: 0pt; margin-left: 0in; margin-right: 0in; margin-top: 0in; vertical-align: baseline;">**<span style="color: black; font-family: 'Cambria','serif'; line-height: 80%;">X = 0.400 mol/L of HCl ** __


 * Redox Reactions :**__ involve the transfer of electrons between 2 species in an aqueous solution; there is never a net change in the number of electrons in the reaction
 * //<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Reduced: //<span style="color: black; font-family: Cambria,serif; font-size: 10pt;"> The species gaining the electrons
 * //<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Oxidized: //<span style="font-family: Cambria,serif; font-size: 10pt;"> The species lost the electrons
 * //<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Reducing agent: //<span style="font-family: Cambria,serif; font-size: 10pt;"> The ion or molecule that donates electrons; it reduces the other species when the reaction occurs and is therefore oxidized.
 * //<span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Oxidizing agent: //<span style="font-family: Cambria,serif; font-size: 10pt;"> The ion or molecule that accepts electrons; it brings out the oxidation of another species and is therefore reduced.

<span style="color: black; font-family: 'Cambria','serif'; font-size: 90%; line-height: 80%;">

__**<span style="font-family: Cambria,serif; font-size: 10pt;">Oxidation Numbers: **__ <span style="font-family: Cambria,serif; font-size: 10pt;">

__**<span style="font-family: Cambria,serif;">Balancing Half Reactions: **__ <span style="font-family: Cambria,serif; font-size: 10pt;">1) <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Balance atoms <span style="font-family: Cambria,serif; font-size: 10pt;">2) <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Balance oxidation number (add electrons) <span style="font-family: Cambria,serif; font-size: 10pt;">3) <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Balance overall charge (add H+ ions in acidic solution or OH- ions in basic solution) <span style="font-family: Cambria,serif; font-size: 10pt;">4) <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Balance hydrogen (add H20 molecules) <span style="font-family: Cambria,serif; font-size: 10pt;">5) <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">Check to make sure the oxygen are balanced

__**<span style="color: black; font-family: Cambria,serif; font-size: 10pt;"> Balancing overall redox reactions : **__ <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">1) Split the equation into the reduction half and oxidation half <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">2) Balance one half-equation <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">3) Balance the other <span style="color: black; font-family: Cambria,serif; font-size: 10pt;">4) <span style="font-family: Cambria,serif;">Combine the half-equations so the electrons cancel out