Aqueous Reactions and Solution Chemistry Written by JoAnne L. Swanson University of Central Florida

DEFINITIONS: Solution – a homogenous mixture of two or more substances. The parts of a solution are the solvent and solutes. Solvent – the substance in greater quantity that the other substances are dissolved in Solute – the substances in lesser quantity that are dissolved in the solvent Hydration – the process whereby water molecules surround solute particles and cause them to dissociate into ions in solution Dipoles – positive and negative poles on a polar molecule caused by an unequal distribution of electrons due to differences in electronegativities of the atoms in the molecule

Electrolytes – substances that when molten or aqueous can conduct electricity because they completely or almost completely dissociate into ions which can carry electrons. Non electrolytes – substances that are molecular in aqueous solution and therefore cannot conduct electricity Weak electrolytes – substances that only partially dissociate into ions and therefore can conduct electricity only slightly electrical current – the flow of electrons.

Strong Electrolytes – include strong acids, strong bases, and soluble ionic compounds. The strong acids are HCl, HClO3, HClO4, HBR, HI, HNO3, H2SO4 These are considered strong acids because they completely dissociate into ions in solution; HCl  H+ + Cl- The strong bases are group IA metal hydroxides and the heavy group IIA metal hydroxides (Ca, Sr, Ba) NaOH  Na+(aq) + OH-(aq) Weak Electrolytes are the other acids and weak bases. Weak bases can be identified as ammonia, NH3 and amines; organic compounds with nitrogen such as CH3NH2 Non Electrolytes are molecular (covalent) compounds. Sugars (like C6H12O6 ), alcohols (CnHxOH), and hydrocarbons are examples of some non electrolytes.

PRECIPITATION REACTIONS – ionic and metathesis rxns Reactions between two soluble ionic compounds in which a double replacement occurs due to the strong electrical attraction between oppositely charged ions. They come together and form a solid ppt. YOU MUST LEARN THE SOLUBILITY RULES IN ORDER TO PREDICT WHETHER OR NOT A PPT REACTION OCCURS. (NH4)3PO4 (aq) + BaNO3 (aq)  NH4NO3 (aq) + Ba3(PO4)2

IONIC EQUATIONS: In a chemical equation that involves aqueous ions, there are ions in the reaction that do not change, or react so we leave them out of the final written equation. The ions that do not react are called spectator ions. The shortened version of the chemical equation is called a NET IONIC EQUATION. Ex. CaCl2 (aq) + Na2CO3 (aq)  NaCl (aq) + CaCO3 IN REALITY IS…. Ca+2(aq) + 2Cl-(aq) + 2Na+ (aq) + CO32- (aq)  2Na+ (aq) + 2Cl-(aq) + CaCO3 Which is written as…. Ca+2(aq) + CO32- (aq)  CaCO3

Some ionic equations will not have a ppt but will have a molecular product or a gas. The neutralization reaction of a strong acid and strong base: HCl +NaOH  HOH (l) + NaCl THE MOLECULAR EQUATION H+ (aq) +Cl- (aq) + Na+ (aq) + OH- (aq)  HOH + Na+ (aq) + Cl- (aq) H+ (aq) + OH- (aq)  HOH (l) THE NET IONIC EQUATION OR A SUBSTANCE MAY JUST GO FROM SOLID TO AQ. Mg(OH)2(s) + HCl  Mg+2(aq) + HOH (l) NET IONIC EQUAT. REMEMBER THAT NON ELECTROLYTES ARE NOT DISSOCIATED WHEN AQUEOUS… HF (aq) + NaOH (aq)  HOH (l) + NaF (aq) HF (aq) + Na+(aq) + OH- (aq)  HOH (l) + Na+(aq) + F- (aq) HF (aq) + OH- (aq)  HOH (l) + F- (aq) THE NET IONIC EQUAT.

CARBONATE, HYDROGEN CARBONATE AND SULFIDE IONS PRODUCE A GAS WHEN REACTED WITH ACID: HCl (aq) + CaS (aq)  H2S + CaCl2(aq) THE NET IONIC EQUATION IS: H+ (aq) + S-2 (aq)  H2S (g) K2CO3 (aq) + HCl (aq)  KCl (aq) + HOH (l) + CO2 THE NET IONIC EQUATION: CO32- (aq) + H+ (aq)  HOH (l) + CO2

Oxidation Reduction Reactions: aka REDOX – where electrons are transferred between substances. In order to know if a redox rxn has occurred, there are two pieces of information that you need to be aware of: THE ACTIVITY SERIES – TELLS WHICH SPECIES IS MORE EASILY OXIDIZED. THE COMMON OXIDATION STATES- IF YOU KNOW THE COMMON OXIDATION STATES, YOU WILL BE ABLE TO PREDICT OXIDATION STATES OF SUBSTANCES THAT MAY NOT BE SO COMMON.

REMEMBER: LOSS OF ELECTRONS IS OXIDIZED “LEO” GAIN OF ELECTRONS IS REDUCED “GER” WHEN A SUBSTANCE IS ITSELF, OXIDIZED, IT IS THE REDUCING AGENT. WHEN A SUBSTANCE IS ITSELF, REDUCED IT IS THE OXIDIZING AGENT.

Rules for oxidation states will be on a separate button to print out Rules for oxidation states will be on a separate button to print out. It made the download much too slow.

REDOX REACTIONS ARE BROKEN INTO “HALF REACTIONS” THESE SHOW SEPARATE EQUATIONS FOR THE SUBSTANCE OXIDIZED AND THE SUBSTANCE REDUCED. Al (s) + HCl (aq)  AlCl3 (aq) + H2(g) Al0  Al+3 + 3e- THE Al WAS OXIDIZED (LEO) 2H+ + 2e-  H20 THE H+ WAS REDUCED (GER) WHEN ELECTRONS ARE GAINED THEY ARE WRITTEN ON THE REACTANT SIDE OF THE HALF REACTION. WHEN ELECTRONS ARE LOST THEY ARE WRITTEN ON THE PRODUCT SIDE OF THE HALF REACTION.

THE ACTIVITY SERIES: See table 4.5 on page 131 in text Ex. Au (s) + CuCl2 (aq)  NR Ex. Fe (s) + CuCl2 (aq)  Cu (s) + FeCl2 (aq) THE REASON EX. 1 SHOWS ‘NR’ (NO REACTION) IS BECAUSE GOLD IS NOT AS EASILY OXIDIZED AS COPPER. THEREFORE, GOLD WILL NOT GO TO Au+3 AND COPPER WILL STAY AS Cu+2 IN EX. 2, Fe IS MORE EASILY OXIDIZED THAN Cu SO THE Fe GOES TO Fe+2 AND THE COPPER GOES TO Cu0 YOU SHOULD BE ABLE TO WRITE A NET IONIC EQUATION FOR REDOX REACTIONS ALSO. (DONE THE SAME WAY)

CONCENTRATIONS OF SOLUTIONS: Molarity, molality, and Normality Molarity – expressed as moles of solute / liters of solution Symbol is: M units are : moles / L Molality – expressed as moles of solute / Kg of solvent Symbol is: m units are: moles / Kg Normality – expressed as equivalents / mole of compound Symbol is: N units are equiv. / mole

Examples follow: Ex. 1 If 2.50 g of NaOH are dissolved in 125 mL of water, what is the molarity of the solution? To solve this we need to convert the grams to moles and then convert the mL of solvent to L of solution (which is solvent + solute) 2.50g NaOH x 1 mole = 0.0625 moles 40.0 g Since water has a density of 1.00 g / mL we see that 125mL = 125 g 125 g water + 2.50 g NaOH = 127.5 g solution and since the solution is dilute, we can use the density of water to say that the solution is about 1 g / mL and therefore, 127.5 g = 127.5 mL = 0.1275 L soln. Now we have 0.0625 moles / 0.1275 L The molarity = 0.490 M

So in summary, if given mL of solvent, you have to convert the solvent to solution. The grams of solute must be converted to moles and then just divide. Ex. 2 How many grams of KOH are required to make 750. mL of a 0.150 M solution? Here we start with a given: 750. mL The 0.150 M is a conversion factor: 0.150 mol / L The goal is to arrive at grams of KOH. (first convert the 750 mL to L) 0.750 L soln x 0.150 mol KOH x 56.1g = 6.31 g KOH 1 L soln 1 mol

Ex. 3. What is the molarity of a solution obtained by dissolving 17.3 g of iron(III) nitrate in enough water to give 250. mL of solution? Change the grams to moles Change the mL to L (its already solution ) Divide the grams by the L 17.3 g Fe(NO3)3 x 1 mol = 0.0715 mole 241.88g 0.0715 mol / 0.250 L = 0.286 M

Ex. 4. How would you prepare 500. mL of a 0.15 M NaCl soln? The given is 500. mL of soln. 0.15 M NaCl soln is the conversion factor. 0.500L soln x 0.15 mol x 58.5 g = 4.39 g NaCl 1 L 1 mol To prepare the solution, 4.39 g of NaCl would be weighed out and diluted up to 500 mL with water. NOTE THAT YOU DO NOT ADD 500 ML OF WATER. THAT WOULD BE 500 ML OF SOLVENT. IF YOU DILUTE THE SOLUTE UP TO THE 500 ML, IT IS 500 ML OF SOLUTION.

Molar Ion Concentrations: Ex. 5 What is the molar concentration of each of the ions in a 0.200 M aqueous solution of Ba(NO3)2? There are 3types of ions in the solution. One Ba+2 ion and 2 NO3- ions. Therefore the solution is 0.200 M in Ba+2 ions and 2 x 0.200 M in NO3- ions, or 0.400 M NO3-

DILUTION PROBLEMS: VC x CC = Vd x Cd WHERE V = VOLUME, C = CONCENTRATION, c = concentrated, d = diluted EX. 6. How would one prepare 500. mL of 0.050 M NaOH solution from a 1.50 M solution? What we are trying to find out is how much of the concentrated solution has to be diluted to make the desired amount of diluted solution. Vd x Cd = Cc x Vc 500. x 0.050 = 1.50 x (Vc) We would need to measure out 16.6 mL of 1.50 M soln and dilute to 500. mL

Dilution problems do not require units to be converted as long as the same units are used on both sides of the equation. Notice how we did not convert the 500. mL to L even though we multiplied by moles / Liters. This is because the numeric ratios work out the same. RECOGNIZE DILUTION PROBLEMS AS HAVING TWO VOLUMES AND ONE CONCENTRATION OF THE SAME SUBSTANCE OR TWO CONCENTRATIONS AND ONE VOLUME OF THE SAME SUBSTANCE. NO CHEMICAL EQUATION IS NEEDED FOR DILUTIONS.

TITRATIONS: RECOGNIZE TITRATION PROBLEMS AS THOSE HAVING TWO CONCENTRATIONS AND ONE VOLUME OF TWO DIFFERENT SUBSTANCES OR TWO VOLUMES AND ONE CONCENTRATION OF TWO DIFFERENT SUBSTANCES. HERE WE TRY TO DETERMINE THE CONCENTRATION OF A SUBSTANCE BY NEUTRALIZING IT (TITRATING IT) WITH ANOTHER SUBSTANCE OR WE TRY TO FIND OUT WHAT VOLUME OF A SUBSTANCE IS NEEDED TO NEUTRALIZE (OR TITRATE) A SUBSTANCE. A BALANCED CHEMICAL EQUATION IS NEEDED.

WHEN SUBSTANCES REACT, WE CAN ADD A CHEMICAL INDICATOR TO THE REACTION VESSEL TO DETERMINE WHEN THE REACTION HAS COME TO COMPLETION. A CHEMICAL INDICATOR IS A SUBSTANCE THAT CHANGES COLOR WHEN THE REACTION REACHES WHAT IS CALLED THE ENDPOINT (ITS COMPLETION). WHEN TITRATING, SMALL AMOUNTS OF ONE REACTANT ARE ADDED TO ANOTHER REACTANT, USING A BURET. WHEN JUST ENOUGH REACTANT IS ADDED, THE INDICATOR WILL CHANGE COLOR. WE CAN THEN DETERMINE HOW MUCH OF THE ONE REACTANT IT TOOK TO REACT COMPLETELY WITH THE OTHER REACTANT. FROM THIS WE CAN CALCULATE THE CONCENTRATION OF THE OTHER REACTANT.

IN ORDER TO DO THE STOICHIOMETRY INVOLVED IN TITRATION PROBLEMS, A BALANCED CHEMICAL EQUATION IS NEEDED. EX. 7 What is the molarity of an H2SO4 soln if 35.0 mL of it were neutralized by 55.0 mL of a 0.51 M NaOH solution? TRICK – when a titration problem has two volumes (instead of two concentrations) , start the problem with the volume of the substance that you know the concentration of, over the volume of the unknown substance. Convert both to liters first. 0.0550 L NaOH x 0.51 mol NaOH x 1 mol H2SO4 = 0.401 M H2SO4 0.0350 L H2SO4 1 L 2 mol NaOH

Ex. 8. A titration with two concentrations and one volume Ex. 8 A titration with two concentrations and one volume. Start with the given, not with a conversion factor. What volume of 0.150 M NaOH is needed to completely neutralize 65.0 mL of 0.0100 M H3PO4 soln.? Note that the molarities are conversion factors. The given is 65.0 mL . 0.0650 L H3PO4 x 0.0100 mol x 3 mol NaOH x 1 L = 1 L 1 mol H3PO4 0.150 mol NaOH = 0.0130 L NaOH soln.

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