Chapter 4 Types of Chemical Reactions and Solution Stoichiometry

Parts of solutions  Solution – homogeneous mixture  Solute – part that dissolves  Solvent – causes the dissolving  Soluble – can be dissolved  Miscible – liquids dissolve in each other

Saturation of Solutions

Aqueous Solutions Dissolved in water Water is a polar molecule The oxygen atoms have a partial negative charge The hydrogen atoms have a partial positive charge The angle is 105 o

Hydration The process of breaking apart ions of a salt. The “+” end of water attracts the anion The “-” end of water attracts the cation

Solubility The ability to dissolve in a given amount of water Usually g/100mL Varies greatly Depends upon ion attraction Will dissolve nonionic substances if they have polar bonds

Electrolytes Electrical current through a substance Ions that are dissolved can move Solutions are classified three ways

Types of Solutions Strong electrolytes –Completely ionized when dissolved in water many ions – conduct well  Weak electrolytes – partially fall apart into ions few ions- conduct electricity slightly  Non-electrolytes – don’t fall apart no ions – don’t conduct electricity

Types of solutions continued Acids – form H + ion when dissolved Strong acids fall apart completely H 2 SO 4 HNO 3 HCl HBr HI HClO 4  Weak acids – do not dissociate completely  Bases – form OH - when dissolved  Strong bases – KOH NaOH

Dissociation Acids HCl  H + (aq) + Cl - (aq) HNO 3  H + (aq) + NO 3 - (aq) H 2 SO 4  H + (aq) + HSO 4 - (aq)

Strong bases NaOH  Na + (aq) + OH - (aq) KOH  K + (aq) + OH - (aq)

Nonelectrolytes Dissolve in water but do not produce any ions Example is ethanol (C 2 H 5 OH) molecules disperse in the water but doesn’t conduct electricity

Composition of Solutions Concentration 1. Molarity (M) – moles of solute per volume of solution in liters 2. M = moles of solute liters of solution

Preparation of Molar Solutions Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution g HCl 1 1 mole HCl 36.5 g HCl = mole HCl 26.8 mL 1 1L 1000 mL =.0268 L M = mole HCl.0268 L = 1.60 M HCl

Concentrations of Ions Give the concentration of each type of ion in 0.50 M Co(NO 3 ) 2 Co(NO 3 ) 2 (s) Co +2 (aq) + 2NO 3 - (aq) Co +2 1 x 0.50 M = 0.50 M Co +2 NO x 0.50 M = 1.0 M NO 3 -

Calculate the number of moles of Cl - ions in 1.75 L of 1.0 x M ZnCl 2. ZnCl 2  Zn +2 (aq) + 2Cl - (aq) 2 x 1.0 x = 2.0 x M Cl L2.0 x mole Cl - L = 3.5 x mole Cl -

Standard solution – a solution where the concentration is accurately known How much solid K 2 Cr 2 O 7 must by weighted out to make a solution? A chemist needs 1.00 L of an aqueous M K 2 Cr 2 O 7 solution.

Dilution 1.Water is added to achieve a particular M 2.Moles of solute after = moles of solute before 3. M 1 V 1 =M 2 V 2 What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H 2 SO 4 solution?

Types of Solution Reactions  Precipitation Reactions 1.A solid forms from two solutions 2. Precipitate – the insoluble solid KNO 3 (aq) + BaCl 2 (aq) 

Three Types of Equations Used to Describe Reactions in a Solution The formula equation gives the overall reaction. The complete ionic equation represents as ions all the reactants and products that are strong electrolytes. The net ionic equation includes only those ions that undergo a change.

Write the formula equation, complete ionic equation, and the net ionic equation for KCl (aq) + AgNO 3 (aq) 

Stoichiometry of Precipitation Reactions Calculate the mass of solid NaCl that must be added to 1.50 L of a M AgNO 3 solution to precipitate all the Ag + ions in the form of AgCl.

When aqueous solutions of Na 2 SO 4 and Pb(NO 3 ) 2 are mixed, PbSO 4 formed when 1.25 L of M Pb(NO 3 ) 2 and 2.00 L of M Na 2 SO 4 are mixed.

Acid-Base Reactions Acid is a proton donor (H + ) Base is a proton acceptor usually OH — accepts a H +  H + (aq) + OH − (aq) 

Cations are surrounded and bound by water molecules protons are also solvated by water molecules Two ways to show this  H + (aq)  H 3 O + (aq) – hydronium ion

Types of acid donors Monoprotic HCl, HNO 3 donates ____ H +  Diprotic H 2 SO 4 donates ____ H +  Triprotic H 3 PO 4 donates ____ H +

Neutralization Reactions What volume of a M HCl solution is needed to neutralize 25.0 mL of M NaOH? HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) H + (aq) + OH − (aq)  H 2 O (l).0250 L mole OH − = L NaOH

Acid-Base Titrations Volumetric analysis Process of determining the amount of a substance by titration Titration Process of delivering solutions to one another Equivalence point (stoichiometric point) Point at which the titration has occurred Endpoint Point at which the indicator changes color

What volume of M HCl, in milliliters, is required to titrate 1.33 g of NaOH to the equivalence point?

Oxidation-Reduction Reactions Electrons are transferred Spontaneous redox rxns can transfer energy Electrons (electricity) Heat LEO the lion says GER

Na + Cl 2  2NaCl oxidation

0 Cl + e __  Cl reduction

Rules for Assigning Oxidation Numbers Rules 1 & 2 1. The oxidation number of any uncombined element is zero 2. The oxidation number of a monatomic ion equals its charge 3. The oxidation number of oxygen in compounds is The oxidation number of hydrogen in compounds is +1

5. The sum of the oxidation numbers in the formula of a compound is The sum of the oxidation numbers in the formula of a polyatomic ion is equal to its charge ex. NO 3 — SO 4 2—

Reducing Agents and Oxidizing Agents The substance reduced is the oxidizing agent The substance oxidized is the reducing agent 0+1 Na  Na + e -- Sodium is oxidized – it is the reducing agent Cl + e --  Cl Chlorine is reduced – it is the oxidizing agent

Can you identify what is being oxidized and what is being reduced? 2AgNO 3 (aq) + Cu (s)  Cu(NO 3 ) 2 (aq) + 2Ag (s) The oxidation number of Ag decreases from +1 to 0 (reduction), copper’s oxidation number increase from 0 to +2 (oxidation)

Look at the reaction between solid copper and silver ions in aqueous solution: Cu (s) + Ag + (aq)  Ag (s) + Cu +2 (aq) Cu + Ag +  Ag + Cu +2 1 e -- gained 2 e - lost