Chapter 4 Solution Stoiciometry

Solutions = Homogeneous Mixtures  Solute – thing being dissolved (lesser part of Homogeneous mixture)  Solvent – medium that there is more of (dissolves solute)

Solubility Ability of substance to be dissolved in solvent (greater than.01 mol/L = soluble) Molarity = moles solute/liters solvent

Calculations: How many grams of sodium hydroxide is necessary to make 2L of 3.5M solution? M=mol solute/L solvent 3.5M = x mol/2L 7mol = x convert x to grams 7mol x 40g/ 1mol = 280 g

Calculations: What volume in mL of water is needed to make 3.6M solution of salt (NaCl) given 4.8g of NaCl? convert to mol 4.8g x 1mol/58.5g =.08mol NaCl 3.6M =.08mol NaCl / X Liters X Liters =.08mol/3.6 mol/L =.022L  22mL

Calculations: Calculate the molarity of each ion in solution of 2M CrCl 3 CrCl 3  Cr Cl - 2M 2M 6M

Calculations: What is the molarity of Fe 3+ ions and SO 4 2- ion in a solution made with 48.05g of Fe 2 (SO 4 ) 3 in Water to make 800 mL of solution? M = moles solute /moles solution Find moles Iron (III) sulfate Find moles Fe 3+ and SO 4 2- Find Molarity

Dilution Adding water to get desired molarity M 1 V 1 = M 2 V 2 How much 18M HCl must be used to make 100 mL of 2M solution? X =11mL

Solubility Rules Memorize table 4.1 on page 118 and handout Used to predict a precipitate (insoluble product) Test on Friday.

Attack the problem 1. Write the products and reactants and balance the equation. 2. Convert to moles if grams or molarity is given 3. Determine the product’s solubility, convert to grams if necessary.

Example Calculate the mass of precipitate produced when 125 mL of.2M AgNO 3 is added to an excess of Na 2 S to produce silver sulfide and sodium nitrate. 1. Balance: 2AgNO 3 + Na 2 S  Ag 2 S + 2NaNO 3 2. Molarity.2M = x mol/.125L 3. Find mol AgNO 3 and then convert to Mass.025 mol AgNO 3 x 1molAgS/2mol AgNO 3 x 247g AgS/1molAgS =2.9g

Net Ionic Equations Write balanced molecular equation Write out ions from strong electrolytes in water (aq) Make sure to balance your charges Look for unchanged ions (Spectator ions) on each side (aq of same charge) and cancel Write the remaining ions in a net ionic equation (charges must balance)

Net Ionic Equations Strong Electrolytes solutes dissociate completely. Strong Acids - amount of H + dissociated determines strength (90-100%) Strong Bases - amount of OH - dissociated determines strength (90-100%) Salts – cation from base, anion from acid Ex: NaCl NaOH + HCl  H 2 O + NaCl Memorize the Strong acids and Bases!!! (p122)

Practice Net Ionic Equations NaOH + HCl  NaCl + H 2 O (molecular eq) Na + (aq) + OH - (aq) + H + (aq) + Cl - (aq)  Na + (aq) + Cl - (aq) + H 2 O (l) (ion eq) Na + + OH - + H + + Cl -  Na + + Cl - + H 2 O Spectator ions, no change in oxidation or state OH - (aq) + H + (aq)  H 2 O (l) net ionic equation

Acids Donate H + Monoprotic = 1H + Diprotic = 2H + Strong acids vs. weak acids Any ionic = strong (anything not in list is weak) Strong acids dissociate completely Weak acids don’t dissociate completely and can become proton acceptors (reaction goes both ways) called a conjugate base

Bases Donate OH - Strong bases vs. weak bases Any ionic = strong (anything not in list is weak) Strong bases dissociate completely Weak bases = NH 3 Forms conjugate acids, can donate H +

Salts Reaction of acids and bases (neutralization) produces metathesis (double replacement) resulting in water and salt Salts – cation from base, anion from acid Ex: NaOH + HCl  H 2 O + NaCl

Practice Net Ionic Equations Write the net ionic equation for the reaction of silver nitrate and potassium phosphate Write the balanced molecular equation: 3AgNO 3 + K 3 PO 4  Ag 3 PO 4 + 3KNO 3 Determine products solubility Ag 3 PO 4(s) Write Ionic equation 3Ag + (aq) + 3NO 3 - (aq) + 3K + (aq) + PO 4 3- (aq)  Ag 3 PO 4(s) + 3NO 3 - (aq) + 3K + (aq)

Cancel spectator ions 3Ag + (aq) + 3NO 3 - (aq) + 3K + (aq) + PO 4 3- (aq)  Ag 3 PO 4(s) + 3NO 3 - (aq) + 3K + (aq) Net ionic equation: 3Ag + (aq) + PO 4 3- (aq)  Ag 3 PO 4(s)

Oxidation numbers (states) Reaction in which electrons are transferred between reactants. Loss of electrons = oxidation Gain of electrons = reduction Oxidation numbers = imaginary set of numbers based on rules (provides a way to keep track of electrons)

Oxidation numbers Whole numbers written as +1, +2, -1 in order to distinguish them from charges (3+, 2+, 2-)  Elemental form of atom, oxidation number is 0  Includes the diatomic gases (ex Cl 2 oxidation number is 0)  Any monatomic ion, oxidation is the charge of the ion  Ex: Na + oxidation number = +1, Cl - = -1

Oxidation numbers Nonmetals usually have negative ox numbers Oxygen = -2 (except in peroxides O 2 2- where Oxygen has an oxidation # of –1) Hydrogen = -1 if bonded to metal and +1 if bonded to nonmetal Fluorine is –1 for all compounds, other halogens are –1 for binary compounds, but if bound to oxygen (oxyanions) they have positive oxidation states

Sum of all oxidation numbers in a neutral compound is zero. In a charged compound, it equals the charge Oxidation numbers

Redox Practice Determine the oxidation number on the following green atoms: H 2 O H 2 SO 4 P 2 O 5 NO 3 - LiH BaO

Redox Practice Determine which reactant has been oxidized, and which has been reduced. Give their initial and final oxidation numbers Cu (s) + 4HNO 3  Cu(NO 3 ) 2(aq) + 2H 2 O (l) + 2NO 2(g) Cu (0  2) N (5  5) N (5  4) Oxidized no change Reduced