Aqueous Reactions & Sol’n Stoichiometry Cheat sheet

Aqueous Reactions & Sol’n Stoichiometry Properties of Aqueous Solutions  Electrolytic Properties    Ionic Compounds in Water    electrolytes dissociate when dissolved in water Molecular Compounds in Water    ionic - conduct electricity (electrolytes) non-ionic - do not conduct electricity (non-electrolytes) non-electrolytes do not dissociate when dissolved in water exceptions: those that react with water (e.g. NH3, HCl) Strong and Weak Electrolytes   strong - dissociate completely weak - dissociate only partly NaCl (s) → Na+ (aq) + Cl- (aq) CH3COOH (s) ↔ H+ (aq) + CH3COO- (aq) (а) in,,'t 1n I ... '" i n· · р-1гн а п 1 �lt г 111 1 l11 L1пoL1ncl tl1 1 n ·. ...(11 11 р Ltncl i · · lv · in lik Пl t \,V' L I, n i n. 'lГ f П1 с[. Precipitation Reactions occur when the mixed solutions contain a combination of ions which form a sparingly soluble (or insoluble) compound Pb(NO3)2 (aq) + 2 KI (aq) → PbI2 (s) + 2 KNO3 (aq) Solubility Guidelines for Ionic Compounds  solubility - amount of substance that can be dissolved in 1 L of water at 25°C  substances with solubility < 0.01 mol/L considered insoluble Predicting Precipitation Reactions when two ionic compounds are mixed in aqueous solution - check the solubilities of the compounds formed when the ions “switch partners”    if either of the new compounds is insoluble (or slightly soluble) - precipitation occurs if both new compounds are insoluble - two precipitation reactions occur if both new compounds are soluble - no precipitation occurs Solubility Rules Solubility Rules (cont’d) Exchange Reactions    also known as metathesis cations exchange with each other driving force for exchange     formation of a precipitate generation of a gas production of a weak electrolyte production of nonelectrolyte AX + BY → AY + BX AgNO3 (aq) + KCl (aq) → AgCl (s) + KNO3 (aq) Example 1 Predict whether or not a precipitate will form when the following two solutions are mixed: (a) AgNO3 (aq) + NaCl (aq) yes, AlCl (s) (b) Pb(NO3)2 (aq) + KI (aq) yes, PbI2 (s) (c) Ba(ClO3)2 (aq) + Li2SO4 (aq) yes, BaSO4 (s) (d) BaCl2 (aq) + NaOH (aq) yes, Ba(OH)2 (s) Ionic Equations molecular equation: Pb(NO3)2 (aq) + 2 KI (aq) → PbI2 (s) + 2 KNO3 (aq) complete ionic equation: Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) + 2 I-(aq) → PbI2(s) + 2 K+ (aq)+ 2 NO3-(aq) net ionic equation: spectator ions Pb2+(aq) + 2 I-(aq) → PbI2(s) net ionic equation shows only ions and molecules directly involved in reaction Example 2 An aqueous solution of sodium carbonate is mixed with an aqueous solution of calcium chloride. A white precipitate immediately forms. Write a net ionic equation to account for this. What are the spectator ions? Na2CO3 (aq) + CaCl2 (aq) → CaCO3 (s) + 2 NaCl (aq) 2 Na+ + CO32- + Ca2+ + 2 Cl- → CaCO3 + 2 Na+ + 2 ClCa2+ (aq) + CO32- (aq) → CaCO3 (s) spectator ions Acid and Base Reactions Acids  substances that ionize or react in water to increase concentration of H+ ions (protons) • HCl and HNO3 - monoprotic acids • H2SO4 - diprotic acid H2SO4 (aq) → H+ (aq) + HSO4- (aq) HSO4- (aq) ↔ H+ (aq) + SO42- (aq)  strong acids HNO3, H2SO4, HClO3, HClO4, HCl, HBr, HI  weak acids all others including (but not limited to) HF, CH3COOH, HCOOH, H2C2O4, H3PO4 Acid and Base Reactions (cont’d) Bases     H+ ion acceptors react with H+ ions to form water H+ (aq) + OH- (aq) → H2O (l) increase [OH-] when dissolved in water NaOH (aq) → Na+ (aq) + OH- (aq) NH3 (aq) + H2O (l) ↔ NH4+ (aq) + OH- (aq) strong bases include Ba(OH)2 and hydroxides of the alkali metals (NaOH, KOH, etc.), the soluble ionic hydroxides  weak bases all slightly soluble or insoluble hydroxides and other compounds like NH3, etc. Reactions of Acids  neutralization reaction (acid + base → salt + water) HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) H+ (aq) + OH- (aq) → H2O (l) net ionic equation  acid + carbonate (or HCO3) → salt + water + CO2 gas 2 HCl (aq) + Na2CO3 (aq) → 2 NaCl (aq) + H2O (l) + CO2 (g)  acid + metal oxide → salt + water 2 HNO3 (aq) + MgO (s) → Mg(NO3)2 (aq) + H2O (l)  acid + metal → salt + H2 gas 2 HCl (aq) + Mg (s) → MgCl2 (aq) + H2 (g) Reactions of Bases  base + ammonium salt → salt + water + NH3 gas NaOH (aq) + NH4Cl (aq) → NaCl (aq) + H2O (l) + NH3 (g)  base + non-metal oxide → salt + water 2 NaOH(aq) + N2O5 (g) → 2 NaNO3 (aq) + H2O (l) Oxidation-Reduction Reactions  characterized by transfer of electrons  oxidation    loss of electrons during reaction oxidation number increases (becomes more positive) reduction   gain of electrons during reaction oxidation number decreases (becomes more negative) 4 Fe (s) + 3 O2 (aq) → 2 Fe2O3 (s) Oxidation-Reduction Reactions Fe (s) + 2 H+ (aq) → Fe2+ (aq) + H2 (g) • loses e• oxidized • reducing agent • gains e• reduced • oxidizing agent Oxidation Numbers determined by following a simple set of rules 1. 2. 3. 4. 5. 6. oxidation number of atoms in neutral molecule must add up to zero; those in an ion must add up to charge on the ion Group I elements --> +1 Group II elements --> +2 Group III elements --> +3 fluorine always -1 in compounds other halogens -1, except in compounds with oxygen or other halogens hydrogen is +1 except in metal hydrides (e.g. LiH) – rule 2 takes precedence here oxygen is -2 in compounds; exceptions: compounds with F (#3) and compounds with O–O bonds (#2 and #4) elemental form --> 0 Example 3 Assign oxidation numbers to the atoms in the following: Na = +1, Cl = -1 (a) NaCl (b) ClO- Cl = +1, O = -2 (c) Fe2(SO4)3 Fe = +3, S = +6, O = -2 (d) SO2 S = +4, O = -2 (e) I2 I=0 (f) KMnO4 K = +1, Mn = +7, O = -2 (g) CaH2 Ca = +2, H = -1 Redox Reactions Revisit reaction between metal and acid (or metal salt) A + BX → AX + B Zn (s) + 2 HBr (aq) → ZnBr2 (aq) + H2 (g) Mn (s) + Pb(NO3)2 (aq) → Mn(NO3)2 (aq) + Pb (s) These are displacement reactions +1 0 Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g) 0 +2 Redox Reactions (cont’d) Metals can be oxidized by aqueous solutions of various salts Fe (s) + Ni(NO3)2 (aq) → Fe(NO3)2 (aq) + Ni (s) Net ionic equation: Fe (s) + Ni2+ (aq) → Fe2+ (aq) + Ni (s) Remember: Whenever one substance is oxidized another must be reduced All metals will not be oxidized by acids or metal salt. How do we determine which will?? Redox Reactions and Activity Series  metals at top most easily oxidized  any metal on list can be oxidized by any metal ion below it Cu (s) + 2 Ag+ (aq) → Cu2+ (aq) + 2 Ag (s) BUT 2 Ag (s) + Cu2+ (aq) → 2 Ag+ (aq) + Cu (s) Example 4 Which of the following metals will be oxidized by Pb(NO3)2: Zn, Cu, Fe? Zn & Fe can be oxidized by Pb2+ since they are both above Pb in the activity series table. Zn (s) + Pb2+ (aq) → Zn2+ (aq) + Pb (s) Fe (s) + Pb2+ (aq) → Fe2+ (aq) + Pb (s) Concentrations of Solutions concentration - amount of solute dissolved in a given quantity of solvent or solution molarity (M) = moles solute volume of sol'n in liters 1.00 M --> 1.00 mol solute / 1 L sol’n dissolve 0.25 mol NaCl in 0.500 L sol’n: Molarity = 0.25 mol / 0.500 L = 0.50 M Example 5 Calculate the molarity of a solution prepared by dissolving 10.0 g of AgNO3 in enough water to make 250.0 mL of solution. ! 1 mol AgNO 3 $ = 10.0 g ( ) mol of AgNO3 #" 169.8731 g &% = 0.05887 mol ! 0.05887 mol $ = 0.235 M molarity = # " 0.2500 L &% Dilution Sometimes you want to take a concentrated solution and make a more dilute solution of it. When you do this, the moles of solute remain constant throughout the process. Mi Vi = Mf Vf Example 6 A flask contains 625 mL of 3.05 M calcium nitrate solution. What volume of 15.8 M Ca(NO3)2 contains the same number of moles of Ca(NO3)2 as this solution? Mi Vi = Mf Vf Vf ! ( 3.05 M) ( 0.625 L ) $ = # &% 15.8 M " = 0.121 L Example 7 What is the molar concentration of nitrate ions in 3.05 M calcium nitrate? 3.05 M Ca(NO3)2 2 NO3- for every 1 Ca(NO3)2 molarity ! 2 mol NO 3- $ = ( 3.05 M Ca(NO 3 )2 ) # " 1 mol Ca(NO 3 )2 &% = 6.10 M NO3- Example 8 How many milliliters of 4.5 M HCl are required to prepare 200 mL of 0.75 M HCl? Mi Vi = Mf Vf (4.5 M) Vi = (0.75 M)(200 mL) Vi = (0.75 M) (200 mL ) 4.5 M = 33 mL Example 9 (a) Describe how to prepare 0.500 L of 0.0250 M aqueous solution of potassium dichromate K2Cr2O7 ! 0.0250 mol $ mol = ( 0.500 L ) # = 0.0125 mol & " % L ! 294.1846 g $ mass = ( 0.0125 mol) # = 3.68 g & " 1 mol % Weigh out 3.68 g of K2Cr2O7 and dissolve in small amount of water. Dilute to 500 mL. Example 9 (cont’d) (b) Describe how to dilute the solution from part (a) to obtain a solution with a final concentration of 0.0140 M. MiVi = MfVf Vf = (0.0250 M) (0.500 L ) 0.0140 M = 0.893 L Dilute solution in (a) to 893 mL. Example 10 When the orange salt potassium dichromate is added to a solution of concentrated hydrochloric acid, it reacts according to the following net ionic equation: K2Cr2O7 + 14 HCl → 2 K+ + 2 Cr3+ + 8 Cl- + 7 H2O + 3 Cl2 Suppose that 6.20 g of K2Cr2O7 reacts with 100.0 ml of concentrated HCl (13.0 M). Calculate the final concentration of Cr3+ ion that results and the number of moles of chlorine gas produced. ! 1 mol $ = 0.021075 mol K2Cr2O7 " 294.1846 g &% mol K2Cr2O7 = ( 6.20 g) # ! 13.0 mol $ = 1.30 mol HCl " 1 L &% mol HCl = ( 0.1000 L ) # ! 3 mol Cl2 $ = 0.0632 mol Cl2 " 1 mol K 2Cr2O7 &% mol Cl2 from K2Cr2O7 = ( 0.021075 mol K 2Cr2O7 ) # ! 3 mol Cl2 $ = 0.279 mol Cl2 " 14 mol HCl &% mol Cl2 from HCl = (1.30 mol HCl) # Example 10 (cont’d) K2Cr2O7 + 14 HCl → 2 K+ + 2 Cr3+ + 8 Cl- + 7 H2O + 3 Cl2 ! 1 mol $ = 0.021075 mol K2Cr2O7 " 294.1846 g &% mol K2Cr2O7 = ( 6.20 g) # mol Cr3+ ! 2 mol Cr 3+ $ from K2Cr2O7 = ( 0.021075 mol K 2Cr2O7 ) # = 0.0422 mol Cr3+ & " 1 mol K 2Cr2O7 % molarity of Cr3+ 0.0422 mol Cr 3+ = 0.100 L = 0.422 M Titrations  chemical reactions of solution of known concentration with solution of unknown concentration HCl + NaOH → NaCl + H2O  point at which stoichiometrically equivalent amounts of HCl and NaOH are brought together is called the equivalence point (endpoint)  typically use an indicator that changes color at the equivalence point Example 11 What is the molarity of a solution of sodium hydroxide if it requires 23.97 mL of that solution to reach the phenolphthalein endpoint when adding it to a solution containing 0.5333 g of KHC8H4O4? KHC8H4O4 + NaOH → NaKC8H4O4 + H2O ! 1 mol $ = 0.0026114 mol KHC8H4O4 " 204.2234 g &% mol KHC8H4O4 = ( 0.5333 g) # ! 1 mol NaOH $ mol NaOH = ( 0.0026114 mol KHC8H4O 4 ) # = 0.0026114 mol NaOH " 1 mol KHC8H4O 4 &% 0.0026114 mol molarity of NaOH = 0.02397 L = 0.1089 M Example 12 The indicator methyl red turns from yellow to red when the solution in which it is dissolved changes from basic to acidic. A 25.00 mL volume of a sodium hydroxide solution is titrated with 0.8367 M HCl. It takes 22.48 mL of this acid to reach a methyl red endpoint. Find the molarity of the sodium hydroxide solution. HCl + NaOH → NaCl + H2O ! 0.02248 L ( ) mol HCl = #" 0.8367 mol $ &% = 0.018809 mol HCl 1L ! 1 mol NaOH $ & = 0.018809 mol NaOH 1 mol HCl % mol NaOH @ endpoint = ( 0.018809 mol HCl) #" 0.018809 mol molarity of NaOH = 0.02500 L = 0.7524 M