1 Compounds in Aqueous Solution Many reactions involve ionic compounds, especially reactions in water KMnO 4 in water K + (aq) ) + MnO 4- (aq)
2 CCR, page 149
3 How do we know ions are present in aqueous solutions? The solutions They are called!"!#$"%#! HCl,, MgCl 2, and NaCl are &. They dissociate completely (or nearly so) into ions.
4 & HCl,, MgCl 2, and NaCl are &. They dissociate completely (or nearly so) into ions.
5 Acetic acid ionizes only to a small extent, so it is a '( it is a '( CH 3 CO 2 H(aq) ---> CH 3 CO 2- (aq) + H + (aq)
6 Acetic acid ionizes only to a small extent, so it is a '( is a '( CH 3 CO 2 H(aq) ---> CH 3 CO 2- (aq) + H + (aq)
7 Some compounds dissolve in water but do not conduct electricity. They are called Examples include: sugar ethanol ethylene glycol
)* 8 Screen 5.4 & Figure 5.1
9 )* Common minerals are often formed with anions that lead to insolubility: sulfide carbonate fluoride oxide Azurite, a copper carbonate Iron pyrite, a sulfide Orpiment, arsenic sulfide
10 An acid -------> > H + in water Some & acids are HCl H 2 SO 4 HClO 4 HNO 3 hydrochloric sulfuric perchloric nitric HNO 3
11 An acid -------> > H + in water HCl(aq) ---> H + (aq) ) + Cl - (aq)
12 #+ * HCl Cl - H 2 O hydronium ion H 3 O +
( 13 WEAK ACIDS = weak electrolytes CH 3 CO 2 H acetic acid H 2 CO 3carbonic carbonic acid H 3 PO 4phosphoric phosphoric acid HF hydrofluoric acid Acetic acid
14 Nonmetal oxides can be acids CO 2 (aq) + H 2 O(liq) ---> > H 2 CO 3 (aq) SO 3 (aq) + H 2 O(liq) ---> > H 2 SO 4 (aq) and can come from burning coal and oil.
,! see Screen 5.9 and Table 5.2 15 Base ---> > OH -- in water NaOH(aq) ---> Na + (aq) ) + OH - (aq) NaOH is a strong base
Ammonia, NH Ammonia, NH 3, 16
17,! Metal oxides are bases CaO(s) ) + H 2 O(liq) --> > Ca(OH) 2 (aq) CaO in water. Indicator shows solution is basic.
18 Know the strong acids & bases!
19! Mg(s) ) + 2 HCl(aq) --> > H 2 (g) + MgCl 2 (aq) We really should write Mg(s) ) + 2 H + (aq) ) + 2 Cl - (aq) ---> H 2 (g) + Mg 2+ (aq) + 2 Cl - (aq) ions are SPECTATOR IONS The two Cl - ions are they do not participate. Could have used NO - 3.
! 20 Mg(s) ) + 2 HCl(aq) --> > H 2 (g) + MgCl 2 (aq) Mg(s) ) + 2 H + (aq) ) + 2 Cl - (aq) ---> > H 2 (g) + Mg 2+ (aq) + 2 Cl - (aq) We leave the spectator ions out Mg(s) + 2 H + (aq) ---> > H 2 (g) + Mg 2+ (aq) to give the NET IONIC EQUATION
+ $ Sections 5.2 & 5.4-5.6 5.6 CD-ROM Ch. 5 21 We will look at!-. /! $!# Pb(NO 3 ) 2 (aq) + 2 KI(aq) ----> > PbI 2 (s) + 2 KNO 3 (aq) AX + BY AY + BX The anions exchange places between cations.
$ 22 The driving force is the formation of an insoluble compound a precipitate. Pb(NO 3 ) 2 (aq) + 2 KI(aq) -----> 2 KNO 3 (aq) + PbI 2 (s) Net ionic equation Pb 2+ (aq) + 2 I - (aq) ---> > PbI 2 (s)
0,$ 23 The driving force is the formation of water. NaOH(aq) ) + HCl(aq) ---> NaCl(aq) ) + H 2 O(liq) Net ionic equation OH - (aq)) + H + (aq) ---> > H 2 O(liq) This applies to ALL reactions of STRONG acids and bases.
Acid-Base Reactions CCR, page 162 24
0,$ 25 A-B B reactions are sometimes called!#$"1# because the solution is neither acidic nor basic at the end. The other product of the A-B A B reaction is a "#, MX. HX + MOH ---> MX + H 2 O M n + comes from base & X n - comes from acid This is one way to make compounds!
/02 &$ 26 This is primarily the chemistry of metal carbonates. CO 2 and water ---> > H 2 CO 3 H 2 CO 3 (aq) + Ca 2+ ---> 2 H + (aq) ) + CaCO 3 (s) (limestone) Adding acid reverses this reaction. MCO 3 + acid ---> > CO 2 + salt
27 /02 & $ CaCO 3 (s) + H 2 SO 4 (aq) ---> 2 CaSO 4 (s) + H 2 CO 3 (aq) Carbonic acid is unstable and forms CO 2 & H 2 O H 2 CO 3 (aq) ---> > CO 2 + water (Antacid tablet has citric acid + NaHCO 3 )
28
29 3 4 * $ Sections 5.8-5.10 5.10
# & 30 In solution we need to define the - "5!# the component whose physical state is preserved when solution forms "#! the other solution component
31 * The amount of solute in a solution is given by its concentration. is given by its concentration Molarity (M) = moles solute liters of solution Concentration (M) = [ ]
32 1.0 L of water was used to make 1.0 L of solution. Notice the water left over. CCR, page 177
$,"!6 4788&* 9:. &+' (7 8 " * Step 1: Calculate moles of NiCl 2 6H 2 O 5.00 g 1 mol 237.7 g = 0.0210 mol Step 2: Calculate molarity 33 0.0210 mol 0.250 L = 0.0841 M [NiCl 2 6 H 2 O ] = 0.0841 M
# +* 34 CuCl 2 (aq) --> Cu 2+ (aq) + 2 Cl - (aq) If [CuCl 2 ] = 0.30 M, then [Cu 2+ ] = 0.30 M [Cl - ] = 2 x 0.30 M
/"$#% 35 What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. ml of a 0.0500 M solution? Because Conc (M) = moles/volume = mol/v this means that ;95
/"$#% 36 What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. ml of a 0.0500 M solution? ;95 Step 1: Calculate moles of acid required. (0.0500 mol/l)(0.250 L) = 0.0125 mol Step 2: Calculate mass of acid required. (0.0125 mol )(90.00 g/mol) = 1.13 g
37 & Weigh out a solid solute and dissolve in a given quantity of solvent. Dilute a concentrated solution to give one that is less concentrated.
$,"!6%+4788 " *<8. '878 8 8. += 38 Add water to the 3.0 M solution to lower its concentration to 0.50 M +
$,"!6%+4788 " *<8. '878 8 8. + = 39 H2 O But how much water do we add? 3.0 M NaOH 0.50 M NaOH Concentrated Dilute
$,"!6%+4788 " *<8. '878 8 8. +? 40 How much water is added? The important point is that --->!"# $!"
$,"!6%+4788 " *<8.. '878. 8. += 41 Amount of NaOH in original solution = M V = (3.0 mol/l)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH Volume of final solution = (0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L or 300 ml
$,"!6%+4788 " *<8. '878 8 8. + = 42 H2 O 3.0 M NaOH 0.50 M NaOH Concentrated Dilute Conclusion: add 250 ml of water to 50.0 ml of 3.0 M NaOH to make 300 ml of 0.50 M NaOH.
&) 43 A shortcut C V initial initial = C V final final
. ph: a way to express acidity -- the concentration of H + in solution. 44 Low ph: high [H + ] High ph: low [H + ] Acidic solution ph < 7 Neutral ph = 7 Basic solution ph > 7
#+. 45 ph = log (1/ [H + ]) = - log [H + ] In a neutral solution, [H + ] = [OH - ] = 1.00 x 10-7 M at 25 o C ph = - log [H + ] = -log (1.00 x 10-7 ) = - (-7) = 7 See CD Screen 5.17 for a tutorial
>.? @. 46 If the [H + ] of soda is 1.6 x 10-3 M, the ph is? Because ph = - log [H + ] then ph= - log (1.6 x 10-3 ) ph = - (-2.80) ph = 2.80
. >.? @ 47 If the ph of Coke is 3.12, it is. Because ph = - log [H + ] then log [H + ] = - ph Take antilog and get [H + ] = 10 - ph [H + ] = 10-3.12 = 7.6 x 10-4 M
"##.!#$% Section 5.10 48 Zinc reacts with acids to produce H 2 gas. Have 10.0 g of Zn What volume of 2.50 M HCl is needed to convert the Zn completely?
/!!$""2$#.!#$% ""# 49 Mass zinc Mass HCl Moles zinc Stoichiometric factor Moles HCl Volume HCl
Zinc reacts with acids to to produce H 2 gas. If If you have 10.0 g of of Zn, what volume of of 2.50 M HCl is is needed to to convert the Zn Zn completely? 50 Step 1: Write the balanced equation Zn(s) ) + 2 HCl(aq) --> > ZnCl 2 (aq) + H 2 (g) Step 2: Calculate amount of Zn 10.0 g Zn 1.00 mol Zn 65.39 g Zn = 0.153 mol Zn Step 3: Use the stoichiometric factor
Zinc reacts with acids to to produce H 2 gas. If If you have 10.0 g of of Zn, what volume of of 2.50 M HCl is is needed to to convert the Zn Zn completely? 51 Step 3: Use the stoichiometric factor 0.153 mol Zn 2 mol HCl 1 mol Zn = 0.306 mol HCl Step 4: Calculate volume of HCl req d 0.306 mol HCl 1.00 L 2.50 mol = 0.122 L HCl
0,!$!# # 52 H 2 C 2 O 4 (aq) + 2 NaOH(aq) ---> acid base Na 2 C 2 O 4 (aq) + 2 H 2 O(liq) Carry out this reaction using a TITRATION. Oxalic acid, H 2 C 2 O 4
*& &' ' +) 53 CCR, page 186
# 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3. Indicator shows when exact stoichiometric reaction has occurred. 4. Net ionic equation H + + OH - --> > H 2 O 5. At equivalence point moles H + = moles OH - 54
LAB PROBLEM #1: Standardize a solution of NaOH i.e., accurately determine its concentration. 55 1.065 g of H 2 C 2 O 4 (oxalic acid) requires 35.62 ml of NaOH for titration to an equivalence point. What is the concentra-tion tion of the NaOH?
1.065 g of of H 2 C 2 O 4 (oxalic acid) requires 35.62 ml of of NaOH for titration to to an an equivalence point. What is is the concentration of of the NaOH? 56 Step 1: Calculate amount of H 2 C 2 O 4 1.065 g 1 mol 90.04 g = 0.0118 mol Step 2: Calculate amount of NaOH req d 0.0118 mol acid 2 mol NaOH 1 mol acid = 0.0236 mol NaOH
1.065 g of of H 2 C 2 O 4 (oxalic acid) requires 35.62 ml of of NaOH for titration to to an an equivalence point. What is is the concentration of of the NaOH? 57 Step 1: Calculate amount of H 2 C 2 O 4 = 0.0118 mol acid Step 2: Calculate amount of NaOH req d = 0.0236 mol NaOH Step 3: Calculate concentration of NaOH 0.0236 mol NaOH 0.03562 L = 0.663 M [NaOH]] = 0.663 M
LAB PROBLEM #2: Use standardized NaOH to determine the amount of an acid in an unknown. 58 Apples contain malic acid, C 4 H 6 O 5. C 4 H 6 O 5 (aq) + 2 NaOH(aq) ---> Na 2 C 4 H 4 O 5 (aq) + 2 H 2 O(liq) 76.80 g of apple requires 34.56 ml of 0.663 M NaOH for titration. What is weight % of malic acid?
76.80 g of apple requires 34.56 ml of 0.663 M NaOH for titration. What is is weight % of malic acid? 59 Step 1: Calculate amount of NaOH used. C V = (0.663 M)(0.03456 L) = 0.0229 mol NaOH Step 2: Calculate amount of acid titrated. 0.0229 mol NaOH = 0.0115 mol acid 1 mol acid 2 mol NaOH
76.80 g of apple requires 34.56 ml of 0.663 M NaOH for titration. What is is weight % of malic acid? 60 Step 1: Calculate amount of NaOH used. = 0.0229 mol NaOH Step 2: Calculate amount of acid titrated = 0.0115 mol acid Step 3: Calculate mass of acid titrated. 0.0115 mol acid 134 g mol = 1.54 g
76.80 g of apple requires 34.56 ml of 0.663 M NaOH for titration. What is is weight % of malic acid? 61 Step 1: Calculate amount of NaOH used. = 0.0229 mol NaOH Step 2: Calculate amount of acid titrated = 0.0115 mol acid Step 3: Calculate mass of acid titrated. = 1.54 g acid Step 4: Calculate % malic acid. 1.54 g 76.80 g 100% = 2.01%
62 A0$$ Thermite reaction Fe 2 O 3 (s) + 2 Al(s) ----> 2 Fe(s) ) + Al 2 O 3 (s) Section 5.7
EXCHANGE: Precipitation Reactions 63 EXCHANGE Gas-Forming Reactions REACTIONS EXCHANGE Acid-Base Reactions REDOX REACTIONS
64 $! -$!# Oxidation 2 H 2 (g) + O 2 (g) ---> > 2 H 2 O(liq) Mg(s) ) + 2 HCl(aq) --> > MgCl 2 (aq) + H 2 (g) All corrosion reactions are oxidations. 2 Al(s) ) + 3 Cu 2+ (aq) ---> > 2 Al 3+ (aq) + 3 Cu(s) Reduction Fe 2 O 3 (s) + 2 Al(s) --> > 2 Fe(s) ) + Al 2 O 3 (s)
65 $! - $!# Cu(s) ) + 2 Ag + (aq) ---> > Cu 2+ (aq) + 2 Ag(s) In all reactions if something has been oxidized then something has also been reduced
Why Study Redox Reactions 66 Batteries Corrosion Manufacturing metals Fuels
67 $! -$!# Redox reactions are characterized by ELECTRON TRANSFER between an electron donor and electron acceptor. Transfer leads to 1. increase in oxidation number of some element = OXIDATION 2. decrease in oxidation number of some element = REDUCTION
- #,!$ 68 The electric charge an element APPEARS to have when electrons are counted by some arbitrary rules: 1. Each atom in free element has ox. no. = 0. Zn O 2 I 2 S 8 2. In simple ions, ox. no. = charge on ion. -11 for Cl - +2 for Mg 2+
- #,!$ 69 3. O has ox. no. = -2 (except in peroxides: in H 2 O 2, O = -1) 4. Ox. no. of H = +1 (except when H is associated with a metal as in NaH where it is -1) 5. Algebraic sum of oxidation numbers = 0 for a compound = overall charge for an ion
- #,!$ 70 NH 3 N = ClO - Cl = H 3 PO 4 P = MnO - 4 Mn = Cr 2 O 2-7 Cr = C 3 H 8 C = Oxidation number of F in HF?
$&B&$A $A $ 71 Corrosion of aluminum 2 Al(s) ) + 3 Cu 2+ (aq) --> > 2 Al 3+ (aq) + 3 Cu(s) Al(s) --> > Al 3+ (aq) + 3 e - Ox. no. of Al increases as e - are donated by the metal. Therefore, Al is OXIDIZED Al is the REDUCING AGENT in this balanced half- reaction.
$&B&$A $A $ 72 Corrosion of aluminum 2 Al(s) ) + 3 Cu 2+ (aq) --> > 2 Al 3+ (aq) + 3 Cu(s) Cu 2+ (aq) + 2 e - --> Cu(s) Ox. no. of Cu decreases as e - are accepted by the ion. Therefore, Cu is REDUCED Cu is the OXIDIZING AGENT in this balanced half- reaction.
$&B&$A $A $ Notice that the 2 half-reactions add up to give the overall reaction if we use 2 mol of Al and 3 mol of Cu 2+ 2 Al(s) --> > 2 Al 3+ (aq) + 6 e - 3 Cu 2+ (aq) + 6 e - --> > 3 Cu(s) ----------------------------------------------------------- 2 Al(s) ) + 3 Cu 2+ (aq) ---> > 2 Al 3+ (aq) + 3 Cu(s) 2+. Final eqn.. is balanced for mass and charge. 73
AB& $&& # # )7 )7 C C 74 Metals (Cu) are reducing agents HNO 3 is an oxidizing agent Cu + HNO 3 --> Cu 2+ + NO 2 Metals (Na, K, Mg, Fe) are reducing agents 2 K + 2 H 2 O --> 2 KOH + H 2
$&B&$A $A $ # )7 C C 75 Reaction Type In terms of oxygen In terms of halogen In terms of electrons Oxidation gain gain loss Reduction loss loss gain
!A *$A $ 76 Metal + halogen 2 Al + 3 Br 2 ---> > Al 2 Br 6
!A *$A $ 77 Nonmetal (P) + Oxygen Metal (Mg) + Oxygen
!A *$A $ 78 Metal + acid Mg + HCl Mg = reducing agent H + = oxidizing agent Metal + acid Cu + HNO 3 Cu = reducing agent HNO 3 = oxidizing agent