acids and bases - uri chemistry chapter 16.pdf · ch 15 summary equilibrium is a balance between...
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CH 15 SummaryEquilibrium is a balance between products and reactants
Use stoichiometry to determine reactant or product ratios, but NOT reactant to product ratios.
Capital K is used to represent the equilibrium constantProducts over reactants raised to their stoichiometric coefficientsCalculated from balanced equation, subscript designates units used, Kc, KpNo units used in final written K
Equlibrium Calculations and Reaction QuotientsICE tables used to manipulate initial and equilibrium concentrations.
Factors influencing K Concentration of chemicals and temperature affect all equilibriaT, P, V changes affect Kp in gases
CatalysisEffect of catalysts and inhibitors, reasons they are used.
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Polyatomic IonsMemorize These!
Ammonium NH4+ Nitrate NO3
-
Hydronium H3O+ Nitrite NO2-
Acetate CH3COO- Phosphate PO43-
Carbonate CO32- Cyanide CN-
Permanganate MnO4- PerchlorateClO4
-
Hydroxide OH- Sulfate SO42-
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Chapter 4: Solution Chemistry
MolarityConversions between moles and liters
DilutionsCalculate concentration of diluted solution from a stock solution M1V1=M2V2Calculate mass of solid needed to make a solution
Acid Base TitrationsIdentifying acids and basesKnow properties of acids and basesKnow list of specific acids both names and formulasDetermine concentration of unknown solutions using titration
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Common Acids to MemorizeStrongHydrochloric Acid: HClSulfuric Acid: H2SO4Nitric Acid: HNO3Perchloric Acid: HClO4Hydrobromic Acid: HBrHydroiodic Acid HI
WeakCarbonic Acid: H2CO3Phosphoric Acid: H3PO4Acetic Acid: CH3COOHHydrofluoric Acid: HF
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Chapter 16
Acids and Bases
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Bronsted Acids and BasesAcid Compound that loses H+ to a base
Base Compound that gains H+ from an acid
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SA WB CA CB
WB WA CA CB
Conjugate Acid-Base PairAcid-Base pair exchanging the H+Conjugate acid: Product with the extra H+
Conjugate base: Product with 1 less H+ ion than reactant
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Carboxylic Acids: -COOHWeak organic acids:
COOH group on molecule is acidicCreates resonance structureStabilizes anion
Never fully dissociate in waterWill always be an equilibrium reaction
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Reactivity of Weak Acid and Bases
Strong Acids and Bases: Full dissociation Conjugate bases and acids form spectator ionsCan use Chem 101 stoichiometry in calculationsNo original reactant or product left in solution
Weak Acids and Bases: Partial dissociationForms an equilibrium: Ka or Kb
Acid/Base strength in aqueous solutionsH3O+ is the strongest acidOH- the strongest baseWater acts as weak acid or base in the reaction
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Acid-Base Properties of Water: Autoionization
Water is slightly conductive due to the following reaction:
2H2O(l) H3O+(aq) + OH–(aq)Process is called Autoionization
Acid-Base Reaction between identical molecules1 molecule acts as an acid, the other a base
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Calculation of [H3O+] and [OH–] in WaterTreat as an equilibrium reaction at 25° C
H2O(l) + H2O(l) H3O+(aq) + OH–(aq) WB WA SA SB
Kw = [H3O+][OH–] = 1.0×10–14 Ion-Product constant Kw is very small: Favors weaker acid-base pair (H2O)
Make ICE tableProducts [H3O+] [OH–]Initial 0 0Change +x +xEquilibrium x x
Solve for [H3O+] and [OH-]1.0 x 10-14 = [H3O+][OH–] = [x][x] = x2
x = [H3O+] = [OH–] = 1.0×10–7 M pH = 7 of pure water
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pHMethod of measuring acidity (p)ower of the (H)ydrogen ion
CalculationspH = -log[H3O+] [H3O+] = 10(-pH)
pOH = -log[OH-] [OH-] = 10(-pOH)
Kw = [H3O+][OH-] = 1x10-14MpKw=pH + pOH = 14
Effects in 1M Strong Acid (pH = 0)[H3O+]= 1M then [OH-]= 1x10-14M
Effects in 1M Strong Base (pH= 14)[OH-]= 1M then [H3O+]= 1x10-14M
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pH and pOHCalculations
Strong Acids and Bases
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pH and pOH CalculationspOH: (p)ower of the (OH-) ion = 14.00-pH1. Find pH and pOH of an 0.0050M HBr at 25°C.
HBr(aq)+ H2O(l)H3O+(aq) + Br–
(aq) [H3O+]= 0.0050M pH = –log[H3O+] = –log(5.0×10–3) = 2.30pOH =14.00- 2.30 = 11.70
2. Find pH and pOH at 25°C of 3.7×10–5 M NaOHNaOH(aq) Na+(aq) + OH– (aq) [OH–] = 3.7×10–5M pOH = -log[OH-]= 4.43pH = 14.00- pOH = 14.00-4.43 = 9.57
Significant FiguresSigfigs in concentration = sigfigs after decimal point in pH
[H3O+] =3.7×10–5 M pH = 4.4313
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Acidity Calculations: Strong Acids & BasesCalculate the H3O+ and OH– concentrations at 25°C
of an aqueous 0.010M solution of nitric acid?Write reactions:HNO3(aq) + H2O(l) H3O+(aq) + NO3
–(aq) H3O+ = 0.010 M2H2O(l) H3O+(aq) + OH– (aq) H3O+ = 1.0×10–7 M
[H3O+] contributed by water is negligible[H3O+] = 0.010 + (1.0×10–7) = 0.0100001 M[OH-] =Kw ÷ [H3O+] = (1.0×10–14) ÷ (0.010) = 1.0×10–12 M
Strong acid increases [H3O+] and suppresses [OH-] pH = -log[H3O+] = -log[0.010M] = 2.00
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Acidity Calculations for Dilute SolutionsCalculate the H3O+(aq) + OH– (aq) of a 1.00×10–6M
solution of NaOH at 25°C?Write reactions:
NaOH(aq) Na+(aq) + OH–(aq) [OH–] = 1.00×10-6M 2H2O(l) H3O+(aq) + OH–(aq) [OH–] = 1.00×10-7M[OH–] < 1.00×10-7M because of Le Chatelier’s Principle
Total concentrations:[OH–] = 1.00×10–6 + 0.10×10-6M (10%)NOT less than 1.00×10–7 not negligibleMust use an ICE table if solution less than 10-6M
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Strength of Acids and Bases
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Strength of Acids and BasesStrong acids and bases are strong electrolytes
Completely ionized in water, no original compound leftGood conductors of electricity.Directional arrow (→) indicates dissociation is complete
Weak acids and bases are weak electrolytesPartial ionization in water, original compound remainingPoor conductors of electricityDouble arrow (↔) indicates dissociation is incompleteGoverned by an equilibrium constant, Ka or Kb
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Strong vs. Weak Acids
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Weak Acids and Bases
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Stronger acids will dominate over weaker acidsHNO2(aq)+ CN-(aq)HCN (aq)+ NO2
- (aq) K>1
If paired with group 1 cation, will be strong base
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Weak Acids and Ka
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Weak Acids, Ka and pKaALWAYS write a reaction of weak acid HA in water:HA(aq) + H2O(l) H3O+(aq) + A–(aq)
The equilibrium constant for this reaction is:
Ka is the acid ionization constant.Quantitative measure of acid strengthLarge Ka :Stronger acid
pKa = -log Ka
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ac KHA
AOHK ==−+
][]][[ 3
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Ka Values of Common Weak Acids
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Ka Calculations0.100 mole of HF is dissolved in 1.00 L of water at 25 °C. The pH at equilibrium was found to be 2.08. Calculate Ka.HF(aq)+ H2O(l) H3O+(aq) + F–(aq)Make table [HF] [H3O+] [F-]
Initial 0.100 0 0Change -x +x +xEquilibrium 0.100 -x +x +x
Find [H3O+], [F-] and [HF][H3O+] = 10(-pH) = 10-2.08 = 8.3 x 10-3 M = [F-]=x[HF] = 0.100 -x = 0.100 – 0.0083M=0 .0917M=.092M
Calculate Ka
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][]][[ 3
HFFOHKa
−+
=
423
3 105.7]092.0[)103.8(
][]][[ −
−−+
=== xxHF
FOHKa
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Percent Ionization and Ka
Measure pH of weak acid of known initial concentration
pH gives [H3O+] at equilibrium. Get equilibrium concentrations from table
Percent ionization (α) of a weak acid
Use stoichiometry of chemical equation to calculate equilibrium concentrations in Ka
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%100][][%100
][][ 3 x
HAAx
HAOH −+
==α
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Ka Calculations
A 0.0100 M solution of HNO2 is 19% ionized at equilibrium. Find Ka.
HNO2(aq)+ H2O(l) H3O+(aq) + NO2–(aq)
Make table [HNO2] [H3O+] [NO2-]
Initial 0.0100 0 0Change -x +x +xEquilibrium 0.0100 -x +x +x
Find [H3O+], [NO2-] & [HNO2]
Calculate Ka
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%100][
][ 3 xHA
OH +
=α MxX 3109.1 X ]0100.0[
][%19 −==
43
23
2
23 106.4)109.10100.0(
)109.1(][
]][[ −−
−−+
=−
== xx
xHNO
NOOHKa
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Determination of Relative Acidity from KaWhich is the stronger acid, HF or HNO2?
HF pKa = –log Ka = –log(7.5×10–4) = 3.12
HNO2pKa = –log Ka = –log(4.6×10–4) = 3.34
Stronger AcidHigher KaSmaller pKa HF stronger acid
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Find the pH of a 0.010 M Solution of Hypochlorous Acid [HClO] at 25 oC.
pH range from strong acid to pure waterStrong acid, pH = -log [0.010] = 2.0Pure water = pH = -log [1.0 x 10-7] = 7.0
Ka of hypochlorous acid = 2.9×10–8
Look up in table, not a calculated numberNot a strong acid: Incomplete dissociationmust use equilibrium calculations
Calculate H3O+ from Ka
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Making Approximations in Equilibrium Calculations
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Find the pH of a 0.010 M Solution of Hypochlorous Acid [HClO] at 25 oC.
Calculate H3O+ from Ka
X = [H3O+] = 0.000017[HClO] = 0.010-000017
~ 0.010 M
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Calculate pH from [H3O+]pH = -log [H3O+]
= -log (1.7 ×10-5) = 4.77
Could have approximated!
[HClO]eq= initial [HClO]
HClO(aq) + H2O(l) H3O+(aq) + ClO–(aq)Initial 0.010 - 0 0Change –x - +x +xEquilibrium 0.010 – x - x x
][]][[ 3
HClOClOOHKa
−+
=
]010.0[]][[109.2 8
xxxx−
=−
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Estimating Equilibrium ConcentrationsPurpose:
Simplify equilibrium concentration calculationsAvoid the quadratic equation
Procedure: Draw up your ICE tableIgnore x in the initial acid concentration Solve for x with the ICE tableDivide x by the initial acid concentration.
If [x]/[HA] x 100< 5%, you can say that [HA] +x = [HA]If not, you cannot make this assumptionThe error in the number will be over 5%
For previous problem estimate [HClO] at equilibrium [x]/[HA] = 1.7x10-5/0.050 = 0.034% so approximate
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HCHO2(aq) + H2O(l) CHO2-(aq) + H3O+(aq)
Initial 0.050 - 0 0Change – x - + x + xEquilibrium 0.050 – x - x x
Initial Data[HCHO2] = 0.050 given in problem[CHO2
-] and [H3O+]= 0 since you are adding nothing at start
Find the pH of a 0.050 M Solution of Formic Acid
Look up Ka in table and solve for x x =3.0x10-3
Calculate pHx = [H3O+ ] = 3.0×10–3 M pH = 2.52
]050.0[]][[
][]][[108.1
2
234xx
HCHOCHOOHxKa ===
−+−
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Calculate the pH of a 3.8×10-5 M solution of acetic acid at 25 oC? pKa = 4.77
Convert pKa to KapKa = 4.77 Ka = 10-4.77= 1.7x10-5
With approximation
X= 2.54 x10-5
pH =4.59
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Solve for [x] with quadratic
x = 1.8×10-5 [H3O+]= 1.8×10-5
Solve for pHpH = –log[H3O+]
pH= –log(1.8×10–5) = 4.74
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2–(aq)
Initial 3.8×10–5 - 0 0Change – x - +x +xEquilibrium 3.8×10–5 – x - x x
]OHC[]OHC][[
232
2323
HOHKa
−+
=
]108.3[]][[107.1 5
5
xxxxxKa −
== −−
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Polyprotic acids
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Polyprotic acidsAcids with more than one ionizable H atom
1. Treat first dissociation as monoprotic acids (1H+)
2. Use successive acid ionization constants 1 for each ionized H atomNumber sequentiallyKa1, Ka2, Ka3, ...
3. Each equilibrium constant is less likely to affect pH than the one before it
You may not need all of them to solve problems
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Calculate the pH of a 0.0050 M sulfuric acid.Sulfuric acid is a strong acid: Full dissociation of first H+
H2SO4(aq) + H2O(l) H3O+(aq) + HSO4–(aq)
0.0050M H2SO4 0.0050M H3O+ + 0.0050M HSO4–
Full dissociation of first H+ only!
Bisulfate ion, HSO4-: Weak acid equilibrium with 2nd H+
HSO4-(aq) + H2O(l) H3O+(aq) + SO4
2–(aq)Ka = 1.1 x 10-2 for HSO4
-(aq) [HSO4–]= [ H3O+ ]= 0.0050M
Estimate of pHBoth H ionize pH =–log(2×0.0050) = 2.00Ionization1 H pH =–log(0.0050) = 2.30
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Calculate the pH of a 0.0050 M sulfuric acid?
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HSO4–(aq) + H2O(l) H3O+(aq) + SO4
2–(aq)Initial 0.0050 - 0.0050 0Change – x -x + x + xEquilibrium 0.0050 –x - 0.0050 + x x
Solve for [H3O+] using quadratic equation
[H3O+] = 0.0050 + x = 0.0050 + 0.0029 = 0.0079 M
Solve for pHpH = – log[H3O+] = –log(0.0079) = 2.10
0.0029M x]0050.0[
]][0050.0[][
]O][[1014
-2432 =
−+
=== −
+−
xxx
HSOSOHxKa
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What is the pH of 0.037M phosphoric acid?H3PO4 has 3 ionizable hydrogen atoms
H3PO4 + H2O3H3O+ + 1PO43-
H3PO4: weak acideach ionization reaction is a separate problemuse the results of previous steps
Possible pH values All three H+ donated, = –log(3×0.037) = 0.961 H ionized = –log(0.037) = 1.43
H2PO4- and HPO4
2- have very small Ka valuespH primarily due to first ionizationLowest pH: –log(0.037)= 1.43 not 0.96
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Ionizable Hydrogens and Ka: Set up ICE table for each step
H3PO4(aq) + H2O(l) H3O+(aq) + H2PO4–(aq)
H3PO4=0.037M-0.013M=0.024MX=H3O+ = H2PO4
–= 0.013MNeed quadratic!
H2PO4-(aq) + H2O(l) H3O+(aq) + HPO4
2-(aq)H3O+ = H2PO4
–= 0.013MX= HPO4
2-= 6.3x10-8MUsed approximation
HPO42-(aq)+ H2O(l) H3O+(aq) + PO4
3–(aq)H3O+ = 0.013MHPO4
2-= 6.3x10-8MPO4
3–= 1.8x10-18M39
107.1]POH[
]POH][[ 3–
43
423 ×==−+OHKa
103.6]POH[
]HPO][[ 8–-
2
23
4
4 ×==−+OH
Ka
102.4]HPO[
]PO][[ 13–-2
-33
4
4 ×==+OH
Ka
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H3PO4(aq) + H2O(l) H3O+(aq) + H2PO4–(aq)
Initial 0.037 - 0 0Change – x - + x + xEquilibrium 0.037 – x - + x xApproximate H3PO4? [x]/[HA] = 0.016/.037 so no approximation
Plug equilibrium concentrations into Ka equation
x = 0.013
Plug x back into table to get equilibrium concentrations[H3PO4] = 0.037 - x = 0.037- 0.013 = 0.024M[H3O+] = x = 0.013M use for second ionization[H2PO4
–] = x = 0.013M use for second ionization
]037.0[]][[
]POH[]POH][[ 107.1
43
4233–
xxxOHKa −
==×=−+
What is the pH of a 0.037 M solution of phosphoric acid at 25˚C?
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H2PO4-(aq) + H2O(l) H3O+(aq) + HPO4
2-(aq)Initial 0.013 - 0.013 0Change – x - + x +xEquilibrium 0.013 – x - 0.013 +x x
What is the pH of a 0.037 M solution of phosphoric acid at 25˚C?
Plug equilibrium concentrations into Ka equation (approximate)Solve for X with quadratic
x = 6.3x10-8
Plug x back into table to get equilibrium concentrations
[H2PO4–] = 0.013 - 6.3x10-8 = 0.013M
[H3O+] = 0.013 + 6.3x10-8 = 0.013M No change in pH!!![HPO4
2–] = x = 6.3x10-8 use for next ionization
]013.0[]][013.0[
]POH[]HPO][[
103.6 -2
238–
4
4 xOHKa ==×=
−+
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HPO42-(aq) + H2O(l) H3O+(aq) + PO4
3–(aq)Initial 6.3x10-8 0.013 0Change – x - + x + xEquilibrium 6.3x10-8 – x - 0.013 + x x
[H3O+] 3rd ionizationKa3 =4.2×10–13< Ka2= 6.3x10-8
If second ionization didn’t change pH, neither will 3rd
Use [H3O+] from last ionization to calculate pH
Calculate pHpH due only to first ionizationpH = –log(1.3×10–2) = 1.89Original estimate was 1.43
What is the pH of a 0.037 M solution of phosphoric acid at 25˚C?
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Weak Bases and Kb
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Weak BasesBrønsted–Lowry reaction for a weak base:
B(aq) + H2O(l) HB+(aq) + OH–(aq)
Equilibrium constant :
Kb is the base ionization constant. Kb defines amount of dissociation
On a log scale: pKb = –log KbStrong bases: High[OH-], low [H3O+]Large Kb & small pKb
pKa + pKb = 14
3 CategoriesMetal hydroxides, Most anions, & Organic amines
44
bc KHHB
K ==−+
]B[]O][[
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Metal HydroxidesStrong bases; Group 1A
ex: NaOH, KOHWeak bases: Anything else withOH-
ex: Ca(OH)2 Ca2+
Fe(OH)3 Fe3+
Dissociation affected by solubilitySparingly soluble saltsWill act as weak bases (Kb)
Use equilibrium chemistry (Ksp)Solubility Product Constant (next chapter.)
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AnionsAct as H+ acceptors in water
The negative charge attracts H+
CO32-, CH3COO-, etc.
Exceptions:Anions of strong acids:
100% dissociation, no equilibrium established
Cl–, Br–, I–, ClO4–, NO3
–, HSO4–
Anions with ionizable hydrogen ionAmphiproticH2PO4
–, etc.
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Organic AminesGeneral structure of R3N:
R contains H,C,O, (main group atoms) Lone pair on nitrogen accepts H+
R3N:(aq) + H2O(l) R3N–H+(aq) + OH–(aq)
Conjugate acid: Ammonium ion, NH4+
Common amines:
47
N
CH3
CH3
H3C
trimethylamine
H2N
aniline
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Common Weak Bases
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Calculations Using Kb
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Relationship between Ka & Kb
Base & waterNH3 + H2O NH4
+ + OH– Kb = 1.8X10-5 pKb = 4.74
Conjugate acid & water:NH4
+ + H2O H3O+ + NH3 Ka= 5.6X10-10 pKa = 9.26Add two reactions together: multiply K valuesMultiple equilibria (Ch. 15)
2H2O H3O+ + OH– Kw =Ka x Kb = 1.00x10-14
pKa + pKb = 14.00 (at 25°C) so pKw = 9.26 + 4.74 = 14.00
Only true for conjugate acid/base pairs in water!
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Reaction: NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)
Initial 0.10 - 0 0Change – x - + x + xEquilibrium 0.10 – x - x x
Find the pH of a 0.10 M solution of ammonia at 25oC.
Fill in Kb expression
Approximate [NH3]
Solve for pH
Mx.][OHxxxNH
HNHxK -
b3
3
45 1031 ]10.0[]][[
][]O][[
108.1 −−+
− =====
11.1189.200.141489.2]103.1log[]log[ 3
=−=−==−=−= −−
pOHpHxOHpOH
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Calculate % Ionization
[OH-] = 1.3 x 10-3 from previous problem
%3.1%10010.0103.1%100
][][O
%3
3
===−−
xxxNH
HIonization
Find the percent ionization of a 0.10 M solution of ammonia at 25oC.
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Molecular Structure and Strength of Acids
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Measuring Acid Strength
Strongest Acid: Weakest bond to acidic H3 Criteria in order of importance
Charge: High negative charge on anion: weaker acidLeaves stronger bond to any remaining
hydrogensStructure:
Larger anions will create longer, weaker H-X bondsPresence of oxoatoms increase electronegativity
Electronegativity: High electronegativity of anion: stronger acid Withdraws electrons from acidic H stronger acid
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Effect of ChargeH3AsO4 H2AsO4
– HAsO42–
Charge: HAsO42– has highest negative
chargeHigher charge: weaker acidHarder to pull H away, stronger bond
Acidity based on charge differenceH3AsO4 > H2AsO4
– > HAsO42–
Structure: No differencesElectronegativity: No differences
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Effect of Structure: Bond Length vs. Electronegativity for Binary acidsDirect bond between H and electronegative atom
56
Two competing forcesLarge Atomic radius
Long bonds are weaker
High Electronegativity High electronegativity pulls electrons to more electronegative atom; weakens bond
Experimentally measured acid strengthHI>HBr>HCl>>HF
Bond length is dominating force
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Oxoacids
More double bonded oxygens: Stronger acidH2SO4 H2SO3
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Two terminal oxo atomsMore resonance stabilizes ionMore e- withdrawal from O-H
H is more positive: more acidic
One terminal oxo atomFewer resonance structuresLess e- withdrawal from O-H H is less positive: less acidic
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Effect of ElectronegativityCharge:
No differenceBoth neutral
Structure: No difference in terminal oxo atoms1 oxo atom each
Electronegativity: S more electronegative than P
H2SO3 more acidic than H3PO4
S more electronegative than SeH2SO4 more acidic than H2SeO4
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Acid Strength
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Bond lengthvs
Electronegativity
Charge Difference
Structure:Terminal O
Electronegativity
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Acid-Base Properties of Salts
Hydrolysis
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Hydrolysis of Ionic Salts
Ionic salts dissolved in water affect pHNaNO3 (s) + H2O(aq) Na+(aq) + NO3
- (aq)
Undergo HydrolysisHydrolysis: Reaction of an ionic salt with waterMay change the pH of the solution
Both cations and anions may undergo hydrolysis Not all ions hydrolyzeExamine both ions to determine acid/base character
3 step process to predict acidity of a salt solution1. Write the reaction that dissociates salt into its ions2. Check the cation for acid hydrolysis: produce H+
3. Check the anion for base hydrolysis: produce OH-
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Is NaCl (aq) Acidic, Basic or Neutral
1. Dissociate the salt into ions:NaCl(aq) Na+(aq) + Cl–(aq)
2. Check the cation for hydrolysis:Na+(aq) + H2O(l) NaOH(aq) + H+(aq) No Hydrolysis: NaOH strong base, not acidic
3. Check the anion for hydrolysis:Cl–(aq) + H2O(l) HCl (aq) + OH-(aq)
No Hydrolysis: HCl is a strong acid, not basicIf neither acidic or basic, solution is neutral
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Is Na2CO3 Acidic, Basic or Neutral1. Dissociate the salt into ions:
Na2CO3 (aq) 2Na+ (aq) + CO32- (aq)
2. Check the cation for hydrolysis:Na+(aq) + H2O(l) NaOH(aq) + H+(aq) No Hydrolysis: NaOH would be a strong baseNo H+ generated: Solution is not acidic
3. Check the anion for hydrolysis:CO3
2-(aq) + H2O(l) HCO3- (aq) + OH- (aq)
Anion hydrolyzes: HCO3- (aq) weak acid
Some OH- is generated: Solution may be basicIf not acidic but possibly basic, solution is basic 63
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Is Fe(NO3)2 Acidic, Basic or Neutral?1. Dissociate the salt into ions:
Fe(NO3)2(aq) Fe2+(aq) + 2NO3–(aq)
2. Check the cation for hydrolysis:Fe2+(aq) + H2O(l) FeOH+(aq) + H+(aq) Reaction occurs: FeOH+ is a weak baseH+ generated: Acidic solution
3. Check the anion for hydrolysis:NO3
–(aq) + H2O(l) HNO3 (aq) + OH-(aq)No OH- generated: Nitric is a strong acid, not
basicOverall: the solution is acidic
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Is ZnF2 Acidic, Basic or Neutral
1. Dissociate the salt into ions:ZnF2(aq) Zn2+(aq) + 2F–(aq)
2. Check the cation for hydrolysis:Zn2+(aq) + H2O(l) ZnOH+(aq) + H+(aq) Reaction Occurs: ZnOH+(aq) is a weak
baseH3O+generated: Acidic Solution
3. Check the anion for hydrolysis:F–(aq) + H2O(l) HF (aq) + OH-(aq)Reaction Occurs: HF is a weak acidOH- is generated: Basic Solution
Can’t tell acidity : Both cation and anion hydrolyze
Both H3O+(aq) and OH-(aq) possible65
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Is ZnF2 Acidic, Basic or Neutral1. Dissociate the salt into ions:
ZnF2(aq) Zn2+(aq) + 2F–(aq)2. Cation Hydrolysis: Ka based
Zn2+(aq) + H2O(l) ZnOH+(aq) + H+(aq)
3. Anion Hydrolysis: Kb basedF–(aq) + H2O(l) HF (aq) + OH-(aq)
Can’t tell acidity : Both cation and anion hydrolyzeBoth H+(aq) and OH-(aq) possibleHigher K value determines acidity
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Determine Acidity
Look up Kas for both acidsKa(Zn2+) = 2.5×10–10 for acid reaction
Ka(HF) = 6.6×10–4 for base reactionCalculate Kb for the anion reaction
Kb (F–) = Kw/Ka = 1.0×10–14/6.6×10–4 = 1.5×10–11
Larger equilibrium constant winsKa(Zn2+)> Kb (F–)
2.5×10–10 > 1.5×10–11
H+ created > OH- created
Cation hydrolysis dominates : Solution is acidic.
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HCOO-(aq) + H2O(l) HCOOH(aq) + OH-(aq)Initial 0.050 - 0 0Change – x - + x + xEquilibrium 0.050 – x - x x
First determine if acidic, basic or neutralAnion hydrolysis: CHOOH is a weak acid, so basicNeed to use Kb
Initial Data[HCOO-] = 0.050 from problem[HCOOH] and [OH-]= 0
Look up Ka in table and calculate Kb
11106.5108.1100.1
][][
4
14−===
−
−
xxx
KKK
a
wb
Find the pH of a solution with 0.050 M CHOONa.
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Solve for x
x =[OH-]= 1.7x10-6
Calculate pHpOH = -log[1.7x10-6 ] = 5.77
pH = 14-pOH = 8.23 69
HCOO-(aq) + H2O(l) HCOOH(aq) + OH-(aq)Initial 0.050 - 0 0Change – x - + x + xEquilibrium 0.050 – x - x x
]05.0[]][[
][]][[106.5 11
xxHCOO
HCOOHOHx == −
−−
Find the pH of a solution with 0.050 M CHOONa.
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Hydrolysis of Anions from Polyprotic AcidsNaHCO3, NaH2PO4, NaHSO4
Both acid and base hydrolysis occur
Cation Hydrolysis: Ka = 4.8x10-11
HCO3- (aq) + 2H2O(l) CO3
2- (aq) + H3O+(aq)
Anion Hydrolysis: Kb = 2.4x10-8
HCO3- (aq) + H2O(l) H2CO3(aq) + OH–(aq)
Kb= Kw/Ka = 1x10-14/4.2x10-7 = 2.4x10-8
Compare equilibrium constants Larger K determines equilibrium reaction
solution is basic70
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Hydrolysis of Complexed Metal IonsMetal Ions form complexes with water (hydrated)
Pull electrons towards metal ionPolarize O-H bond in attached watersH+ ions dissociate from oxygen
Cation hydrolysis onlyAl(H2O)6
3+(aq) + H2O(l) Al(OH)(H2O)52+(aq) + H3O+(aq)
Ka = 1.3x10-5
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AlH
H
O
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Lewis Acids and Bases
BF3 (g)+ NH3(g) F3B-NH3(g)
Lewis Acid: Accepts a pair of electronsBF3No ionizable HNot bronsted acid
Lewis Base: Donates a pair of electrons:NH3No H to acceptNot bronsted base
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F B F
F
H N H
H