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.

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-

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

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

Chapter 16

Acids and Bases

Bronsted Acids and BasesAcid Compound that loses H+ to a base

Base Compound that gains H+ from an acid

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

Carboxylic Acids: -COOHWeak organic acids:

COOH group on molecule is acidicCreates resonance structureStabilizes anion

Never fully dissociate in waterWill always be an equilibrium reaction

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

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

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

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

pH and pOHCalculations

Strong Acids and Bases

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

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

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

Strength of Acids and Bases

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

Strong vs. Weak Acids

Weak Acids and Bases

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

Weak Acids and Ka

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

ac KHA

AOHK ==−+

][]][[ 3

Ka Values of Common Weak Acids

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

][]][[ 3

HFFOHKa

3 105.7]092.0[)103.8(

][]][[ −

−−+

=== xxHF

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

%100][][%100

][][ 3 x

HAOH −+

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

%100][

][ 3 xHA

=α MxX 3109.1 X ]0100.0[

][%19 −==

23 106.4)109.10100.0(

)109.1(][

]][[ −−

−−+

NOOHKa

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

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

Making Approximations in Equilibrium Calculations

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

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−

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

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

HCHOCHOOHxKa ===

−+−

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

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][[

]108.3[]][[107.1 5

xxxxxKa −

== −−

Polyprotic acids

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

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

Calculate the pH of a 0.0050 M sulfuric acid?

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 =

=== −

HSOSOHxKa

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

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–

423 ×==−+OHKa

103.6]POH[

]HPO][[ 8–-

4 ×==−+OH

102.4]HPO[

]PO][[ 13–-2

4 ×==+OH

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

4233–

xxxOHKa −

==×=−+

What is the pH of a 0.037 M solution of phosphoric acid at 25˚C?

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

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 xOHKa ==×=

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

Weak Bases and Kb

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

bc KHHB

K ==−+

]B[]O][[

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.)

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.

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:

trimethylamine

aniline

Common Weak Bases

Calculations Using Kb

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!

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 -

45 1031 ]10.0[]][[

][]O][[

108.1 −−+

− =====

11.1189.200.141489.2]103.1log[]log[ 3

=−=−==−=−= −−

pOHpHxOHpOH

Calculate % Ionization

[OH-] = 1.3 x 10-3 from previous problem

%3.1%10010.0103.1%100

===−−

HIonization

Find the percent ionization of a 0.10 M solution of ammonia at 25oC.

Molecular Structure and Strength of Acids

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

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

Effect of Structure: Bond Length vs. Electronegativity for Binary acidsDirect bond between H and electronegative atom

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

Oxoacids

More double bonded oxygens: Stronger acidH2SO4 H2SO3

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

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

Acid Strength

Bond lengthvs

Electronegativity

Charge Difference

Structure:Terminal O

Electronegativity

Acid-Base Properties of Salts

Hydrolysis

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-

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

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

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

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

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

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.

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

14−===

Find the pH of a solution with 0.050 M CHOONa.

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.

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

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

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

Acids and Bases - URI Chemistry Chapter 16.pdf · CH 15 Summary Equilibrium is a balance between...

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