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Acids and Bases

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electrolytes review• Substances which produce ions in a solution!

e.g. NaCl (s) ⇔ Na+ (aq) + Cl- (aq)

Strong electrolytes • produce a relatively large amt. of ions • good conductors of electricity Weak electrolytes • produce a relatively small amt. of ions • poor conductors of electricityNonelectrolyte• Substances that do not produce ions in solution! e.g.

sugar

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Acid properties review• Acids are electrolytes• Acids cause indicators to turn a

characteristic color• Water solutions of acids taste sour and feel

“wet” (note: do not taste or feel acids in the lab)

• Acids neutralize bases:! HCl + NaOH → NaCl + H2O! neutralization – chemical reaction between

an acid and a base to produce an ionic salt and water

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Base properties review• Bases are electrolytes• Bases cause indicators to turn a

characteristic color• Water solutions of bases taste bitter and feel

slippery (note: do not taste or feel bases in the lab)

• Bases neutralize acids! NaOH + HCl → NaCl + H2O

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Arrhenius acids• Acids are any substances that dissociate to produce

hydrogen ions (H+) when dissolved in water.• e.g. nitric acid • HNO3(aq) → H+(aq) + NO3-(aq)

Arrhenius bases• Bases are any substances that dissociate to produce

hydroxide ions (OH-) when dissolved in water. e.g. !NaOH (aq) → Na+ (aq) + OH− (aq)

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Ionization of covalently bonded electrolytes

• formation of ions caused by the reaction of molecules of H2O and a molecular compound

e.g. HC2H3O2 + H2O ⇔ H3O+ + C2H3O2−

acetic acid ! hydronium ion acetate ion ( same as HC2H3O2 (aq) ⇔ H+ (aq) + C2H3O2

- (aq) )

• weak acid, therefore equilibrium favors reactants

• H3O+ is the acid portion of the ionization• H3O+ simplified is H+ (H2O removed)• we use H+ for most examples

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ka

• The mass action expression for acids in equilibrium is Ka (ionization constant for acids)

! HC2H3O2 (aq) ⇔ H+ (aq) + C2H3O2− (aq)

!

! Ka = [H+] [C2H3O2−] = 1.8 x 10-5

! [HC2H3O2] (at 25 oC)

Assign: p. 605 #108

ka

e.g. #2! HCl (aq) H+ (aq) + Cl- (aq)

• strong acid, equilibrium favors products! Ka = [H+] [Cl-] = very large! !! [HCl]• Ka indicates to what extent ions are

formedstrong acid Ka = values “large or very large”weak acid Ka = values less than 1

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Calculating Ka from H+ • Question: In a 0.100 M solution of formic acid

(HCHO2) the [H+] = 0.0042 M at equilibrium. Calculate Ka.

• HCHO2 → H+ + CHO2−

• Ka = ! [H+] [CHO2−]

! ! [HCHO2]

• The 0.100 M value is an initial value and the [H+] is an equilibrium value therefore we use I.C.E.

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Calculating Ka from H+HCHO2 → ! H+ + ! CHO2

−

I! ! 0.100!! 0! ! 0C! −0.0042!! +0.0042! +0.0042E! ! 0.096!! 0.0042! 0.0042

Ka = (0.0042)(0.0042) = 1.8 x 10−4 at 25 oC! ! 0.096

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h+ conc for strong acids1. Find moles2. Find molarity3. Assume 100% ionization, which

means [H+] directly related to [acid]

Assign: p. 601 #6, 7! !! ! p. 604 #10

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h+ conc for weak acids *Question: Calculate the [H+] of a 0.75 M

solution of acetic acid (HC2H3O2) with a Ka of 1.8 x 10-5

technically 0.75M is an initial value so ...! ! HC2H3O2 ! → ! H+ + C2H3O2

−

Initial! ! 0.75! ! ! 0! 0Change! − x! ! ! + x! + x Equilibrium! 0.75 − x ! ! x! x

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h+ conc for weak acids! Ka = [H+] [C2H3O2

−] ! ! [HC2H3O2]1.8 x 10-5 = (x)(x) ! 0.75 – x

Therefore:! Ka = (x)(x)! ! 1.8 x 10-5 = x2! 0.75! ! ! ! 0.75! !! x = 3.67 x 10-3 M = [H+]

solution requires quadratic equation,for a weak acid this value is very small,

disregard this amount

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simplified (use this one)h+ conc for weak acids• Find moles• Find molarity• Substitute initial molarity of acid into

Ka expression and solve for unknown term

• note: weak acids are not 100% ionized (i.e. not all molecules break apart to form ions)

* usually start with acid conc

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h+ for weak acidsE.g. Determine [H+] for hydrofluoric acid

(HF) when 6.00g is dissolved in 2.0L of water. Ka = 3.5 x 10-4

6.00g 1 mol! =! 0.3 mol HF 20g! !(MM HF = 1(1) + 1(19) = 20g/mol)

M = n/V = 0.3 / 2.0 = 0.15 M

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h+ for weak acidsKa = [H+] [F−]

! !! ! [HF] let x = [H+] and [F−]3.5 x 10-4 = (x)(x)

0.15 x = 7.2 x 10-3 M = [H+]

percent ionization% ionization = [H+] x 100

[initial acid]

intial acid conc

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Bronsted Lowry Acids• Acids are proton (H+) donors.• e.g. ionization of hydrochloric acid! !HCl (g) + H2O (l) → H3O+ (aq) + Cl− (aq)! !acid! base(proton donor) (proton acceptor)

note: hydronium is a hydrogen ion which has been hydrated:! H+ + H2O → H3O+ (hydronium ion)

Bronsted Lowry bases• Bases are proton (H+) acceptors.• e.g. ionization of NH3 gas! !NH3 (g) + H2O → NH4

+ (aq) + OH− (aq)! !base ! acid(proton acceptor) (proton donor)

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Conjugate Acid Base Pairs • transfer of protons (donation, acceptance)

occurs in pairsHCl (g) + !H2O (l) → H3O+ (aq) + Cl− (aq)acid ! ! base !! acid ! ! base! !! conjugates

• HCl is the conjugate acid of the base Cl− • H2O is the conjugate base of the acid H3O+ • H3O+ is the conjugate acid of the base H2O• Cl− is the conjugate base of the acid HCl

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Conjugate Acid Base Pairs! !! acid! conjugates baseNH3 (g) + H2O → NH4

+ (aq) + OH− (aq)base! conjugates acid• always an acid/base pair on opposite

sides of the reaction.Assign: p.599 #2

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amphoteric substances• Substances, such as water, which can act as both

acids and bases are said to be amphoteric.• Amino acids and proteins are amphoteric, as they

both contain a basic amino group ( -NH2), and an acid carboxyl group ( -COOH).

• There are trends in the amphoteric properties of oxides of metals and non-metals on the periodic table.

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amphoteric substancesGroup 5A (15) • As the elements become more metallic down a group,

their oxides become more basic. • e.g. N2O5 least basic (most acidic)

!

! Bi2O3 most basic

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amphoteric substancesPeriod 3 • As elements become less metallic across a period, their

oxides become less basic. • e.g. most basic (least acidic) Na2O to least basic Cl2O7

Na! Mg! Al! Si! P! S! ClNa2O! MgO! Al2O3! SiO2! P4O10! SO3! Cl2O7

! !! Ionic! ! ! Molecular

basic! basic amphoteric! acidic

NetworkCovalent

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Self ionization of water (KW)• H2O + H2O → H3O+ + OH− • Using the Arrhenius definition:• H2O ⇔ H+ + OH−

• Keq = ! [H+] [OH−]! ! [H2O]

• Multiply equation by the molar concentration of water (a constant because it is pure water): [H2O]•Keq = [H+] [OH−]

• ion product for water:• Kw = [H+] [OH−] p.608

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Kw• experimentally the Kw at 25 oC has been

determined as 1.0 x 10-14

• in the equation: 1.0 x 10-14 = [H+] [OH−]

• let x = [H+] and [OH−] ! 1.0 x 10-14 = (x)(x)! x = 1.0 x 10-7 M

• [H+] and [OH−] = 1.0 x 10-7 M for water (neutral) What is the pH...?

• Assign: p. 609 #18

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pH and poh of solutions• pH = −log[H+]• See Sample Problem 19-2, p. 610• [H+] = 1.0 x 10-7 M • pH = −log (1.0 x 10-7) = +7.0• See pH scale, p. 611

• pOH = −log[OH-]• Assign: p.611 #19

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KW and PH• Acidic

pH less than 7 ! [H+] > [OH−]

• BasicpH more than 7! [H+] < [OH−]

• Neutral!pH equal to 7! ! [H+] = [OH−]

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Calculating Ka from PH • Question: In a 0.100 M solution of formic

acid (HCHO2) the pH is 2.38 at 25 oC. Calculate Ka.

• HCHO2 → H+ + CHO2−

• Ka = ! [H+] [CHO2−]

! ! [HCHO2]

• [H+] = inverse log(−pH) ! ! = inverse log(−2.38) ! ! = 0.0042 M at equilibrium

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Calculating Ka from PHHCHO2 → ! H+ + ! CHO2

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I! ! 0.100!! 0! ! 0C! −0.0042!! +0.0042! +0.0042E! ! 0.096!! 0.0042! 0.0042

Ka = (0.0042)(0.0042) = 1.8 x 10−4 at 25 oC! ! 0.096

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Calculate H+ and OH− , pH and pOH 0.367 M HNO2 with Ka = 7.1 x 10-4

! Ka = ! [H+] [NO2−]!

! ! ! [HNO2]! 7.1 x 10-4 = (x) ( x) !! ! ! 0.367! x = 1.61 x 10-2 M = [H+]

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! Kw = [H+] [OH−] = 1.0 x 10-14

! 1.0 x 10-14 = 1.61 x 10-2 [OH−]

! [OH−] = 6.2 x 10-13 M

• pH = −log [H+]! pOH = −log [OH-]! pH = 1.8! ! pOH = 12.2

! pH + pOH = 14

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complete Neutralization• Use ! NAVA = NBVB

• if an acid and a base are neutralized they must have equal H+ and OH-

• some acids have more than one H+ and some bases more than one OH-

• to account for this we use normalityWhere: NA = acid normality

= MA • (number of H+ per molecule of acid)VA = acid volume (L)NB = base normality

= MB • (number of OH− per molecule of base)VB = base volume (L)

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E.g. What volume of 2.0 M H2SO4 is required to neutralize 0.500 L of 0.50 M Al(OH)3?

N = (M)(# ions)NA = (2.0)(2) = 4.0 N! N (normal)VA = ?NB = (0.50)(3) = 1.5 NVB = 0.500 L

VA = NBVB = (1.5)(0.5) ! ! NA! 4.0! ! ! ! = 0.19 L or 190 mlAssign: p. 621 #30 - 32

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indicators• All indicators are weak acids. • Their dissociated form produces an anion

that is different color from the undissociated acid in aqueous form.

• Adding acids or alkali shifts the equilibrium and brings about color change.

• At the equivalence point, roughly equal amounts of HIn and In- are present.

• HIn (aq) ⇔ H+ (aq) + In- (aq)(red)! litmus! (blue)

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Titration LabDetermine the NaOH concentration when 10.0 ml of 0.1 M HCl neutralizes 30.0 ml of base.Use ! NAVA = NBVB

NA = (M)(# ions)= (0.1)(1)= 0.1N HCl

NB= (0.1)(10.0) (30.0)NB= 0.03N NaOH

NB = (M)(# ions) M = NB = 0.03 = 0.03M # ions 1 NaOH

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Acids bases cleanup ; )alkali - refers to bases soluble in water

Kb is base dissociation constant e.g. NH3 ! Kb = [NH4+][OH-]! ! ! [NH3]

Equations:• Kw = [H+] [OH−] = 1.0 x 10-14 @ 25 oC• pH = −log [H+]• [H+] = 10-pH

• pOH = −log [OH-]! ! [OH-] = 10-pOH

• pH + pOH = 14 (at 25 oC) • NAVA = NBVB • N = M(# ions)• % ionization = [H+]/[initial acid] x 100

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