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Lecture 7- Complex Acid - Chemistry Laboratory | Trường Đại học Quốc tế, Đại học Quốc gia Thành phố HCM
Lecture 7- Complex Acid - Chemistry Laboratory | Trường Đại học Quốc tế, Đại học Quốc gia Thành phố HCM được sưu tầm và soạn thảo dưới dạng file PDF để gửi tới các bạn sinh viên cùng tham khảo, ôn tập đầy đủ kiến thức, chuẩn bị cho các buổi học thật tốt. Mời bạn đọc đón xem!
Chemistry Laboratory (CH012IU) 59 tài liệu
Trường Đại học Quốc tế, Đại học Quốc gia Thành phố Hồ Chí Minh 696 tài liệu
Lecture 7- Complex Acid - Chemistry Laboratory | Trường Đại học Quốc tế, Đại học Quốc gia Thành phố HCM
Lecture 7- Complex Acid - Chemistry Laboratory | Trường Đại học Quốc tế, Đại học Quốc gia Thành phố HCM được sưu tầm và soạn thảo dưới dạng file PDF để gửi tới các bạn sinh viên cùng tham khảo, ôn tập đầy đủ kiến thức, chuẩn bị cho các buổi học thật tốt. Mời bạn đọc đón xem!
Môn: Chemistry Laboratory (CH012IU) 59 tài liệu
Trường: Trường Đại học Quốc tế, Đại học Quốc gia Thành phố Hồ Chí Minh 696 tài liệu
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Analytical Chemistry 1 Lecture 7
Complex Acid-Base Titration
Instructor:*Nguyen*Thao Trang Complex(Acid-base(titration
• (1)(two(acids(or(two(bases(of(different(strengths,
• (2)(an(acid(or(a(base(that(has(two(or(more(acidic(or(basic( functional(groups,(or •
(3)(an(amphiprotic substance,(which(is(capable(of(acting(as( both(an(acid(and(a(base. 2
Mixtures)of)Strong)and)Weak)Acids)or)Strong)and)Weak)Bases
• The(dissociation(constant(for(the(weak(acid(or(base(is(
somewhat(less(than(about(1.00(×10–4
Ex.*pH*of*0.1200*M*HCl and*0.0800*M*weak*acid*HA* (Ka=*1.00*×10–4) [H ](=( 3O+ CHCl +([A– – ](=(0.1200(+([A ]
If([A–](<<(0.1200(M,(([H ]( 0.1200( 3O+ ≈ M à pH(=(0.92 3
Mixtures of Strong and Weak Acids or Strong and Weak Bases
• The(dissociation(constant(for(the(weak(acid(or(base(is(
somewhat(less(than(about(1.00(×10–4. Check(assumption:
Ka=([H3O+]([A–]((/([HA](=(1.00(×10–4
[A–]((/([HA](=(Ka /([H3O+](=(1.00(×10–4( /(0.1200(=(8.33(×10–4(
[HA](=([A–]((/(8.33(×10–4(
CHA =([HA](+([A–]((=(0.0800(M(=({[A–]((/(8.33(×10–4(}(+([A–]((≈ (1.20×103()([A–](=(0.0800(M
à [A–](=(6.7(×10–5(M( <<(0.1200(M 4
Mixtures)of)Strong)& Weak)Acids)or)Strong)&)Weak)Bases
• When*25.00*mL*mixture*of*HCl*and*HA*is*titrated*with*a*strong* base:
– In)the)presence)of)5.00mL)OH- 0.1000)M: H+ +))))))))OH – ⇌ H O 2
5.00(x(0.1000(((((((((((((((((((((((((((((((((((((((mol
CHCl =((25.00(x(0.1200(- 5.00(x(0.1000)/(25.00(+(5.00)(=(0.0833(M
[H3O+]( =(CHCl +([A-](~(0.0833(M(à pH(=(1.08
We(can(check(if(the(assumption( is(still(valid:
[HA]( =(0.0800(x(25.00/30.00(=(0.0667(M
à [A-](=(8.0(x10-5 M(<<(0.0833(M 5
Mixtures)of)Strong)& Weak)Acids)or)Strong)&)Weak)Bases
– In)the)presence)of)29.00mL)OH- 0.1000)M: H+ +))))))))OH – ⇌ H O 2
29.00(x(0.1000(((((((((((((((((((((((((((((((((((((((mol
CHCl =((25.00(x(0.1200(- 29.00(x(0.1000)/(25.00(+(29.00)(=(1.85(x(10-3 M
[H3O+]( =(CHCl +([A-](~(1.85(x(10-3 M
We(can(check(if(the(assumption(is(still(valid:
CHA=(0.0800(x(25.00/54.00(=(3.7(x(10-2 M
à [A-](=(1.90(x10-3 M,(no(longer(<<(CHCl. 6
Mixtures)of)Strong)& Weak)Acids)or)Strong)&)Weak)Bases
– In)the)presence)of)29.00mL)OH- 0.1000)M:
[H3O+]( =(CHCl +([A-](=(1.85(x(10-3 +([A-]( (1)
CHA =([HA]( +([A-]( =(3.7(x(10-2 M( (2) [HA]( =([H3O+][A-]/K [H ][A a( =( 3O+ -]/(1.00( x(10-4 )( (3)
Substitute( (3)(into((2)(yields: [H ][A ]/K [A 3O+ - a( +( -]( =(3.7(x(10-2 M(
à [A-]( =(3.7(x(10-2 M(/([H3O+]( +(1.00(x(10-4 )( (4)
Substitute( (4)(into((1)(yields:([H ](
3O+ =( 3.03(x(10-3 M(à pH(=2.52 7
Mixtures)of)Strong)& Weak)Acids)or)Strong)&)Weak)Bases
• When*25.00*mL*mixture*of*HCl*and*HA*is*titrated*with*a*strong* base:
– When(the(amount(of(the(base( NaOH(=(amount(of(HCl(present:( the(
solution(is(identical(to(the(solution(containing(the(weak(acid(HA(and( NaCl.
– The(remainder( of(the(titration(curve( is(identical(to(that(for(a(diluted( HA.
– The(shape( of(a(titration(curve(varies( (depending( on(the(strength(of(the( weak(acid). 8
Mixtures)of)Strong)& Weak)Acids)or)Strong)&)Weak)Bases Curves for the titration of strong acid /weak acid mixture (with different Ka of the waek acid) with 0.1000 M NaOH. Two( Each titration is on 25.00 equivalence( mL of a solution that is points 0.1200 M in HCl and Small/nonexistent( 0.0800 M in HA. equivalence( point à The composition of a mixture of a strong acid with a weak acid can be determined if the Ka of the weak acid lies between 10-4 - 10-8. 9
Polyfunctional Acids and Bases
The)phosphoric)acid)system • H O ⇌ H – + – + 3PO4 +(H + H O ; K ][H O ]( /([H 2 2PO4 3 a1 =( [H2PO4 3 3PO4]( =(7.11 ×10–3( • H – 2( – 2( – + – 2PO4 +(H O ⇌ HPO + H O+ ; K ][H O ]( /([H 2 4 3 a2 =([HPO4 3 2PO4 ]( =(6.32×10–8( • HPO 2(– 3(– 3(– 2( – 4 +(H O ⇌ PO + H O+ ; K =([PO ][H O+]( /([HPO ]( 2 4 3 a3 4 3 4 =(4.5 ×10–13 with(K K K a1 >( a2 >( a3 10
Polyfunctional Acids and Bases
The)phosphoric)acid)system
Equilibrium(constant(Keq for(the(overall(reaction: H 2H 2– 3PO4 +( O ⇌ HPO + 2H O+ 2 4 3
àTwo(adjacent(stepwise(equilibria:(K K 2(– a1 a2 =([HPO4 ][H O+]2 /( 3 [H –8(
3PO4](=((7.11 ×10–3()× (6.32×10 )=(4.49×10–10( H 3H 3– 3PO4 +( O ⇌ PO + 3H O+ 2 4 3
à Three(adjacent(stepwise(equilibria: K 3(– a1Ka2Ka3 =([PO4 ](
[H O+]3/[H PO ](=((7.11 ×10–3()× (6.32×10–8 ( –
) × 4.5 ×10 13) =(2.0×10–22 3 3 4 11
Polyfunctional Acids and Bases
The)carbon)dioxide/carbonic)acid)system • CO2((aq)(+(H O ⇌ H (1) 2 2CO3 Khyd =([H2CO ](/([CO 3 2(aq)](=(2.8 ×10–3( à [CO2(aq)](≫ [H2CO ]( 3 • H – 2CO +(H O ⇌ HCO + H O+ (2) 3 2 3 3 K – 1 =([HCO ]([H O+](/([H ](=(1.5×10–4( 3 3 2CO3
• HCO – +(H O ⇌ CO 2(– + H O+ (3) 3 2 3 3 K 2(– – 2 =([CO
][H O+](/([HCO ](=(4.69 ×10–11 3 3 3 12
Polyfunctional Acids and Bases
The)carbon)dioxide/carbonic)acid)system
Since([CO2(aq)](≫ [H2CO ](,(the(acidity(of(solutions(of(carbon( 3
dioxide(is(obtained(by(combining( 1)(&((2): • CO2(aq)(+(2H O ⇌ HCO – + H O+(((( 2 3 3 K – a1 =([HCO ]([H O+](/([CO × 4() 3 3 2(aq)]( (=((2.8 ×10–3 – ) × (1.5 10 =(4.2(×10–7( and
• HCO – +(H O ⇌ CO 2(– + H O+ 3 2 3 3 K 2(– – a2 =([CO
][H O+](/([HCO ](=(4.69 ×10–11 3 3 3 13
Polyfunctional Acids and Bases
Ex.)Calculate)the)pH)of)a)solution)that)is)0.02500M)CO2
The(mass(balance( for(CO2 containing(species: C ( – 2(– CO2 =([CO2 aq)]( +([H ]( +([HCO ]( +( 2CO [CO ](=(0.02500(M 3 3 3 Since([CO – 2( – 2(aq)]( ≫ [H ]( ,([HCO ],( ([CO ],( 2CO3 3 3 C ( CO2 ~([CO2 aq)]( =(0.02500(M
The(charge(balance( (equation(is: [H O+]( = [
( HCO –]( +(2 [CO 2(– ](+([OH – ] 3 3 3
Assume( [HCO –]( ≫ 2 [CO 2(– ](+([OH – ],(then:([H O+]( (~([HCO –]( 3 3 3 3 K – a1 =([HCO ]( [H O+]( /([CO 3 3 2(aq)]( =(4.2(×10–7(
à [H O+]( =(0.02500(× 4.2(×10–7( =(1.02(×10–4( M( 3 àpH(=(3.99 (Determine( [H ],( ([CO 2(– ],( 2CO
[OH – ](to(verify( the(assumption.) 3 3 14
Buffer involving polyprotic acids
- Buffer(composed(of(a(weak(diprotic(acid(H2A(and(its(salts( (NaHA(and(Na2A):
• 1st)system:(the(acid(H2A(and(its(conjugate(base(NaHA
– The(dissociation( of(HA- to(yield(A2- can(be(neglected.
– The(calculation(is(based( on(the(dissociation( of(H2A.(
• 2nd)system:(the(acid(HA- (or(NaHA)(and(its(conjugate(base( Na2A
– The(dissociation( of(HA- to(yield(A2- is(dominated.
– [H2A]( <<([HA-]( or([A2-]
• pH(of(system(2(>(pH(of(system(1(as(Ka( (H2A)>(Ka((HA-) 15
Buffer involving polyprotic acids
Ex.$Calculate$[H3O+]$of$$mixture$of$2.00$M$phosphoric$acid$H3PO4$+$1.50$M
potassium$dihydrogen phosphate$ H2PO – 4 • H O ⇌ H – 3PO4 +(H + H O+ 2 2PO4 3 K – + a1 =([H2PO4 ]( [H O ]( /([H 3 3PO4]( =(7.11 ×10–3( [H O+] =(K [H – 3 a1 × 3PO4]( /([H ]( 2PO4
=(7.11 ×10–3(× (2.00(/(1.50) =(9.48(×10–3(M • H – 2( – 2PO4 +(H O ⇌ HPO + H O+ 2 4 3 K 2( – – a2 =([HPO4 ][H O+]( /([H ]( =(6.32×10–8( 3 2PO4
Ka1 >>(Ka2:(Neglect(2nd dissociation •
Check(the(assumption:( [HPO 2(– – 4 ]=(K [H ]( / [H a2 2PO4 O+]( =1.00×10–5 M( 3 à valid! 16
Buffer involving polyprotic acids
Ex.$Calculate$[H O+]$of$$a$buffer$that$is$0.0500$M$is$KHP$and 3 0.150$M$in$K2P. • H2P(+(H2O(( ⇌ HP – + H O+ 3
Ka1 =([HP–]([H O+](/([H ]( 1.12 10–3 3 2P =( × • HP – +(H O ⇌ P2– + H O+ 2 3
Ka2 =([P2–]([H O+](/([HP – ](=(3.91 ×10–6 3
Assume([HP – ](~([KHP](=(0.0500(M;([P2–](~([K2P](=(0.150(M à [H O+]( =(1.30(×10–6 3 Check(the(assumption: [H P + 2 ]=([HP – ][H O ]/(K
6.00 10–5 M(<<([HP –],([P2–]( 3 a1 = × à valid! 17
pH of an amphiprotic salt NaHA
Calculate the pH of solutions of salts that have both acidic and
basic properties ~ amphiprotic salts formed during titration of
polyfunctional acids and bases. Ex: NaOH + H A ⇌ NaHA + H 2 2O
The pH of the solution is determined by: HA- + H2O ⇌ A2– + H O+ , K 3 a2 or HA- + H2O ⇌ H2A+ OH- , Kb2
The relative magnitudes of the Ka2 and Kb2 will determine if the solution is acidic or basic: [H + 2- 3O ] [A ] Ka2 = [HA-] K - w [H2A] OH ] Kb2 = = Ka1 [HA-] 18
pH of an amphiprotic salt NaHA • Mass(balance( equation:
CNaHA =([HA –](+([H2A]( +(([A2–] (1) • Charge(balance( equation: [Na+](+([H [A ](+([OH 3O+](=( [HA –]( +(2( 2– –] Since([Na+]( =(C +([H ]((+(2([A ](+([OH NaHA à CNaHA( 3O+]( =([HA – 2– –] (2) (2)(– (1)(yields:([H –
3O+]( =([A2–](+( [OH ]( - [H2A]( (3)
Ka1 =([H3O+] [HA –](/ [H2A]( à [H A]( 2 =([H ]( 3O+]( [HA – / Ka1
Ka2 =([H3O+]( [A 2–](/([HA –]à [A 2–](= K ]( a2 [HA – /[H3O+](
Kw =([H3O+]([OH–]( à [OH–]( =(Kw / [H3O+]( From((3): [H ](/[H + + 3O+]( =(Ka2 [A 2– 3O+]( +(Kw /[H3O ]( - [H O 3 ]([HA –](/(Ka1 (–]("#K à [H w 3O+]( = Ka2HA ( $#"#[HA –]/Ka1 19
pH of an amphiprotic salt NaHA
pH(of(NaHA(is(determined(by(assuming([HA –](~ CNaHA [H3O+](= Ka2CNaHA(("#Kw $#"#CNaHA/Ka1 When(C ≫ ≫ NaHA/Ka1 1(and(Ka2CNaHA Kw : [H3O+](= Ka1Ka2 20
pH of an amphiprotic salt NaHA 21
pH of an amphiprotic salt NaHA 22
Titration curves for polyfunctional acids
• Compounds with two or more acid functional groups yield
multiple end points in a titration curve.
• If the ratio Ka1/Ka2 > 103, the theoretical titration curve for the
polyprotic acids can be calculated as that for the monoprotic acids. •
2 usable end points appear if the ratio of dissociation
constants > 104 and if the weaker acid or base has K K a or b > 10-8 23
Titration curves for polyfunctional acids
Ex.(Construct(the(titration(curve(for(the(titration(of(25.00(mL(of(
0.1000(M(maleic(acid((HOOC-CH=CH-COOH,(Ka1 =((1.3(x(10-2;(Ka2 =((
5.9(x(10-7)(with(0.1000(M(NaOH.
Ka1/Ka2 >(103,(the(titration(curve( can(be(calculated(as(for(simple(monoprotic(
weak(acids (except( for(the(1st(equivalence( point). Titration(reactions: H2A +(NaOH ⇌ NaHA +(H2O((1st)
NaHA +(NaOH ⇌ Na2A +(H2O(((2nd) 24
Titration curves for polyfunctional acids
Ex. Construct the titration curve for the titration of 25.00 mL of
0.1000 M maleic acid (HOOC-CH=CH-COOH, Ka1 = 1.3 x 10-2; K = a2
5.9 x 10-7) with 0.1000 M NaOH.
Ka1/Ka2 > 103, the titration curve can be calculated as for simple monoprotic
weak acids (except for the 1st equivalence point). • Before the titration:
The 1st dissociation dominates and the 2nd dissociation is negligible: à [H [A 3O+] ~ [HA –] ; 2–] ~ 0 Mass balance:
CH2A = [H2A] + [HA –] + [A2–] ~ [H2A] + [HA –] = [H2A] + [H ] 3O+ à [H2A] = 0.1000 - [H3O+] Ka1 = [HA –] [H = 3O+] / [H2A]
[H3O+]2 /(0.1000 - [H3O+]) = 1.3 x 10-2
à [H3O+] = 3.01 x 10-2 M à pH = 1.52 25
Titration curves for polyfunctional acids •
Start the titration VOH- = 5.00 mL : buffer region H st 2A + NaOH ⇌ NaHA + H2O (1 )
NaHA + NaOH ⇌ Na nd 2A + H2O (2 )
Buffer mainly consists: H2A and HA – (1st dissociation dominates and the 2nd dissociation is negligible):
CNaHA ~ [HA –] = (5.00 mL)(0.1000 M)/(25.00 + 5.00 mL) = 1.67 x 10-2 M
[H2A] = (25.00 - 5.00 mL)(0.1000 M)/ (25.00 + 5.00 mL) = 6.67 x 10-2 M
Ka1 = [HA –] [H3O+] / [H2A] = 1.3 x 10-2
à [H3O+] = 5.2 x 10-2 M , which is not ≪ CH2A and CHA– Then,
[H2A] = CH2A - [H3O+] + [OH-] ~ CH2A - [H3O+] = 6.67 x 10-2 - [H3O+] ;
[HA –] = CNaHA + [H3O+] - [OH-] ~ CNaHA - [H3O+] = 1.67 x 10-2 +[H3O+] . à [H3O+] = 1.81 x 10-2 M à pH = 1.74
Adding more OH-, the pH is calculated in a similar way! 26
Titration curves for polyfunctional acids •
Right before the equivalence point, VOH- = 24.90 mL H2A +(NaOH ⇌ NaHA +(H2O((1st)
NaHA +(NaOH ⇌ Na2A +(H2O(((2nd)
[H2A] is so small, comparable to [A2–], the 2nd dissociation should be considered:
CNaHA ~ [HA –] = (24.90 mL)(0.1000 M)/(25.00 + 24.90 mL) = 4.99 x 10-2 M
[H2A] = (25.00 – 24.90 mL)(0.1000 M)/ (25.00 + 24.90 mL) = 2.00 x 10-4 M
Mass balance: CH2A + CNaHA = [H2A] + [HA –] + [A2–] (a ; ) Charge balance: [Na+] + [H [A 3O+] = [HA –] + 2 2– - ] + [OH ] (b)
Close to Ve, the solution mainly consists of HA – (OH- can be negelected), from (b): [Na+ – +
] ~ CNaHA ~ [HA ] + 2[A2–] - [H3O ] (c) Substitute (c) into (a): [H [A 3O+] = CH2A + 2–] - [H2A] = C + K H2A
a2 [HA –] / [H3O+] + [H3O+] [HA –] / Ka1 à [H3O+] = 1.014 x 10-4 M à pH = 3.99 27
Titration curves for polyfunctional acids •
At)the)1st equivalence) point,) VOH- =)25.00)mL H2A +(NaOH ⇌ NaHA +(H2O((1st)
NaHA +(NaOH ⇌ Na2A +(H2O(((2nd) C [
NaHA( ~(( HA –]( =(((25.00(mL)(0.1000(M)/(25.00( +(25.00(mL)(=( 5.00(x(10-2(M [H3O+]( = Ka2C (( NaHA "#Kw ( $#"#[HA –]/Ka1 ) $.00(x(10−$+ à [H3O+]( = 5.'(x(10−( (5.00(x(10−2 (+ =(7.80(x(10-5(M
$#"(5.'(x(10−()($.-(x(10−.) à pH(=(4.11 28
Titration curves for polyfunctional acids •
Just)after)the)equivalence) point,) VOH- =)25.01)mL H2A +(NaOH ⇌ NaHA +(H2O((1st)
NaHA +(NaOH ⇌ Na2A +(H2O(((2nd)
nH2A( =(nNaHA((in(1)(=((25.00(mL)(0.1000(M)=(2.500(mmol
nOH- excess(=((25.01(- 25.00)(0.1000(M)=( 0.001000(mmol nNaHA((in(2)(=( N
n a2A(=(nOH- excess(=(0.001000(mmol
à nNaHA(excess(=(nNaHA((in(1)(- nNaHA((in(2)(=(2.499(mmol
à cNaHA(excess(=(((2.499(mmol)/(25.00(+(25.01(mL)(=(4.997(x(10-2(M
cNa2A((=(((0.001(mmol)/(25.00( +(25.01(mL)(=(2.000(x(10-5(M 29
Titration curves for polyfunctional acids •
Just)after)the)equivalence) point,) VOH- =)25.01)mL H2A +(NaOH ⇌ NaHA +(H2O((1st)
NaHA +(NaOH ⇌ Na2A +(H2O(((2nd)
The(solution(is(mainly(composed(of(NaHA)and)Na2A ,( Mass(balance:( C [H A 2– Na2A +( C =( NaHA( 2 ]( +([HA –](+( [A
](=(4.997(x(10-2(+ 2.000(x(10-5( =(4.999(x(10-2(M Charge(balance:( [Na+]+[H ]+2[A [OH 3O+]( =([HA – 2–]( +( -]à OH- can(be(negelected.
[Na+](=nOH-/V( =((25.01mL)((0.1000(M)/(25.00(+(25.01(mL)((=(5.001(x(10-2(M
Solving(([H3O+](from(the(mass(and(charge(balance(eq.( : [H3O+] =([A2–]( - [H2A]( +(C +(C Na2A NaHA - [Na+] =((K / +( a2 [HA –]( [H3O+]( [H ](
3O+]( [HA – / Ka1((+(CNa2A +(CNaHA - [Na+] Substitute( the(values(of(K +
a1(,(Ka2(,(CNa2A +(CNaHA and([Na ]: à [H -5( 3O+] =(7.60((x(10 M( à pH(=(4.12 30
Titration curves for polyfunctional acids •
Second)buffer)region,) VOH- =)25.50)mL) H2A +(NaOH ⇌ NaHA +(H2O((1st)
NaHA +(NaOH ⇌ Na2A +(H2O(((2nd)
Buffer( mainly(consists:(HA – and(A2 – (the(reaction( HA – +(H H 2O(⇌ 2A(+(OH – can( be(neglected )
nOH- excess(=((25.50(- 25.00)(0.1000(M)=( 0.05000(mmol
nNaHA((in(2)(=(nNa2A(=(nOH- excess(=(0.005000(mmol à nNaHA(excess(=( N
n aHA((in(1)(- nNaHA((in(2)(=(0.04500(mmol
à cNaHA(excess(=(((0.04500(mmol)/(25.00(+(25.50(mL)(=(8.911(x(10-4(M
cNa2A((=(((0.005(mmol)/(25.00( +(25.50(mL)(=(9.901(x(10-5(M Ka2(=([H ][ ](=5.9( 3O+ A2 – ]/ [HA – x(10-7(M
à [H3O+] =2.89(x(10-5(M(,(which(is(≪ CA2 – and(CHA– à pH(=(4.54(
Adding(more(OH-,(the(pH(is(calculated(in(a(similar(way! 31
Titration curves for polyfunctional acids •
Right)before)the)2nd equivalence) point,) VOH- =)49.90)mL and) more H2A +(NaOH ⇌ NaHA +(H2O((1st)
NaHA +(NaOH ⇌ Na2A +(H2O(((2nd) [A2–] [ / HA –](becomes( large!( C (
NaHA(=( 25.00(mL)(0.1000(M)(– (24.90(mL)(0.1000(M)/(25.00( +(49.90(mL)( =(1.335(x(10-4(M
CA2– =((24.90)(0.1000(M)/(25.00(+(49.90(mL)(=(3.324(x(10-2(M
The(primary(equilibrium( now(is: A2– +((((((((((H2O((⇌ HA – +((((OH –
(3.324(x(10-2(– x)((((((((((((((((((((((((((((((x(((((((((((((x
Kb1 =Kw/Ka2=[HA –].[OH –]/ [A2–]( =(1.335(x(10-4(+(x)x/(3.324(x(10-2(– x)=1.69(x10-8
à x(=([OH –](=(4.10(x(10-6 M(à pOH((=(5.39 à pH(=(8.61 32
Titration curves for polyfunctional acids •
At)the)2nd equivalence)point,) VOH- =)50.00)mL H2A +(NaOH ⇌ NaHA +(H2O((1st)
NaHA +(NaOH ⇌ Na2A +(H2O(((2nd)
All(HA – reacts(completely(with(NaOH(to(produce( A2–
CA2– =((25.00)(0.1000(M)/(25.00(+(50.00(mL)(=(3.333(x(10-2(M
The(primary(equilibrium( now(is: A2– +((((((((((H2O((⇌ HA – +((((OH –
(3.333(x(10-2(– x)((((((((((((((((((((((((((((((x(((((((((((((x K 2–
b1 =Kw/Ka2=[HA –].[OH –]/[A
]( =x2/(3.333(x(10-2(– x)=1.69(x(10-8
à x(=([OH –](=(2.37(x(10-5 M(à pOH((=(4.62 à pH(=(9.38 33
Titration curves for polyfunctional acids •
At)just) beyond)the)2nd equivalence)point,) VOH- =)50.01)mL H2A +(NaOH ⇌ NaHA +(H2O((1st)
NaHA +(NaOH ⇌ Na2A +(H2O(((2nd)
CA2– =((25.00)(0.1000(M)/(25.00(+(50.01(mL)(=(3.332889 x(10-2(M
[OH –]excess =((50.00(- 50.01(mL)/(25.00(+(50.01(m ) L (=(1.333(x(10-5(M
The(primary(equilibrium( now(is: A2– +((((((((((H2O((⇌ HA – +((((OH –
(3.332889 x(10-2(– x)((((((((((((((((((((((((((x(((((((((((((x K 2– -5( b1=Kw/Ka2=[HA –].[OH –]/[A
]( =x (x(+(1.333(x(10 )/(3.332889 x(10-2(– x) à x(=(1.807(x(10-5(M
à [OH –](=(1.807(x(10-5(+(1.333(x(10-5((=(3.140(x(10-5( à pOH((=(4.50( à pH(=(9.50 34
Titration curves for polyfunctional acids •
At)beyond) the)2nd equivalence) point,) VOH- =)51.00)mL H2A +(NaOH ⇌ NaHA +(H2O((1st)
NaHA +(NaOH ⇌ Na2A +(H2O(((2nd)
CA2– =((25.00)(0.1000(M)/(25.00(+(51.00(mL)(=(3.289(x(10-2(M
[OH –]excess =((51.00(- 50.00(mL)(0.1000)/(25.00(+(51.00(mL)(=(1.316(x(10-3(M
The(primary(equilibrium( now(is: A2– +((((((((((H2O((⇌ HA – +((((OH –
(3.289(x(10-2(– x)((((((((((((((((((((((((((((((x(((((((((((((x K 2– -3( b1=Kw/Ka2=[HA –].[OH –]/[A
]( =x (x(+(1.316(x(10 )/(3.289(x(10-2(– x) à x((=(1.807(x(10-5(
à [OH –](=(1.807(x(10-5(+(1.316(x(10-3((~ 1.316(x(10-3( M à pOH((=(2.88 à pH(=(11.12 35