Unit 13: Acids & Bases

ChemistryChapter 19

Welcome

To The

GowerHour

I. Bronsted-Lowry Acids and Bases:A. Acid: Substance that _________ a ___________ to another

substance.

B. Base: Substance that ________ a __________ from another substance.

C. Acid – Base reaction: ___ is transferred from the ____ to the _____.

  Ex.

 

D. Conjugate acid – base pair: The pair of acid and base that differ by a ___. Ex. ACID BASE

donates H+, proton

accepts H+, proton

H+ acidbase

↔Acid BaseAcid Base

HX + Y– HY + X–

H+

HF F–

HNO2HC2H3O2

NO2–

C2H3O2–

Conjugate Acid add H+

Conjugate Base subtract H+

Memorize: NH3 = Base

Memorize: NH4+

= Acid

E. Amphoteric substance: Substance that can either _______ or _______ a hydrogen ion.

  Ex. H2O

 

F. Example: Label the acids and bases, draw lines to connect the conjugate pairs.

(1) NH3 + H2O NH4+ + OH- (2) HF + H2O F- + H3O

+

G. Categorize each of the following as an Acid, Base, or Amphoteric and below each, write the conjugate acid or base.

  H2SO4 H2PO4- NH3 NO2

- HSO4-

donateaccept

H2O↔– H+ + H+

↔OH–

hydroxideH3O+

hydronium

SO42– ↔

– H+ + H+

↔ H2SO4HSO4– CO3

2– ↔– H+ + H+

↔ H2CO3HCO3–

(CA) (CB)Acid Base (CA)(CB)Acid Base

AcidHSO4

–

AmphoH3PO4

HPO42–

Base NH4

+

BaseHNO2

AmphoH2SO4

SO42–

Generally: H and

★Memorize★ ★

II. Water Dissociation Constant (Ion product)

A. The acidic / basic properties of _________ solutions are dependent upon the ____________ that involves the solvent, _______.

 

1. Reaction:

 

2. Equilibrium expression:

 

  3. Kw = ___________ @ 25 C. ___________ are favored.

aqueousequilibrium

water

H2O (l) + OH– (aq)↔ H+

(aq)

Acid Base

Keq = [H+][OH–] = Kw

1.0 x 10−14 Reactants

H20 very little conc. of ions

B. In pure water:      [H+] = [OH-] =  C. In impure water (contains an acidic or basic substance):   If [H+] > 1.0 x 10-7 M, solution is _______.

If [H+] < 1.0 x 10-7 M, solution is _________________.If [H+] = 1.0 x 10-7 M, solution is _________.

H2O (l) + OH– (aq)↔ H+

(aq)

Kw = 1.0 x 10−14 = [H+][OH–]x xx2 = 1.0 x 10−14 Mx = 1.0 x 10−7 M

x = 1.0 x 10−7 M x = 1.0 x 10−7 M Kw = 1.0 x 10−14 = (1.0 x 10−7 M)(1.0 x 10−7 M)

ICE

Kw = 1.0 x 10−14 = [H+][OH–] As [H+] [OH–] acidicbasic or alkaline

neutral

1.0 x 10−7 = pH 7

pH < 7 = acidpH > 7 = basepH = 7 = neutral

III. pH and pOH

A. Because [H+] and [OH-] are generally very small numbers, a _____ based system of measuring acidity is used.

  Practice: What is the pH of a solution with: [H+] = 10-5 M?

[H+] = 10-11 M? 1. pH and pOH are _________, because we cannot take the

logarithm of a unit. 2. For each pH change of 1, the [H+] changes by a factor of ____.

Ex.

3. [H+][OH-] = 1.0 x 10-14

OrpH + pOH = 14

pH = – log [H+] pOH = – log [OH–]

log

unitless

10

pH of 1 is 100 x’s more concentrated than pH of 3.Richter scale

(102 = 100)pH = 1 pH = 5 104 = 10,000

Hp

511

B. Example calculations

1. Calculate the pH of the following solutions, and indicate if the solution is acidic or basic:

(a) [H+] = 1.0 x 10-11 M

  

(b) [H+] = 2.11 x 10-2 M

 

 

(c) [OH-] = 3.98 x 10-7 M

pH = 11

pH = – log (2.11 x 10-2 ) pH = 1.68

pOH = – log (3.98 x 10-7 )pOH = 6.40

basic or alkaline

acidic

pH = – log [H+]

pH = 7.60 basic

2. Calculate the [H+] and [OH-] of the following and indicate if they are acidic or basic:

(a) pH = 9.35

(b) pH = 1.10

(c) pOH = 2.98

basicpH = – log [H+]

9.35 = – log [H+]– 9.35 = log [H+]

10– 9.35 = [H+][H+] = 4.47 x 10– 10 M

[OH–] = 2.24 x 10– 5 M

10x = antilog (opposite of log)

Divide both side by log[H+][OH-] = 1.0 x 10-14

acidic

[H+] = 7.94 x 10– 2 M

[OH–] = 1.26 x 10– 13 M

[OH–] = 1.05 x 10– 3 M

[H+] = 9.55 x 10– 12 M

10x Key

C. pH Scale:

0 7 14

Neutral

Acidic Alkaline/Basic

1 2 3 4 5 6 8 9 10 11 12 13

High quality

H2O

H2SO4

NaOHLye

Acid Rain

IV. Properties of Acids and Bases

A. Properties of Acids

1. _________ taste (ex. _________)

2. Reacts with some metals to form ___ (ex.___________)

3. __________electricity (__________ in solution)

4. React with _________

B. Properties of Bases

1. _______taste (ex.________)

2. _____________ (ex._____________)

3. ______________electricity (___________ in solution)

Sour VinegarH2 Magnesium

Conducts electrolytes

bases

Bitter Soap

Slippery Soap, NaOH

Conducts electrolytes

V. Strong vs. WeakA. Strong Acids: Completely ______ in solution (Not

_________reaction).Ex. HCl

(1)

(2)

*****MEMORIZE*****

HCl H2SO4 HNO3

HBr HI HClO4

ionizereversible

HClH2O H+

(aq)+ Cl(aq)

HCl H3O+ + Cl+ H2OChemists use H+ & H3O+ interchangeably. H+ is often used for simplicity, but H3O+ more closely represents reality.

(H3O+ = Hydronium ion)

hydrochloric sulfuric nitric

hydrobromic hydroiodic perchloric

B. Strong Bases: Completely _______ in solution. (Group I and II hydroxides)

Ex. NaOH

Practice:

Ca(OH)2

*****Memorize***** LiOH NaOH KOH

Ca(OH)2 Ba(OH)2 Sr(OH)2

ionize

H2O Na+ + OH

Ca2+ + OHH2O

NaOH + H2O

+ OHNa+ + H2O

2

C. Weak Acids: Ionize __________ (_________reaction). Generally less than _____ of the molecules ionize.

Ex. H2CO3 (1)

(2)

Example of weak acids: HC2H3O2, H2CO3

partially reversible

H2CO3 H+ + HCO3

H2CO3 H3O+ + HCO3+ H2O

If it is not a strong acid then it is a weak acid!!

1 %

H2O

D. Weak Bases: Ionize __________ (_________reaction).

Partial dissociation of an weak base in water:Ex. NH3 (1)

Example of weak bases: NH3, NH2CH2CH3

partially reversible

NH3 NH4+ + OH+ H2O

If it is not a strong base then it is a weak base!!

(aminoethane)

VI. Calculations with weak acids and weak bases.

A. Weak Acids (Ka): As Ka decreases, less ____ is formed in solution, so acid is _______.

B. Weak Bases (Kb): As Kb decreases, less _____ is formed in solution, so base is _______.

[HA]

]][A[H K K

-

aeq

H+

weaker

HA (aq) H+(aq) + A

(aq) Ka = acid-dissociation constant

OH

weaker

B (aq) BH+(aq) + OH

(aq) Kb = base-dissociation constant

↔+ H2O(l)

[B]

]][OH[BH K K

-

beq

↔H2O

4-23

a 10 x 3.61 (0.0175)

)10 x 51.2( K

C. Examples

1. Example: Aspirin is a weak organic acid whose molecular formula is HC9H7O4. An aqueous solution of aspirin is prepared by

dissolving 3.60 g/ L. The pH of this solution is found to be 2.60. Calculate Ka for aspirin.

Initial (M)

Change (M)

Equilibrium (M)

HC9H7O4 ↔ H+ + C9H7O4-

2.00 x 10-2 0 0

0.0175 2.51 x 10-3 2.51 x 10-3

– x + x + x

= 2.00 x 10-2 M = I3.60 g

Lx

1 mol

180.15 g pH = 2.60

[H+] = 2.51 x 10-3

]OH[HC

]OH][C[H K

479

-479

a

3. Compare the pH of 0.100 M HCl and 0.100 M HCN (Ka = 4.90 x 10-10)

[HCN]

]][CN[H K

-

a x)- (0.100

x 10 x 4.90

210-

0.100

x 10 x 4.90

210-

x = 7.0 x 10–6 M

pH = 5.15

0.100

- x

0.100 - x

0 0

+ x + x

x x

HCN H+ + CN ↔

I

C

E

HCl H+ + Cl [H+] = 0.100 M

pH = 1.00

Assume x is small

VII. Multiple Equilibria: When two reactions are added to give a third (net reaction) the equilibrium constants are multiplied (______________).

  Ex. HC2H3O2 Ka =

 

C2H3O2- Kb =

  

  Kw = _______________ • ___________

Equation:

K1 • K2 = Knet

↔ H+ + C2H3O2 1.8 x 10 – 5

HC2H3O2 + OH+ H2O ↔ 5.6 x 10 – 10

H2O ↔ H+ + OH–

[H+] [C2H3O2]

[HC2H3O2][OH][HC2H3O2]

[C2H3O2]

Kw = [H+][OH]

Kw = 1.0 x 10−14

Ka • Kb = Kw

A. As Ka increases, Kb ________ (_______ relationship).

 

B. The stronger the acid, the ________ its conjugate base.

 

Ex. If Ka = 1.0 x 1030, determine Kb.

decrease inverse

weaker

Kb = 1.0 x 10 – 44

Ka = strong acid; Kb = weak base

VIII. Polyprotic Acids: Certain weak acids contain more than one ionizable ___. These acids dissociate in multiple steps.

 A. Diprotic acid: Dissociates to form ___ hydrogen ions.

  Ex.

 

B. Triprotic acid: Dissociates to form ___ hydrogen ions.

  Ex.

 

 

 C. The _________ formed in one step (e.g. _______________) dissociates in the next step.

 D. The dissociation constant (___________________) becomes smaller with each successive step: Ka1 >

 E. The acids formed in successive steps become progressively ________.

H+

2

H2CO3 H+ + HCO3H2O

↔ Ka1 = 4.2 x 10 – 7

H+ + CO32-H2O

↔ Ka2 = 4.8 x 10 – 11HCO3

3

H3PO4 H+ + H2PO4H2O

↔ Ka1 = 7.5 x 10 – 3

H2PO4 H+ + HPO4

2-H2O↔ Ka2 = 6.2 x 10 – 8

HPO42- H+ + PO4

3-H2O↔ Ka3 = 4.8 x 10 – 13

substance HCO3 , H2PO4

equilibrium constantKa2 > Ka3

weaker

F. Example: Write the dissociation reactions of sulfurous acid, H2SO3.

 

 

 

 

 G. Example: Write the dissociation reactions of citric acid, H3C5H5O7.

H2SO3 H+ + HSO3H2O

↔

H+ + SO32-H2O

↔HSO3

H3C5H5O7 H+ + H2C5H5O7 H2O

↔

H2C5H5O7 H+ + HC5H5O7

2-H2O↔

HC5H5O7 2- H+ + C5H5O7

3-H2O↔

IX. Acid / Base Properties of salt solutions:A. A salt is an ____ compound not containing ___ or____.B. When a salt dissolves in water, the ___ are formed.Ex. NaCl(s)

Ex. K2CO3(s) C. Cations: Weak acids or “spectator” ions?1. Cations derived from ___________ are _________ ions.

(Do not react with water, therefore have _______ on pH.)2. Other cations are slightly _____. (Lewis acid = ______acceptor).D. Anions: Weak bases or “spectator” ions?1. Anions derived from __________are ________ ions.2. Other anions are slightly ____.

ionic H+ OH

ionNa+ + Cl

K+ + CO32-2

strong bases spectatorno effect

acidic e - pair

strong acids spectatorbasic

H2O

H2O

E. Spectator ions:

 1. Cations:

 2. Anions:

F. Example: Consider water solutions of these four salts:

(a) NH4I, (b) Zn(NO3)2, (c) KClO4, (d) Na3PO4

Classify each salt solution as acidic, basic, or neutral. (Show the dissociation reactions of each.)

Li+ ; Na+ ; K+ ; Ca2+ ; Sr2+ ; Ba2+(All others weak acids)

NO3 ; SO4

2- ; ClO4 ; Cl ; Br ; I (All others

weak bases)

a) NH4IH2O

NH4+ + I¯

(w. acid) (Spec)Acidic

b) Zn(NO3)2

H2OZn2+ + NO3¯

(w. acid) (Spec)Acidic

c) KClO4

H2OK+ + ClO4¯

(Spec) (Spec)Neutral

d) Na3PO4

H2ONa+ + PO4

3-

(Spec) (w. base)Basic

2

3

Ex. HCl H+ + Cl–

Which will undergo hydrolysis?The salts that react w/water to change the pH!

A. Strong acid + Strong base --> ____ + ____

 

Example: HCl(aq) + NaOH(aq) -->

 

Example: HNO3(aq) + Ca(OH)2(aq) -->

 

 

Net reaction:

+ Ca2+2 H+ + 2 NO3¯ + 2 OH– + 2 NO3¯Ca2+

X. Acid – Base Reactions

salt H2O

NaCl (aq)

2 Ca(NO3)2 (aq)

+ H2O (l)

+ H2O (l)2

H2OH+ + OH–

(neutralization)

+ 2 H2O (l)

B. Weak acid + Strong base ______+__________Note: Strong Base

Example: HC2H3O2 + OH-   Example: HF + OH-

C. Strong acid + Weak base __________ Note: Strong acid

Example: H+ + NH3

  Example: H+ + ClO-  

 D. Weak acid + Weak base _________+__________ 

Example: HC2H3O2 + NH3

  Example: HF + ClO-

water weak base

+ C2H3O2H2O

+ FH2O

weak acid

NH4+

HClO

weak baseweak acid

↔↔

+ C2H3O2NH4

+

+ FHClO

↔

weak base (Soln. will still be basic)

weak acid (Soln. will still be acidic)

XI. Titration: A process in which one reagent is added to another with which it reacts; an ________ is used to determine the point at which _____ quantities of the two reagents have been added.

A. Equivalence point: the point at which the _____________ reaction is complete.

B. End point: The point at which the ________ changes color.

1. Indicators change at different ______.

2. In doing a titration, one must choose an _________ where the equivalence point and the _________ coincide.

3. Example Indicators:

  Indicator Color change pH at end point

Methyl Orange ___________ ___________

Bromothymol Blue ___________ ___________

Phenolphthalein ___________ ___________

indicatorequal

neutralization

indicator

pH’s

indicatorend point

orange - red

yellow - blue

clear - pink

5

7

9

nA = nB

C. Acid – Base Indicators: Produce a _____ change in an acid-base reaction.

1. Example: Phenolphthalein, bromothymol blue

2. Natural indicators: purple cabbage, hydrangeas

 

3. Reversible reactions: (Phenolphthalein)

H-In H+ + In-

CLEAR PINK

Add acid, rxn shifts ____, solution turns _____.

Add base, rxn shifts _____, solution turns _____.

color

left clear

right pink

D. Titration Curves:

1. Strong acid – Strong Base (Equivalence point = _______)

0.100 M HCl is titrated with 0.100 M NaOH

~ pH 7

2. Weak Acid – Strong Base (Equivalence point = _______)

  0.100 M CH3COOH is titrated with 0.100 M NaOH

~pH 8.8

3. Strong Acid – Weak Base (Equivalence point = _______)

  0.100 M NH3 is titrated with 0.100 M HCl

~ pH 5.3

E. Calculations with titrations:

1. STOICHIOMETRY

  2. At equivalence point: nA = nB (mol acid = mol base)

B

BB

A

AA

n

VM

n

VM

coefficients

NaCl + H2O+ NaOHHCl

1 mol HCl

1 mol NaOH

0.2500 mol NaOH

1 L NaOH

10-3 L

1 ml

35.00 ml NaOH 1 L HCl

0.4375 mol HCl 10-3 L

1 ml

= 20.00 ml HCl

(a)

(b) (i)

1. Example titration problem: 35.00 mL of 0.2500 M sodium hydroxide is titrated with 0.4375 M HCl. (a) Write the balanced chemical equation. (b) Determine the volume of HCl added at the equivalence point (i) using stoichiometry and (ii) using the titration equation.

x x x x x

Problems:

A

BBA M

VMV

B

BB

A

AA

n

VM

n

VM

M) (0.4375

mL) M)(35.00 (0.2500VA VA = 20.00 mL HCl

(b) (ii)

2. Example: A 15.0 mL sample of a solution of H2SO4 with an

unknown molarity is titrated with 32.4 mL of 0.145 M NaOH to the bromothymol endpoint. What is the molarity of the sulfuric acid solution?

B

BB

A

AA

n

VM

n

VM

Na2SO4 + H2O+ NaOHH2SO4 2 2

BA

BBA nV

VMM

mL) 2(15.0

mL) M)(32.4 (0.145MA

MA = 0.157 M H2SO4

3. A Ca(OH)2 solution was used to titrate 15.0 mL of a 0.125 M

H3PO4 solution. If 12.4 mL of Ca(OH)2 are used to reach the

endpoint, what is the concentration of the Ca(OH)2?

Ca3(PO4)2 + H2O+ Ca(OH)2H3PO4 3 62

AB

BAAB nV

nVMM

B

BB

A

AA

n

VM

n

VM

mL) 2(12.4

mL) M)(15.0 3(0.125MB

MB = 0.227 M Ca(OH)2

IQ 21. Show the dissociation rxn of each of the following salts and

determine if each solution is acidic, basic, or neutral.

2. Predict the products and balance the following:

a. HNO3 + Sr(OH)2

b. HClO2 + OH–

a) Ca(ClO3)2

H2O Ca2+ + ClO3¯(w. base)(Spec)

Basic2

a) FeCl2

H2O Fe2+ + Cl¯(spec)(w. acid)

Acidic2

Sr(NO3)2

+ ClO2¯H2O

+ H2O22

IQ 1

1. Write formulas for two salts that:

(a) contain CO32- and are basic:

(b) contain Li+ and are neutral:

2. Predict the products and balance the following. Write the net rxn for each.

a. HNO3 + Ca(OH)2

b. HClO2 + LiOH

Net Rxn:

Ca(NO3)2

+ LiClO2H2O

+ H2O22

Ca2+ NO3-NO3

- OH-Ca2+H+

H+ + OH- H2O

LiBr Li2SO4

Li2CO3 CaCO3

Net Rxn:

ClO2- OH-Li+H+ ClO2

-Li+

H2O+ OH-HClO2 + ClO2-

Titration of a Strong Acid w/ a Strong Base1. Clean burette w/ NaOH; Drain into “waste beaker”.

2. Fill burette with NaOH to 0.00 mL (hundredths)(meniscus). Record initial vol.

3. Measure out 10.00 mL (hundredths)(meniscus) of acid using the graduated cylinder (Use the pipet). Record volume of acid; add to the Erlenmeyer flask.

4. Add 2 drops of phenolphthalein to the acid. IMPORTANT!!!!

5. Add a little water to Erlenmeyer flask (Rinse acid off sides).

6. Start adding NaOH slowly; swirl flask as you add.

7. At the endpoint (when soln. remains light pink) stop adding base and record volume of base added. (Check: rinse flask)

8. Dispose titrated solution in sink; rinse flask and repeat steps 3-8.

9. The final vol. of the base is your initial vol. for the next trial.

Reminders1. Wear safety glasses/goggles.

2. Only use the pipette for acid!

3. Base in burette

4. Clean up after 3rd trial.

5. Must get a clean up stamp before leaving.

6. Leave excess acid and base in appropriate beakers.

Titration of a Strong Acid w/ a Strong Base

NaCl + H2O+ NaOHHCl

A

BBA V

VMM

B

BB

A

AA

n

VM

n

VM

MA = ?

VA = 10.00 mL

nA = 1

MB = 0.0915 M

VB = ? mLnB = 1

Titration of Vinegar Lab

1. Fill burette with NaOH to 0.00 mL (hundredths)(meniscus). Record initial vol.

2. Measure out 1.00 mL (hundredths)(meniscus) of vinegar using the graduated cylinder (Use the pipet). Record volume of vinegar; add to the Erlenmeyer flask.

3. Add 2 drops of phenolphthalein to the vinegar. IMPORTANT!!!!

4. Add a little water to Erlenmeyer flask (Rinse acid off sides).

5. Start adding NaOH slowly; swirl flask as you add.

6. At the endpoint (when soln. remains light pink) stop adding base and record volume of base added.

7. Dispose titrated solution in sink; rinse flask and repeat steps 2-7.

8. The final vol. of the base is your initial vol. for the next trial.

Reminders1. Wear safety glasses/goggles.

2. Only use the pipette for acid!

3. Base in burette

4. Clean up after 3rd trial.

5. Must get a clean up stamp before leaving.

6. Leave excess acid and base in appropriate beakers.

Vinegar Calculations

1. Vinegar = Acetic acid (solute) in water (solvent).

2. Average number of moles of acetic acid:

3. Mass (g) of acetic acid (HC2H3O2):

4. % mass of acetic acid:

Average [ ] of acid

H2O+ NaOHHC2H3O2 + C2H3O2–

(g) acid acetic of mass OHHC of

massmolar x OHHC mol

232232

100 x solution g

solute g % 100 x

vinegarg

OHHC g 232

Density of vinegar = 1.02 g/mL

232A OHHC mol vinegar L M

Average vol. of vinegar (i.e. 1 mL)

1. Vinegar = Acetic acid (solute) in water (solvent).

2. Average number of moles of acetic acid:

3. Mass (g) of acetic acid (HC2H3O2):

4. % mass of acetic acid:

Vinegar Calculations

Lin volaverage x vinegarL

OHHC molM 232

A Average [ ]

of acid

H2O+ NaOHHC2H3O2 + NaC2H3O2

(g) acid acetic of mass OHHC of

massmolar x OHHC mol

232232

100 x solution g

solute g % 100 x

vinegarg

OHHC g 232

Density of vinegar = 1.02 g/mL

Average vol. of vinegar (i.e. 1 mL)

Neutralization Capacity of an Antacid1. Fill burette with 0.0953 M NaOH to 0.00 mL (hundredths)

(meniscus). Record initial vol.

2. Weigh ~ 0.10 g of Tums (CaCO3). Record and add to Erlenmeyer flask.

3. Add water to E. flask and swirl to dissolve powder.4. Measure out 5.00 mL (hundredths)(meniscus) of 0.336 M HCl

using the graduated cylinder (Use the pipet). Record volume of acid; add to the E. flask containing antacid and swirl.

5. Add a little water to E. flask (Rinse acid off sides).6. Add 2 drops of phenolphthalein to the acid.

IMPORTANT!!!!7. Start titrating; swirl E. flask as you add NaOH.8. At the endpoint (when soln. remains light pink) stop adding

base and record volume of base added.9. Dispose titrated solution in sink; rinse flask and repeat steps 2-7.10. The final vol. of the base is your initial vol. for the next trial.

Reminders1. Wear safety glasses/goggles.

2. Only use the pipette for acid!

3. Base in burette

4. Clean up after 3rd trial.

5. Must get a clean up stamp before leaving.

6. Leave excess acid and base in appropriate beakers.

Antacid CalculationsAcid (HCl) (VA)

VA1 = Volume of acid neutralized by the antacid.

Antacid (CaCO3)(VA1)

B

BB

A

A2A

n

VM

n

VM

Step 1: Solve for VA2

NaOH (VA2)

VA2 = Volume of acid neutralized by the NaOH.VA = Total volume of acid (5 mL). VA = VA1 + VA2

Step 2: Solve for VA1

VA = VA1 + VA2

Antacid Calculations

Step 3: Solve for VA1/ g.

g

VA1

Step 4: Solve for VA1/ Tablet

Antacid : 1 Tablet = 1.30 g

Mass of antacid used in the trial (i.e. 0.10)

Tablet

g 1.30 x

g

VA1

Step 5: Solve for VA1/ centAntacid : 72 Tablet = $5.49

cents

Tablet x

Tablet

VA1

Labs Due: Tuesday 05/15

Tablet

V A1

cents

V A1

Acid Nomenclature Review

1. Acetic acid (w) 11. Carbonic acid (w)

2. Oxalic acid (w) 12. Perchloric acid (s)

3. Hydrocyanic acid (w) 13. Hypobromous acid (w)

4. Cyanic acid (w) 14. Nitric acid (s)

5. Sulfurous acid (w) 15. Chloric acid (w)

6. Sulfuric acid (s) 16. Hydrofluoric acid (w)

7. Hypochlorous acid (w) 17. Phosphorous acid (w)

8. Bromous acid (w) 18. Hydroiodic acid (s)

9. Periodic acid (w) 19. Nitrous acid (w)

10. Phosphoric acid (w) 20. Hydrochloric acid (s)