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)
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