Acids and Bases Ch 15
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Ch 15: Acids and Bases
Homework:
Read Chapter 15 Work out sample/practice exercises in the sections,
Bonus problems: 39, 41, 49, 63, 67, 83, 91, 95, 99, 107, 111, 115, 117, 123, 139
Check for the MasteringChemistry.com assignment and complete before due date
Acids, Bases, Salts:
Acids, bases and salts are very important and perform many essential functions.
Digestive juices (0.1 M HCl; needed to kill bacteria, break down food and
activate enzymes)
pH buffers; Households and Industrial uses
Drain cleaner (NaOH)
fertilizer (NH4NO3)
Car battery acid (40% H2SO4)
Table salt as a food preservative or for flavor (NaCl)
Example 1: Come up with more common acids, bases, or salts and their uses.
Acids and Bases Ch 15
A few common acids, their uses and relative strength:
A few common bases, their uses and relative strength:
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Acids and Bases Ch 15
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Review Electrolytes and Double Displacement Reactions:
Electrolytes:
 Nonelectrolyte: A molecule or substance that remains whole in aqueous
solutions, it cannot split apart. Examples: any insoluble solid, gas (CO2, O2,
SO2), and molecules (sugar, CH4, H2O)
 Weak Electrolyte: An ionic substance that will partially ionize into its separate
ions in aqueous solution. Examples: Weak acids (HF, HC2H3O2) and Weak
bases (NH4OH, CH3NH2 ), and slightly soluble solids (PbCl2) . Partial ionization
is an equilibrium reaction in which the reactant is favored, K<1 ; HF (aq) 
H+1 (aq) + F-1 (aq).
 Strong Electrolyte: An ionic substance that completely dissociates into its ions
in aqueous solution. Examples: Strong Acids (HCl), Strong Bases (NaOH),
Soluble Salts (KBr); K>>1; HCl (aq)  H+1 (aq) +Cl-1 (aq).
Titration: Reacting a solution of unknown concentration with a known (standard)
concentration, stopping the titration when an indicator (phenolphthalein) changes
color.
Example 2:
Write the dissociation reactions in the solvent water for the following
substances. Strong electrolytes will have (), weak electrolytes have (), and
nonelectrolytes are no reaction (NR).
FeCl3 (aq)
HNO3 (aq)
Sr(OH)2 (aq)
CH3OH (aq)
HF(aq)
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Double Displacement Reactions:
Double Displacement reactions have two ionic reactants. Reactants will
exchange ions in making products… AB + CD  AD + CB
A reaction occurs if a nonelectrolyte (Solid, Liquid (H2O), Gas), or Weak
Electrolyte is formed as one or more of the products. If all the reactants and the
products are strong electrolytes, then no reaction takes place. Review the
Solubility Rules.
a) Whole or Molecular equation
2 AgNO3 (aq) + CaCl2 (aq)  Ca(NO3)2 (aq) + 2 AgCl (s)
b) Total Ionic Equation with Spectator Ions
2 Ag+1 (aq) + 2 NO3-1 (aq) + Ca+2 (aq) +2 Cl-1 (aq) 
Ca+2 (aq) + 2 NO3-1 (aq) + 2 AgCl (s)
c) Net Ionic Reaction
Ag+1 (aq) + Cl-1 (aq)  AgCl (s)
Example 3:
For the Double Displacement reactions, write the whole, total ionic (circling
the spectator ions) and net ionic equations given the reactants. Identify the type of
reaction as Precipitation, Neutralization, or No Reaction.
a) HCl (aq) + NaOH (aq)
b) NH4OH, same as NH3 (aq) + H2S (aq)
c) Al(NO3)3 (aq) + KCl (aq)
d) K2S (aq) + Zn(ClO3)2 (aq)
Acids and Bases Ch 15
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Review Acids, Bases and Salts :
Acids:
Properties:
 Taste Sour
 Reacts with “active” metals to liberate H2:
Zn (s) + 2 HCl(aq)  H2(g)+ ZnCl2(aq)
 Reacts with carbonates to liberate CO2:
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)
 React with bases to form ionic salts (neutralize):
NaOH (aq) + HCl (aq)  NaCl (aq) + H2O (l)
 Conduct electrical current
 Certain dyes change color with acids (litmus-red)
 Acids ionize in water to increase the H+1 ion concentration
Nomenclature:
a) Binary acids, those anions ending in ide: Hydro root ic acid;
(HCl), Hydrochloric Acid; (H2S), hydrosulfuric acid
b) Ternary oxyacids: if anion ends with ate, root ic acid; (HNO3),
nitric acid, if anion ends with ite, root ous acid, (HNO2), nitrous
acid
Bases (also known as alkalis):
Properties:
 Taste bitter
 Slimy to the touch
 Conducts electricity
 Certain dyes change color with bases (litmus-blue)
 React with acids to form ionic salts (neutralize):
 Ionize in water to increase the OH-1 ion concentration
Nomenclature: metal name + hydroxide; (NaOH), Sodium Hydroxide; or
common name (NH3), ammonia, organic bases with N are amines,
(CH3NH2), methyl amine
Salts:
The ionic substances not readily identified as an acid or base, some are very
soluble, others only slightly soluble in water.
Nomenclature: cation name + anion name; (CuCl2) copper (II) chloride,
(NH4)2SO4 , ammonium sulfate, (NaCl), sodium chloride.
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Example 4: Fill in the Table with names or formulas.
Name
Formula
Hydrosulfuric acid
Ethyl amine
Potassium fluoride
MgHCO3
H2C2O4
Ba(OH)2
Acids and Bases Defined :
Arrhenius Definition (1884): Most limited definition requires water.
Acid: Substance that will increase the H+1 ion concentration in an aqueous
solution. (HF)
Base: Substance that will increase the OH-1 ion concentration in an aqueous
solution. (KOH)
Neutralization is the combination of an acid with a base to form water and a
salt.
Bronsted Lowry Definitions (1923): broader definition, more base possibilities
Acid: Donates one H+1 ion. (HA), NH4+1
Base: Accepts one H+1 ion. (A-1), NH3
Conjugate Acid/Base Pairs: These are different by only a single H+1. The
acid has one more H+1 compared to the base in a conjugate pair. NH4+1 is
the conjugate acid for NH3 the conjugate base. HF is the conjugate acid
for F-1 the conjugate base.
Amphiprotic Substance: One that can both accept and donate H+1 and can be
either acid or base dependent on the environment. H2O can accept H+1
and become H3O+1 or donate H+1 and become OH-1. HCO3-1 is also
amphiprotic.
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Amphoteric Substance is another term used. Although all amphiprotic species
must be amphoteric, not all amphoteric substances are amphiprotic. Amphoteric
substances are able to react with an acid or a base. For example, the amphoteric
metal oxide, ZnO, contains no hydrogen and cannot donate a proton. ZnO acts as a
Lewis acid or Lewis base which accepts or donates electron pairs.
Amphoteric ZnO reacts with both acids and with bases:
In acid: ZnO + 2H+ → Zn2+ + H2O
In base: ZnO + H2O + 2OH- → [Zn(OH)4]2Example 5:
a) Write the formulas for the conjugate bases given the acids:
Acid
NH4+1
HF
HNO2
H2SO3
H2O
Conjugate
base
b) Write the formulas for the conjugate acids given the bases:
Base
C2H3O2-1
(CH3)2NH
ClO-1
SO3-2
Conjugate
acid
c) Which of the species in (a) and (b) above are amphiprotic?
H2O
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Lewis Definitions: This definition does not require water or aqueous reactions.
Acid: Accepts a share of a nonbonding electron pair. (BH3)
Base: Donates a share of a nonbonding electron pair. (NH3)
Generally results in a covalent bond forming, the product is called an adduct
Example 6:
Identify the Lewis acid and base for the reactants below.
a)
(CH3)3N + BF3  (CH3)3NBF3
b)
FeBr3 + Br-1  FeBr4 -1
c)
Zn+2 + 4 NH3  Zn(NH3)4+2
d)
SO2 + H2O  H2SO3
Acids and Bases Ch 15
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Structure of Acids:
 Binary acids have the hydrogen attached to a nonmetal atom: HF
 Oxy acids or ternary acids have the hydrogen attached to an oxygen atom:
 Carboxylic acids have COOH group the hydrogen
attached to the COO group is acidic:
*ADD THE LONE PAIR ELECTRONS TO THE PICTURES OF
SULFURIC ACID, NITRIC ACID AND CARBOXYLIC ACID GROUP.
Structure of Bases:
 Most ionic bases contain OH-1 ions
 Some contain CO3-2 ions
 Molecular bases contain structures that would like to add an H+1 ion,
mostly NH3 and amine groups.
Relative strengths:
A stronger acid will have a weaker conjugate base and vice versa. Strong
acids have a negligible conjugate base. Stronger bases have weaker conjugate
acids. Reactions always favor having a greater amount of the weaker acid
and base at equilibrium.
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Example 7:
a) Which acid (HI or HF) has the weaker conjugate base?
b)
Which base (C2H3O2-1 or OH-1) has a weaker conjugate acid?
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Autoionization of Water:
Pure water will ionize just slightly; it is generally considered a nonelectrolyte since
the amount it ionizes is so small.
Kw = 1.0 x 10-14 at 25°C for the reaction…
H2O (l) + H2O (l)  H3O+1 (aq) + OH-1 (aq)
The hydronium ion (H3O+1) is often written as a proton in water, H+1 (aq), even
though the H+1 is so reactive it cannot exist alone in water. H+1 is chemically
bonded to one or more water molecules in an aqueous solution connected by
hydrogen bonding. H(H2O)n+1
H2O (l) + H2O (l)

H3O+1 (aq) + OH-1 (aq)
or
H2O (l)

H+1 (aq) + OH-1 (aq);
Kw = [H+1][OH-1]
Temperature: As the temperature changes, so will Kw
0°C
Kw = 1.1 x 10-15
25°C
Kw = 1.0 x 10-14
37°C
Kw = 2.5 x 10-14
60°C
Kw = 9.6 x 10-14
pH Scale:
pH is a value that helps in determining the acidity of a solution. pH can be
determined using pH meter, pH paper, or indicators (intense colored organic dyes
that change color at different pH values).
pH = -log[H+1]
[H+1] = 10-pH
Kw = 1.0 x 10-14 = [H+1][OH-1] at 25°C
pOH = -log[OH-1]
[OH-1] = 10-pOH
pKw = -log[Kw]
pKw = pH + pOH = 14 at 25°C
Acid has a pH < 7
Neutral has pH = 7
Base has a pH > 7
pX = -log[X], pKa = -logKa, pKb = -logKb; Ka = 10-Ka , Kb = 10-Kb
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Significant Figures and Logs:
• When you take the log of a number written in scientific notation, the digit(s) before
the decimal point come from the exponent on 10, and the digits after the decimal
point come from the decimal part of the number
log(2.0 x 106)
= log(106) + log(2.0)
= 6 + 0.30303… = 6.30303...
• Because the part of the scientific notation number that determines the significant
figures is the decimal part, the sig figs are the digits after the decimal point.
Example 8:
Given that the pH of a solution is 4.88, solve for the [H+1], [OH-1], and pOH using
appropriate significant digits.
Strong Acids and Bases:
Strong Acids: HCl, HBr, HI, HNO3, HClO4, HClO3, H2SO4
Strong Soluble Bases: LiOH, NaOH, KOH, RbOH, CsOH, FrOH, Ca(OH)2,
Sr(OH)2, Ba(OH)2, Ra(OH)2
Strong Acids and Strong Bases dissociate into ions nearly completely, so the
dissociation reaction equilibrium constant K is very, very large.
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Example 9:
Calculate the concentration of hydrogen and hydroxide ions, and the pH and pOH.
a) 0.060 M HCl; [H+1] = [Acid] for monoprotic strong acids.
b) 0.012 M Ba(OH)2 ; [OH-1] = (number OH-1)x[Base] for strong soluble bases
Equilibrium Involving Weak Acids:
Weak acids are much more common and numerous compared to strong acids.
Weak acids (WA or HA) only will partially ionize. We use the equilibrium
constant, Ka, in which Ka<1, the smaller the Ka, the weaker the acid. The book
appendix has Ka values.
The weak acid equilibrium reaction is generally of the form…
HA  H+1 + A-1 or HA + H2O  H3O +1 + A-1
Approximation in Calculations: When using RICE you may avoid the quadratic
equation if x is added or subtracted from a relatively large number compared to the
value of Ka. The addition or subtraction of the x can be considered negligible when
Ka is 1000 times smaller than the concentration. Approximation values can be used
if Ka is between 100 to 1000 times smaller than the concentrations.
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Example 10:
In 0.120 M solution, a weak monoprotic acid (HA) is 5.00% ionized. Calculate
using the RICE equation the equilibrium concentrations, pH and Ka.
Example 11:
The pH of a 0.100 M solution of a monoprotic acid (HA) is 2.97. Using RICE,
calculate Ka.
Example 12:
Calculate the equilibrium concentrations and pH for the following solutions. (a)
0.150 M HC2H3O2 (b) 0.150 M HCN
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Equilibrium Involving Weak Bases:
Weak soluble bases are generally going to be amines, ammonia, and conjugate base
ions of acids. Weak bases (B or A-1) need to have water as an additional reactant to
provide the H+1 and only will partially ionize. We use the equilibrium constant, Kb,
in which Kb<1, the smaller the Kb, the weaker the base. Appendix D has K values
The weak base equilibrium reaction is generally of the form…
B + H2O  BH+1 + OH-1
Or
A-1 + H2O  HA + OH-1
every anion can potentially act as a base and accept an H+1.
Example 13:
Calculate the % ionization, pOH, pH and the equilibrium concentrations in 0.15 M
NH3 (aq). Kb = 1.8 x 10-5
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Example 14:
The pH is 11.37 for a NH3 (aq) solution. Calculate the Molarity of NH3.
Polyprotic Acid Equilibria:
 Ionization of polyprotic acids occur stepwise. Ka is different for each step,
decreasing as each H+1 is lost (Ka1 > Ka2 > Ka3).
 Generally, the difference in Ka values is enough so that the second ionization
does not happen to a large extent… except for extremely dilute solutions, the
[H+1] can be assumed to come from the first step alone.
 [A-2] = Ka2 as long as the second ionization is negligible.
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Example 15:
Calculate the concentrations of all species in 0.25 M H2SO4 solution.
Given: Ka1 >>1 (assume complete ionization), Ka2 = 1.2 x 10-2
Example 16:
Calculate the concentrations of all species in 0.40 M H3AsO4 solution.
Given: Ka1 = 5.5 x 10-3, Ka2 = 1.7 x 10-7, Ka1 = 5.1 x 10-12
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Relating Ka and Kb for Conjugate Acid/Base Pairs:
Ka
HF  H+1 + F-1
Ka = 3.5 x 10-4
Kb
F-1 + H2O  HF + OH-1
Kb = ?
+1
-1
Kw
H2O  H + OH
Kw = 1.0 x 10-14
Kw = Ka x Kb
Example 17:
Demonstrate how Kw = Ka x Kb and calculate the Kb value for F-1
Salt Solutions (Acid-Base Properties):
Salts are ionic compounds with cation and anion that may be conjugates of a
base and acid.
The reaction…
Acid + Base

Water + Soluble Salt
…has several General Categories forming soluble salts:
SA + SB  soluble neutral salt + water
WA + SB  soluble basic salt + water
SA + WB  soluble acidic salt + water
WA + WB  salt of unknown acidity + water
The reaction… Acid + Base  Water + Salt
Can be reversed to create the reaction for the hydrolysis of a salt…
Salt + Water  Acid + Base
Writing the whole, total ionic equation and net ionic equations:
Balance all atoms and charges
SA, SB and SS (Soluble salts) are written as charged ions
WA, WB, gases, solids, liquids are written whole
Include phases
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Example 18:
Write the whole, total ionic, and net ionic reactions for the hydrolysis of the
following salts. Identify spectator ions. Predict acidity of the salt, (Is it neutral,
acidic or basic)
a) CaBr2
b) NaNO2
c) NH4NO3
d) NH4C2H3O2
e) CH3NH2F
Example 19:
Calculate [OH-1], pH, % hydrolysis for 0.10 M NaClO solution found in Clorox
bleach. Ka of HClO = 3.5 x 10-8
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Example 20:
Write the whole, total and net ionic hydrolysis reactions for NH4Br. Calculate
[H+1], pH, % hydrolysis for 0.20 M NH4Br solution. Look up the appropriate
equilibrium constant you will require in the calculations.
Hydrated Metal Cations can act as Weak Acids:
Alkali metal and alkaline earth metal cations are pH neutral, negligible
counter-ions of strong bases. They do not act as weak acids.
Small highly charged metals can coordinate with water and release H+1 ion
from water to reduce the charge.
Cu(H2O)6+2 (aq)  H+1 (aq) + Cu(H2O)5(OH)+1(aq) Ka = 3 x 10-8
Smaller metals with higher positive charges more acidic
Acid Strength:
 The stronger an acid is at donating H+1, the weaker the conjugate base is
at accepting H+1
 Cation makes a stronger acid than neutral molecule which is more acidic
than anion… H3O+1 > H2O > OH-1 or NH4+ > NH3 > NH2−
 Larger Ka = stronger acid
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Chemical Structure Influences the Acid/Base Strength:
BinaryAcids:
Bond Strength (Greatest factor in same vertical column),
weaker bond (larger anion)  more acidic
Bond strengths: HF << HCl < HBr < HI
Electronegativity difference: (factor when comparing anions in the same
period, horizontal row)
greater difference  more acidic.
H2S is a weaker acid than HCl
-1
F ion (special case) causes an increased ordering in water molecules
creating the unfavorable lowering of entropy. This helps to explain why
HF is a weak acid.
Ternary Oxyacids:
Electronegativity of Center Atom, higher electronegativity  more acidic
H2SO4 > H2SeO4 > H2TeO4
Oxidation State of Center,
Larger oxidation number (more oxygens) more acidic.
HClO4 > HClO3 > HClO2 > HClO
Carboxylic Acids: The ability for the conjugate base to have resonance
structures will stabilize the base and it is more likely to have the
hydrogen ion lost. R-COOH
Polyprotic Acids:
The fewer H+1 a species has, the weaker the acid becomes.
H3PO4 is a stronger acid than H2PO4-1 and both are stronger than the
HPO4-2 acid since each successive Ka gets much smaller.
Example 21:
Explain the following observations:
a) H3PO4 is a stronger acid than H3AsO4
b) H2SO3 is a stronger acid than HSO3-1
Acid Rain:
Over 85% of U.S. fuel is from fossil fuels producing CO2, SO2, and NO2 which are linked
to acid rain and damages to ecosystems and structures. Natural processes as volcanoes
also add to it.
Nonmetal oxides and water create acids.
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More Practice:
1.
Give the conjugate base of the following Bronsted-Lowry acids…
a) NH4+1 b) H2PO4-1 c) HC7H5O2
2.
Give the conjugate acid of the following Bronsted-Lowry bases…
a) CN-1
b) H2PO4-1 c) C2H5NH2
3.
Designate the Bronsted-Lowry acid and base on the reactant side of each equation
and the conjugate acid and base on the product side: (all aqueous)
a) NH4+1 + CN-1  HCN + NH3
b) (CH3)3N + H2O (l)  (CH3)3NH+1 + OH-1
c) HCHO2 + PO4-3  CHO2-1 + HPO4-2
4.
The hydrogen oxalate ion, HC2O4-1, is amphiprotic. Write the balanced chemical
equation showing how it acts as an acid and how it acts as a base in water.
5.
Which of the following is the stronger acid, HBrO or HBr?
6.
Which is the stronger base, F-1 or Cl-1?
7.
Calculate [H+1], pH and pOH for the following and determine if acidic or basic…
a) [OH-] = 0.00040 M
b) 2.5 x 10-2 M HCl
c) solution where 100x[H+] =[OH-]
8.
By what factor does [H+] change for a pH change of a) 2.00 units, b) 0.50 units ?
9.
Predict the products of the following acid-base reactions, and also predict whether
the reactants or the products are preferred when at equilibrium.
a) NH3 (aq) + HBr (aq)

-1
-1
b) HCO3 (aq) + F (aq)

-1
c) H2O (l) + NO3 (aq)

10.
The average pH of normal arterial blood is 7.40 at body temperature (37˚C), at
which Kw = 2.4 x 10-14. Calculate [H+1], [OH-1] and pOH for this temperature.
11.
A 0.100 M solution of lactic acid (HC3H5O3) has a pH of 2.44. Calculate Ka.
12.
A 0.100 M solution of chloroacetic acid is 11.0% ionized. Solve for the equilibrium
concentrations of its ions and the Ka.
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13.
Saccharin (HNC7H4SO3) has a pKa of 2.32 at 25˚C. Solve for the pH of a 0.100 M
solution of saccharin.
14.
Phosphoric acid is triprotic (H3PO4). Calculate the pH and equilibrium
concentrations involved in phosphoric acid for a 0.100 M solution.
Given: Ka1 =7.5 x 10-3, Ka2 = 6.2 x 10-8, Ka3 = 4.2 x 10-13
15.
Calculate the pH of 0.075 M ethylamine (C2H5NH2), Kb = 6.4 x 10-4
16.
Ephedrine (C10H15ON) is used in nasal sprays as a decongestant. A 0.035 M
solution has a pH of 11.33. Solve for the equilibrium concentrations and Kb.
17.
Predict wheter the following aqueous salts will be acidic, basic, or neutral…
a) NH4Br b) NaC2H3O2 c) KClO4
18.
An unknown salt is either NaOCl or NaF. When 0.050 moles of the salt is
dissolved in 0.500 L the pH of the solution is 8.08. What is the salt?
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Answers:
1)
a) NH3 b) HPO4-2 c) C7H5O2-1
2)
a) HCN b) H3PO4 c) C2H5NH3+1
3)
a) NH4+1 + CN-1  HCN + NH3
acid
base
acid
base
b) (CH3)3N + H2O (l)  (CH3)3NH+1 + OH-1
base
acid
acid
base
-3
-1
-2
c) HCHO2 + PO4  CHO2 + HPO4
acid
base
base
acid
-1
-2
4)
Acid: HC2O4 + H2O  C2O4 + H3O+1
Base: HC2O4-1 + H2O  H2C2O4 + OH-1
5)
HBr
6)
F-1
7)
a) [H+1]= 2.5 x 10-11, pH = 10.60, pOH = 3.40
b) [H+1]= 2.5 x 10-2 , pH = 1.60 , pOH = 12.40
c) [H+1]= 1.0 x 10-8 , pH = 8.00, pOH = 6.00
8)
a) 100x b) 3.16x
9)
a) Products: NH3 (aq) + HBr (aq)  NH4+1 (aq) + Br-1 (aq)
b) Reactants: HCO3-1(aq) + F-1 (aq)  CO3-2(aq) + HF (aq)
c) Reactants: H2O (l) + NO3-1 (aq)  OH-1 (aq) + HNO3 (aq)
10) [H+1]= 4.0 x 10-8, [OH-] = 6.0 x 10-7, pOH = 6.22
11) Ka = 1.4 x 10-4
12) [H+1] = [ClCH2COO-1] = 0.0110M, [ClCH2COOH] = 0.0890M, Ka = 1.4 x 10-3
13) [H+1]= 2.0 x 10-2, pH = 1.71
14) [H3PO4] = 0.073 M, [H+1]= 0.027 M, [H2PO4-1] = 0.027 M, [HPO4-2] = 6.2 x 10-8,
[PO4-3] = 3x10-20 , [OH-] = 3.7 x 10-13 , pH=1.56
15) pH =11.82
16) [OH-1] = [C10H15ONH+1] = 0.0021M, [C10H15ON] = 0.033M, Ka = 1.4 x 10-4
17) a) acid, b) base, c) neutral
18) NaF