Acids and Bases

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Acids andBases
RNA uses amino-acids to build proteins/enzymes
Digestive Acids help to break down food into
reusable molecular fragments
It is the acids in citrus fruits that give them
the sour taste and allows the fruit to stay in a
state of preservation till germination
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Properties of Acids

• sour taste
• react with active metals (Al, Zn, Fe), but not
  Cu, Ag, or Au
• 2 Al(s) 6 HCl(aq) ? 2 AlCl3(aq) 3 H2(g)
• Corrosive
• react with carbonates, producing CO2
• marble, baking soda, chalk, limestone
• CaCO3(s) 2 HCl(aq) ? CaCl2(aq) CO2(g)
  H2O(l)
• change color of vegetable dyes
• blue litmus turns red
• react with bases to form ionic salts
• HCl(aq) NaOH(aq)?NaCl(aq) H2O(l)

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Common Acids
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Structure of Acids

• binary acids have acid hydrogens attached to a
  nonmetal atom
• HCl, HF

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Structure of Acids

• oxy acids have acid hydrogens attached to an
  oxygen atom
• H2SO4, HNO3

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Structure of Acids

• carboxylic acids have COOH group
• HC2H3O2, H3C6H5O7
• only the first H in the formula is acidic
• the H is on the COOH

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Properties of Bases

• also known as alkalis
• taste bitter
• alkaloids plant product that is alkaline
• often poisonous
• solutions feel slippery
• change color of vegetable dyes
• different color than acid
• red litmus turns blue
• react with acids to form ionic salts
• Neutralization
• HCl(aq) NaOH(aq)?NaCl(aq) H2O(l)

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Common Bases
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Structure of Bases

• most ionic bases contain OH- ions
• NaOH, Ca(OH)2
• some contain CO32- ions
• CaCO3 NaHCO3
• molecular bases contain structures that react
  with H
• mostly amine groups

Amino acids have a base at one end and an acid at
the other, neighboring amino acids can neutralize
to form a polypeptide
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Indicators

• chemicals which change color depending on the
  acidity/basicity
• many vegetable dyes are indicators
• anthocyanins
• litmus
• from Spanish moss
• red in acid, blue in base
• phenolphthalein
• found in laxatives
• red in base, colorless in acid

Anthocyanins give these pansies their dark purple
pigmentation and are the pigment in red cabbage
that is so sensitive to acidity
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acids and bases Arrhenius Theory

• bases dissociate in water to produce OH- ions and
  cations
• ionic substances dissociate in water
• NaOH(aq) ? Na(aq) OH(aq)
• acids ionize in water to produce H ions and
  anions
• HCl(aq) ? H(aq) Cl(aq)
• HC2H3O2(aq) H(aq) C2H3O2(aq)

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Arrhenius Theory
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Hydronium Ion

• the H ions produced by the acid are so reactive
  they cannot exist in water
• H ions are protons
• instead, they react with a water molecule(s) to
  produce complex ions, mainly hydronium ion, H3O
• H H2O ? H3O ? H(aq)
• there are also minor amounts of H with multiple
  water molecules, H(H2O)n

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Arrhenius Acid-Base Reactions

• the H from the acid combines with the OH- from
  the base to make a molecule of H2O
• it is often helpful to think of H2O as H-OH
• the cation from the base combines with the anion
  from the acid to make a salt
• acid base ? salt water
• HCl(aq) NaOH(aq) ? NaCl(aq) H2O(l)
• H(aq)Cl-(aq)Na(aq)OH-(aq)?Na(aq)Cl-(aq)H2O
  (l)
• H(aq) OH-(aq)? H2O(l)
• All acid base reactions have this same net ionic
  equation in the Arrhenius idea of the acid and
  base

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Limitations of the Arrhenius Theory

• does not explain why molecular substances, like
  NH3, dissolve in water to form basic solutions
  even though they do not contain OH ions
• does not explain how some ionic compounds, like
  Na2CO3 or Na2O, dissolve in water to form basic
  solutions even though they do not contain OH
  ions
• does not explain why molecular substances, like
  CO2, dissolve in water to form acidic solutions
  even though they do not contain H ions
• does not explain acid-base reactions that take
  place outside aqueous solution

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Acids and bases Brønsted-Lowry

• in a Brønsted-Lowry Acid-Base reaction, an H is
  transferred
• does not have to take place in aqueous solution
• broader definition than Arrhenius
• An acid is H donor, base is H acceptor
• base structure must contain an atom with an
  unshared pair of electrons
• in an acid-base reaction, the acid molecule gives
  an H to the base molecule
• HA B ? A HB

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Brønsted-Lowry Acids

• Brønsted-Lowry acids are H donors
• any material that has H can potentially be a
  Brønsted-Lowry acid
• because of the molecular structure, often one H
  in the molecule is easier to transfer than others
• HCl(aq) is acidic because HCl transfers an H to
  H2O, forming H3O ions
• water acts as base, accepting H

HCl(aq) H2O(l) ? Cl(aq) H3O(aq) Acid
base
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Brønsted-Lowry Bases

• Brønsted-Lowry bases are H acceptors
• any material that has atoms with lone pairs can
  potentially be a Brønsted-Lowry base
• because of the molecular structure, often one
  atom in the molecule is more willing to accept H
  transfer than others
• NH3(aq) is basic because NH3 accepts an H from
  H2O, forming OH(aq)
• water acts as acid, donating H

NH3(aq) H2O(l) ? NH4(aq) OH(aq) base acid
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Amphoteric Substances

• amphoteric substances can act as either an acid
  or a base
• have both transferable H and atom with lone pair
• Example
• water acts as base, accepting H from HCl
• HCl(aq) H2O(l) ? Cl(aq) H3O(aq)
• water acts as acid, donating H to NH3
• NH3(aq) H2O(l) ? NH4(aq) OH(aq)

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Brønsted-Lowry Acid-Base Reactions

• one of the advantages of Brønsted-Lowry theory is
  that it allows reactions to be reversible
• HA B A HB
• the original base has an extra H after the
  reaction so it will act as an acid in the
  reverse process
• and the original acid has a lone pair of
  electrons after the reaction so it will act as
  a base in the reverse process
• A HB HA B

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Conjugate Pairs

• In a Brønsted-Lowry Acid-Base reaction, the
  original base becomes an acid in the reverse
  reaction, and the original acid becomes a base in
  the reverse process
• each reactant and the product it becomes is
  called a conjugate pair
• the original base becomes the conjugate acid and
  the original acid becomes the conjugate base
• NH3(aq) H2O(l) NH4(aq)
  OH(aq)
• Base Acid
  Conjugate Acid Conjugate Base

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Brønsted-Lowry Acid-Base Reactions
HA B A HB
acid base conjugate conjugate
base acid
HCHO2 H2O CHO2
H3O acid base conjugate conjugate
base acid
H2O NH3 HO NH4
acid base conjugate conjugate
base acid
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Conjugate Pairs
In the reaction H2O NH3 HO NH4
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Identify the Brønsted-Lowry Acids and Bases and
Their Conjugates in the Reaction
H2SO4 H2O HSO4 H3O
When the H2SO4 becomes HSO4-, it lost an H so
H2SO4 must be the acid and HSO4- its conjugate
base
When the H2O becomes H3O, it accepted an H so
H2O must be the base and H3O its conjugate acid
H2SO4 H2O HSO4 H3O acid
base conjugate conjugate
base acid
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Identify the Brønsted-Lowry Acids and Bases and
Their Conjugates in the Reaction
HCO3 H2O H2CO3 HO
When the HCO3? becomes H2CO3, it accepted an H
so HCO3- must be the base and H2CO3 its conjugate
acid
When the H2O becomes OH-, it donated an H so H2O
must be the acid and OH- its conjugate base
HCO3 H2O H2CO3 HO base
acid conjugate conjugate
acid base
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Practice Write the formula for the conjugate
acid of the following
H2O NH3 CO32- H2PO4-
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Practice Write the formula for the conjugate
acid of the following
H2O H3O NH3 NH4 CO32- HCO3- H2PO41- H3PO4
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Practice Write the formula for the conjugate
base of the following
H2O NH3 CO32- H2PO4-
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Practice Write the formula for the conjugate
base of the following
H2O HO- NH3 NH2- CO32- since CO32- does not
have an H, it cannot be an acid H2PO41- HPO42-
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Arrow Conventions

• chemists commonly use two kinds of arrows in
  reactions to indicate the degree of completion of
  the reactions
• a single arrow indicates all the reactant
  molecules are converted to product molecules at
  the end
• a double arrow indicates the reaction stops when
  only some of the reactant molecules have been
  converted into products

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Strong or Weak

• a strong acid is a strong electrolyte
• practically all the acid molecules ionize, ?
• a strong base is a strong electrolyte
• practically all the base molecules form OH ions,
  either through dissociation or reaction with
  water, ?
• a weak acid is a weak electrolyte
• only a small percentage of the molecules ionize,
• a weak base is a weak electrolyte
• only a small percentage of the base molecules
  form OH ions, either through dissociation or
  reaction with water,

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Strong Acids

• The stronger the acid, the more willing it is to
  donate H
• use water as the standard base
• strong acids donate practically all their Hs
• 100 ionized in water
• strong electrolyte
• H3O strong acid

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Weak Acids

• weak acids donate a small fraction of their Hs
• most of the weak acid molecules do not donate H
  to water
• much less than 1 ionized in water
• H3O ltlt weak acid

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Polyprotic Acids

• often acid molecules have more than one ionizable
  H these are called polyprotic acids
• the ionizable Hs may have different acid
  strengths or be equal
• 1 H monoprotic, 2 H diprotic, 3 H triprotic
• HCl monoprotic, H2SO4 diprotic, H3PO4
  triprotic

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Polyprotic Acids

• polyprotic acids ionize in steps
• each ionizable H removed sequentially
• removing of the first H automatically makes
  removal of the second H harder
• H2SO4 is a stronger acid than HSO4-

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Increasing Basicity
Increasing Acidity
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Strengths Acids and Bases

• commonly, Acid or Base strength is measured by
  determining the equilibrium constant of a
  substances reaction with water
• HA H2O A-1 H3O1
• B H2O HB1 OH-1
• the farther the equilibrium position lies to the
  products, the stronger the acid or base
• the position of equilibrium depends on the
  strength of attraction between the base form and
  the H
• stronger attraction means stronger base or weaker
  acid

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General Trends Acidity

• the stronger an acid is at donating H, the weaker
  the conjugate base is at accepting H
• higher oxidation number stronger oxyacid
• H2SO4 gt H2SO3 HNO3 gt HNO2
• cation stronger acid than neutral molecule
  neutral stronger acid than anion
• H3O1 gt H2O gt OH-1 NH41 gt NH3 gt NH2-1
• base trend opposite

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Acid Ionization Constant, Ka

• acid strength measured by the size of the
  equilibrium constant when react with H2O
• HA H2O A-1 H3O1
• the equilibrium constant is called the acid
  ionization constant, Ka
• larger Ka stronger acid

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Autoionization of Water

• Water is actually an extremely weak electrolyte
• therefore there must be a few ions present
• about 1 out of every 10 million water molecules
  form ions through a process called autoionization
• H2O H OH
• H2O H2O H3O OH
• all aqueous solutions contain both H3O and OH
• the concentration of H3O and OH are equal in
  water
• H3O OH 10-7M _at_ 25C

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Ion Product of Water

• the product of the H3O and OH concentrations is
  always the same number
• the number is called the ion product of water and
  has the symbol Kw
• H3O x OH Kw 1 x 10-14 _at_ 25C
• if you measure one of the concentrations, you can
  calculate the other
• as H3O increases the OH must decrease so
  the product stays constant
• inversely proportional

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Acidic and Basic Solutions

• all aqueous solutions contain both H3O and OH
  ions
• neutral solutions have equal H3O and OH
• H3O OH 1 x 10-7
• acidic solutions have a larger H3O than OH
• H3O gt 1 x 10-7 OH lt 1 x 10-7
• basic solutions have a larger OH than H3O
• H3O lt 1 x 10-7 OH gt 1 x 10-7

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Calculate the OH- at 25C when the H3O 1.5
x 10-9 M, and determine if the solution is
acidic, basic, or neutral
H3O 1.5 x 10-9 M OH- if OH- gt H3O
basic if H3OgtOH- acid
Given Find
Concept Plan Relationships
Solution
Check
The units are correct. The fact that the H3O
lt OH- means the solution is basic
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Complete the TableH vs. OH-
H 100 10-1 10-3 10-5 10-7
10-9 10-11 10-13 10-14
OH-
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Complete the TableH vs. OH-
Acid
Base
H 100 10-1 10-3 10-5 10-7
10-9 10-11 10-13 10-14
OH-10-14 10-13 10-11 10-9 10-7
10-5 10-3 10-1 100
Even though it may look like it, neither H nor
OH- will ever be 0
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pH

• the acidity/basicity of a solution is often
  expressed as pH
• pH -logH3O, H3O 10-pH
• Since in water H3O 10-7M
• pHwater -log10-7 7
• pH lt 7 is acidic pH gt 7 is basic, pH 7 is
  neutral

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pH

• the lower the pH, the more acidic the solution
• the higher the pH, the more basic the solution
• 1 pH unit corresponds to a factor of 10
  difference in acidity
• normal range 0 to 14
• pH 0 is H 1 M, pH 14 is OH 1 M
• pH can be negative (very acidic) or larger than
  14 (very alkaline)

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Sig. Figs. 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...
• since 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 in the log
• log(2.0 x 106) 6.30

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Calculate the pH at 25C when the OH- 1.3 x
10-2 M, and determine if the solution is acidic,
basic, or neutral
Given Find
OH- 1.3 x 10-2 M pH
Concept Plan Relationships
Solution
Check
pH is unitless. The fact that the pH gt 7 means
the solution is basic
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pOH

• another way of expressing the acidity/basicity of
  a solution is pOH
• pOH -logOH-, OH- 10-pOH
• pOHwater -log10-7 7
• need to know the OH- concentration to find pOH
• pOH lt 7 is basic pOH gt 7 is acidic, pOH 7 is
  neutral

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pH and pOH Complete the Table
pH
H 100 10-1 10-3 10-5 10-7
10-9 10-11 10-13 10-14
OH-10-14 10-13 10-11 10-9 10-7
10-5 10-3 10-1 100
pOH
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pH and pOHComplete the Table
pH 0 1 3 5 7
9 11 13 14
H 100 10-1 10-3 10-5 10-7
10-9 10-11 10-13 10-14
OH-10-14 10-13 10-11 10-9 10-7
10-5 10-3 10-1 100
pOH 14 13 11 9 7
5 3 1 0
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Relationship between pH and pOH

• the sum of the pH and pOH of a solution 14.00
• at 25C
• can use pOH to find pH of a solution

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pK

• a way of expressing the strength of an acid or
  base is pK
• pKa -log(Ka), Ka 10-pKa
• pKb -log(Kb), Kb 10-pKb
• the stronger the acid, the smaller the pKa
• larger Ka smaller pKa
• because it is the log

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Finding the pH of a Strong Acid

• there are two sources of H3O in an aqueous
  solution of a strong acid the acid and the
  water
• for the strong acid, the contribution of the
  water to the total H3O is negligible
• for a monoprotic strong acid H3O HA
• 0.10 M HCl has H3O 0.10 M and pH 1.00

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Finding the pH of a Weak Acid

• there are also two sources of H3O in and aqueous
  solution of a weak acid the acid and the water
• however, finding the H3O is complicated by the
  fact that the acid only undergoes partial
  ionization
• calculating the H3O requires solving an
  equilibrium problem for the reaction that defines
  the acidity of the acid

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Find the pH of 0.200 M HNO2(aq) solution _at_ 25C
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations assuming the H3O from water is 0
HNO2 H2O NO2- H3O
HNO2 NO2- H3O
initial
change
equilibrium
HNO2 NO2- H3O
initial 0.200 0 0
change
equilibrium
since no products initially, Qc 0, and the
reaction is proceeding forward
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Find the pH of 0.200 M HNO2(aq) solution _at_ 25C
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
HNO2 NO2- H3O
initial 0.200 0 0
change
equilibrium
x
x
-x
x
x
0.200 -x
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Find the pH of 0.200 M HNO2(aq) solution _at_ 25C
Ka for HNO2 4.6 x 10-4
HNO2 NO2- H3O
initial 0.200 0 0
change -x x x
equilibrium 0.200 x x
determine the value of Ka from Table 15.5
since Ka is very small, approximate the HNO2eq HNO2init and solve for x
0.200 -x
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Find the pH of 0.200 M HNO2(aq) solution _at_ 25C
Ka for HNO2 4.6 x 10-4
HNO2 NO2- H3O
initial 0.200 0 0
change -x x x
equilibrium 0.200 x x
check if the approximation is valid by seeing if x lt 5 of HNO2init
x 9.6 x 10-3
the approximation is valid
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Find the pH of 0.200 M HNO2(aq) solution _at_ 25C
Ka for HNO2 4.6 x 10-4
HNO2 NO2- H3O
initial 0.200 0 0
change -x x x
equilibrium 0.200-x x x
substitute x into the equilibrium concentration definitions and solve
HNO2 NO2- H3O
initial 0.200 0 0
change -x x x
equilibrium 0.190 0.0096 0.0096
x 9.6 x 10-3
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Find the pH of 0.200 M HNO2(aq) solution _at_ 25C
Ka for HNO2 4.6 x 10-4
HNO2 NO2- H3O
initial 0.200 0 0
change -x x x
equilibrium 0.190 0.0096 0.0096
substitute H3O into the formula for pH and solve
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Find the pH of 0.200 M HNO2(aq) solution _at_ 25C
Ka for HNO2 4.6 x 10-4
HNO2 NO2- H3O
initial 0.200 0 0
change -x x x
equilibrium 0.190 0.0096 0.0096
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
though not exact, the answer is reasonably close
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What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka 1.4 x 10-5 _at_ 25C?
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What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka 1.4 x 10-5 _at_ 25C?
HC6H4NO2 H2O C6H4NO2- H3O
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations assuming the H3O from water is 0
HA A- H3O
initial 0.012 0 0
change
equilibrium
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What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka 1.4 x 10-5 _at_ 25C?
HC6H4NO2 H2O C6H4NO2- H3O
HA A- H3O
initial 0.012 0 0
change
equilibrium
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
x
-x
x
x
x
0.012 -x
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What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka 1.4 x 10-5 _at_ 25C?
HC6H4NO2 H2O C6H4NO2- H3O
determine the value of Ka
since Ka is very small, approximate the HAeq HAinit and solve for x
HA A2- H3O
initial 0.012 0 0
change -x x x
equilibrium 0.012 x x
0.012 -x
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What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka 1.4 x 10-5 _at_ 25C?
Ka for HC6H4NO2 1.4 x 10-5
check if the approximation is valid by seeing if x lt 5 of HC6H4NO2init
HA A2- H3O
initial 0.012 0 0
change -x x x
equilibrium 0.012 x x
x 4.1 x 10-4
the approximation is valid
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What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka 1.4 x 10-5 _at_ 25C?
HA A2- H3O
initial 0.012 0 0
change -x x x
equilibrium 0.012-x x x
substitute x into the equilibrium concentration definitions and solve
x 4.1 x 10-4
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What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka 1.4 x 10-5 _at_ 25C?
substitute H3O into the formula for pH and solve
HA A2- H3O
initial 0.012 0 0
change -x x x
equilibrium 0.012 0.00041 0.00041
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What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka 1.4 x 10-5 _at_ 25C?
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
HA A2- H3O
initial 0.012 0 0
change -x x x
equilibrium 0.012 0.00041 0.00041
the values match