Acids and Bases

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Acids and Bases

• Acids taste sour (citric acid, acetic acid)
• Bases taste bitter (sodium bicarbonate)
• There are 3 ways to define acids and bases, you
  will learn 2 of these
• Arrhenius
• - Acids form H3O in water (HCl H2O ? H3O
  Cl-)
• - Bases form OH- in water (NaOH ? Na OH-)
• Brønsted-Lowry (B-L)
• - Acids donate H and Bases accept H
• HCl NaOH ? H2O NaCl
• HCl is the acid, it donates H to OH- (the base)

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B-L Acids and Formation of H3O

• In an acid-base reaction, there is always an
  acid/base pair (the acid donates H to the base)
• H is not stable alone, so it will be transferred
  from one covalent bond to another
• Example formation of H3O from an acid in water
• HBr H2O ? H3O Br-

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Identifying B-L Acids and Bases

• Compare the reactants and the products
• - The reactant that loses an H is the acid
• - The reactant that gains an H is the base
• Examples
• HCl H2O ? H3O Cl-
• Acid HCl and Base H2O (HCl gives H2O an H)
• NH3 H2O ? NH4 OH-
• Acid H2O and Base NH3 (H2O gives NH3 an H)
• CH3CO2H NH3 ? CH3CO2- NH4
• Acid CH3CO2H and Base NH3 (CH3CO2H gives NH3
  an H)

4
Conjugate Acids and Bases

• When a proton is transferred from the acid to the
  base (in a B-L acid/base reaction), a new acid
  and a new base are formed
• HA B ? A- HB
• acid base ? conjugate base conjugate
  acid
• The acid (HA) and the conjugate base (A-) that
  forms when HA gives up an H are a conjugate
  acid/base pair
• The base (B) and the conjugate acid (HB) that
  forms when B accepts an H are another conjugate
  acid/base pair

5
Identifying Conjugate Acid/Base Pairs

• Identify the acid and base for the reactants
• Identify the acid and base for the products
• Identify the conjugate acid/base pairs
• acid conjugate base
• 
  
• base conjugate acid

HF
H3O
F-
H2O
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Acid and Base Strength

• Strong acids give up protons easily and
  completely ionize in water
• HCl H2O ? H3O Cl-
• Weak acids give up protons less easily and only
  partially ionize in water
• CH3CO2H H2O ? CH3CO2- H3O
• Strong bases have a strong attraction for H and
  completely ionize in water
• KOH(s) ? K (aq) OH-(aq)
• NaNH2 H2O ? NH3 NaOH
• Weak bases have a weak attraction for H and only
  partially ionize in water
• HS- H2O ? H2S OH-

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Direction of an Acid/Base Equilibrium

• In general, theres an inverse relationship
  between acid/base strength within a conjugate
  pair
• - strong acid ? weak conjugate base
• - strong base ? weak conjugate acid
• (and vice-versa)
• The equilibrium always favors the direction that
  goes from stronger acid to weaker acid
• Example 1 HBr H2O ? H3O Br-
• stronger acid (HBr) ? weaker acid (H3O)
• (equilibrium favors products)
• Example 2 NH3 H2O ? NH4 OH-
• weaker acid (H2O) ? stronger acid (NH4)
• (equilibrium favors reactants)

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Dissociation Constants

• Since weak acids dissociate reversibly in water,
  we can write an equilibrium expression
• HA H2O ? H3O A-
• Keq H3OA-/HAH2O
• But, since H2O remains essentially constant we
  can write
• Ka Keq x H2O H3OA-/HA
• The acid dissociation constant (Ka) is a measure
  of how much the acid dissociates (A higher Ka a
  stronger acid)
• Example CH3CO2H H2O ? CH3CO2- H3O
• Ka H3OCH3CO2-/CH3CO2H 1.8 x 10-5
• Can also write dissociation constants for weak
  bases
• NH3 H2O ? NH4 OH-
• Kb NH4OH-/NH3 1.8 x 10-5

9
Ionization of Water

• Since H2O can act as either a weak acid or a weak
  base, one H2O can transfer a proton to another
  H2O
• H2O H2O ? H3O OH-
• Keq H3OOH-/H2OH2O
• Since H2O is essentially constant, we can
  write
• Kw Keq x H2O2 H3OOH-
• (where Kw the ion-product constant for water)
• For pure water H3O OH- 1.0 x 10-7 M
• So, Kw H3OOH- (1.0 x 10-7 M)2 1.0 x
  10-14
• (units are omitted for Kw as for Keq and Ka)

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Using Kw

• If acid is added to water, H3O goes up
• - for an acidic solution H3O gt OH-
• If base is added to water, OH- goes up
• - for a basic solution OH- gt H3O
• Kw is constant (1.0 x 10-14) for all aqueous
  solutions
• Can use Kw to calculate either H3O or OH- if
  given the other concentration
• Example if H3O 1.0 x 10-4 M, what is the
  OH-?
• Kw H3OOH-
• OH- Kw/ H3O 1.0 x 10-14/1.0 x 10-4 1.0
  x 10-10 M
• Is this an acidic or a basic solution?
• Since H3O gt OH-, its an acidic solution

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The pH Scale

• pH is a way to express H3O in numbers that are
  easy to work with
• H3O has a large range (1.0 M to 1.0 x 10-14 M)
  so we use a log scale
• pH - log H3O
• The pH scale goes from 0 - 14
• Each pH unit a ten-fold change in H3O
• pH 7 neutral, pH lt 7 acidic, pH gt 7 basic
• Can use an indicator dye (on paper or in
  solution) that changes color with changes in pH,
  or a pH meter, to measure pH

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Calculating pH and pOH

• Can calculate pH from H3O
• If H3O 1.0 x 10-3 M, what is the pH?
• pH - log H3O - log(1.0 x 10-3) 3.00
• Note sig. figs. in H3O decimal places in
  pH
• Can also calculate H3O from pH
• If pH is a whole number, H3O 1 x 10-pH
• So, if pH 2, then H3O 1 x 10-2
• Can calculate pOH from OH-
• pOH - logOH-
• Also, since Kw H3OOH-
• then pKw - log Kw - log (1.0 x 10-14)
  14.00
• And, pKw pH pOH 14.00
• So, if pH 3.00, then pOH 14.00 - 3.00 11.00

15
Reactions of Acids and Bases

• Acids and bases are involved in a variety of
  chemical reactions (well study 3 types here)
• Acids react with certain metals to produce metal
  salts and H2 gas, for example
• Mg(s) 2HCl(aq) ? MgCl2(aq) H2(g)
• Acids react with carbonates and bicarbonates to
  produce salts, H2O and CO2 gas, for example
• NaHCO3(aq) HCl(aq) ? NaCl(aq) H2O(l)
  CO2(g)
• Acids react with bases (neutralization reactions)
  to form salts and H2O, for example
• HBr(aq) LiOH(aq) ? LiBr (aq) H2O(l)
• Neutralization reactions are balanced with
  respect to moles of H and moles of OH-, for
  example
• H2SO4 2NaOH ? Na2SO4 2H2O

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Acidity of Salt Solutions

• Salts dissolved in water can affect the pH
• When salts dissolve, they dissociate into their
  ions
• NaCl ? Na Cl-
• If one of those ions can donate a proton to H2O,
  or accept one from H2O, the pH will change
• Na2S ? 2Na S2-
• S2- H2O ? HS- OH-
• S2- is a weak base that can accept an H from
  H2O
• Since OH- is increased, the solution is basic

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Salts that form Neutral Solutions

• When a strong acid dissolves in water, a weak
  conjugate base is formed that cant remove a
  proton from water
• When a strong base dissolves in water, the metal
  that dissociates cant form H3O
• So, salts containing ions that come from strong
  acids and bases do not affect the pH of the
  solution
• Example
• KBr ? K Br- (KOH strong base, HBr strong
  acid)
• (KOH HBr ? KBr H2O)
• K has no proton to donate, so cant form H3O
• Br- is too weak of a base to pull a proton off
  of H2O, so cant form OH-
• So, the solution remains neutral

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Salts that form Basic Solutions

• When a weak acid dissolves in water, the
  conjugate base formed is usually strong enough to
  remove a proton from H2O to form OH-
• So, salts that contain ions that come from a weak
  acid and a strong base form basic solutions
• Example
• NaCN ? Na CN- (HCN is a weak acid)
• CN- H2O ? HCN OH-
• (HCN NaOH ? NaCN H2O)
• (Na doesnt affect the pH, its from a strong
  base)

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Salts that form Acidic Solutions

• When a weak base dissolves in water, the
  conjugate acid formed is usually strong enough to
  donate a proton to H2O to form H3O
• So, salts that contain ions from a weak base and
  a strong acid form acidic solutions
• Example
• NH4Br ? NH4 Br- (from NH3 and HBr)
• (NH3 HBr ? NH4Br)
• NH4 H2O ? NH3 H3O
• (Br- doesnt affect the pH)

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Buffer solutions

• A small amount of strong acid or base added to
  pure water will cause a very large change in pH
• A buffer is a solution that can resist changes in
  pH upon addition of small amounts of strong acid
  or base
• Body fluids, such as blood, are buffered to
  maintain a fairly constant pH
• Buffers are made from conjugate acid/base pairs
  (either a weak acid and a salt of its conjugate
  base or a weak base and a salt of its conjugate
  acid)
• Thus, they contain an acid to neutralize any
  added base, and a base to neutralize any added
  acid
• Buffers cant be made from strong acids or bases
  and the salts of their conjugates since they
  completely ionize in H2O

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How to Make a Buffer Solution

• An acetate buffer is made from acetic acid and a
  salt of its conjugate base
• CH3CO2H and CH3CO2Na
• The salt is used to increase the concentration of
  CH3CO2- in the buffer solution
• Recall CH3CO2H H2O ? CH3CO2- H3O
• (the equilibrium favors reactants, so the
  concentration of CH3CO2- is low)
• But, CH3CO2Na ? CH3CO2- Na
• CH3CO2H CH3CO2Na H2O ? 2 CH3CO2- H3O
  Na
• If acid is added CH3CO2- H3O ? CH3CO2H H2O
• If base is added CH3CO2H OH- ? CH3CO2- H2O
• Buffer capacity how much acid or base can be
  added and still maintain pH (depends on buffer
  type and concentration)

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Calculating pH of a Buffer

• The pH of a buffer solution can be calculated
  from the acid dissociation constant (Ka)
• Example (for acetate buffer)
• CH3CO2H H2O ? CH3CO2- H3O
• Ka H3OCH3CO2-/CH3CO2H 1.8 x 10-5
• H3O Ka x CH3CO2H/CH3CO2-
• What is the pH of an acetate buffer that is 1.0 M
  CH3CO2H and 0.50 M CH3CO2Na?
• H3O 1.8 x 10-5 x 1.0 M/0.50 M 3.6 x 10-5 M
• pH - logH3O - log(3.6 x 10-5) 4.44

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Dilutions

• Often solutions are obtained and stored as highly
  concentrated stock solutions that are diluted for
  use (i.e. cleaning products, frozen juices)
• When a solution is diluted by adding solvent, the
  volume increases, but amount of solute stays the
  same, so the concentration decreases
• Mol solute concentration (mol/L) x V (L)
  constant
• So, C1V1 C2V2
• For molarity, it becomes M1V1 M2V2

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Dilution Calculations

• Example 1
• What volume (in mL) of 8.0 M HCl is needed to
  prepare 1.0 L of 0.50 M HCl?
• M1V1 M2V2 V1 M2V2/ M1
• V1 0.50 M x 1.0 L/ 8.0 M 0.0625 L 63 mL
• Example 2
• How many L of water do you need to add to dilute
  0.50 L of a 10.0 M NaOH solution to 1.0 M ?
• V2 M1V1/ M2
• V2 10.0 M x 0.50 L/ 1.0 M 5.0 L
• volume of water needed 5.0 L - 0.50 L 4.5 L

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Acid-Base Titration

• Molarity of an acid or base solution of unknown
  concentration can be determined by titration
• A measured volume of the unknown acid or base is
  placed in a flask and a few drops of indicator
  dye (such as phenolpthalein) are added
• A buret is filled with a measured molarity of
  known base or acid (the titrant) and small
  amounts are added until the solution changes
  color (neutralization endpoint)
• At neutralization endpoint H3O OH-
• Molarity of unknown is calculated from moles of
  titrant added (mole ratio comes from balanced
  chemical equation)

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Example Titration of H2SO4 with NaOH

• What is the molarity of a 10.0 mL sample of H2SO4
  if the neutralization endpoint is reached after
  adding 15.0 mL of 1.00 M NaOH?
• Calculate moles NaOH added
• 15.0 mL x (1 L/ 1000 mL) x (1.00 mol/ 1 L)
  0.0150 mol NaOH
• Write the balanced chemical equation
• H2SO4 2NaOH ? Na2SO4 2H2O
• Calculate moles H2SO4 neutralized
• 0.0150 mol NaOH x 1 mol H2SO4/ 2 mol NaOH
  0.00750 mol H2SO4
• Calculate molarity of H2SO4
• 0.00750 mol H2SO4/ 0.0100 L 0.750 M H2SO4