acids and bases
DESCRIPTION
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 H 3 O + in water (HCl + H 2 O H 3 O + + Cl - ) - PowerPoint PPT PresentationTRANSCRIPT
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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+)
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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
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Identifying Conjugate Acid/Base Pairs
1. Identify the acid and base for the reactants 2. Identify the acid and base for the products3. Identify the conjugate acid/base pairs
acid conjugate base + +
base conjugate acid
HF H2O H3O+ F-
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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, there’s 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 = [H3O+][A-]/[HA][H2O]
• But, since [H2O] remains essentially constant we can write:
Ka = Keq x [H2O] = [H3O+][A-]/[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 = [H3O+][CH3CO2-]/[CH3CO2H] = 1.8 x 10-5
• Can also write dissociation constants for weak bases:NH3 + H2O NH4
+ + OH-
Kb = [NH4+][OH-]/[NH3] = 1.8 x 10-5
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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 = [H3O+][OH-]/[H2O][H2O]
• Since [H2O] is essentially constant, we can write:
Kw = Keq x [H2O]2 = [H3O+][OH-]
(where Kw = the ion-product constant for water)
• For pure water: [H3O+] = [OH-] = 1.0 x 10-7 M
So, Kw = [H3O+][OH-] = (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+] > [OH-]
• If base is added to water, [OH-] goes up- for a basic solution [OH-] > [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 = [H3O+][OH-][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+] > [OH-], it’s 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 < 7 = acidic, pH > 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 = - log[OH-]
• Also, since Kw = [H3O+][OH-]
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
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Reactions of Acids and Bases• Acids and bases are involved in a variety of chemical reactions (we’ll
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 can’t remove a proton from water• When a strong base dissolves in water, the metal that dissociates
can’t 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 can’t form H3O+
Br- is too weak of a base to pull a proton off of H2O, so can’t 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+ doesn’t affect the pH, it’s 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- doesn’t 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 can’t 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 = [H3O+][CH3CO2-]/[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 = - log[H3O+] = - 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:1. 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
2. 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-]
3. 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