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Acids & Bases. Properties of Acids. Taste sour Contain H + ion pH less than 7 React with bases to form a salt and water React with some metals to produce hydrogen gas. Properties of Acids. Turn litmus paper red Phenolphthalein is colorless in the presence of an acid - PowerPoint PPT Presentation
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Properties of Acids
Taste sour Contain H+ ion pH less than 7 React with bases to form a
salt and water React with some metals to
produce hydrogen gas
Properties of Acids
Turn litmus paper red Phenolphthalein is colorless in
the presence of an acid Bromothymol blue is yellow in
the presence of an acid Found in citrus fruits in the
form of citric acid
Properties of Acids
Found in soured milk and in sore muscles in the form of lactic acid
Found in vitamin C in the form of ascorbic acid
Found in carbonated beverages in the form of carbonic acid…that’s also what you exhale
Properties of Bases
Taste bitter Contain OH- ion pH greater than 7 React with acids to form a salt
and water React with organic material
Properties of Bases
Feel slippery because they immediately begin to dissolve the outer layer of skin tissue
Turn litmus paper blue Phenolphthalein is fuchsia in
the presence of a base Bromothymol blue is blue in
the presence of a base
Properties of Bases
Found in drain cleaners usually in the form of sodium hydroxide
Found in ammonia-based cleaners like Windex
Lye (NaOH) is used to make soaps
Classifying Acids and Bases
Svante Arrhenius—Swedish guy who put forth his definitions of acids and bases in 1884 at the age of 25Worked with our buddy van’t
HoffReceived 1903 Nobel Chemistry
Prize for electrolytic dissociation discoveries
Arrhenius Acids
are substances that will dissociate in water to yield hydrogen ion (H+)
Arrhenius Bases
are substances that will dissociate in water to yield hydroxide ion (OH-)
Classifying Acids and Bases
Johannes Brønsted (Danish) and Thomas Lowry (English) came up with a new way to classify acids and bases and their conjugates (pairs that have features in common but are opposites) in the 1920’snever received a Nobel for
furthering these acid/base concepts, and Arrhenius never accepted them!
Brønsted-Lowry Acid
A reactant that donates a proton in a chemical reactionThe proton is actually the
hydrogen ion…since a hydrogen atom has 1 proton and 1 electron and the ion with a 1+ charge indicates that it has lost an electron
Brønsted-Lowry Base
A reactant that accepts a proton in a chemical reaction
Brønsted-Lowry Conjugate Acid
A product thatis formed when a base accepts a
protonin the reverse reaction, will
donate a proton
Brønsted-Lowry Conjugate Base
A product thatis formed when an acid donates
a proton (what’s left after the donation occurs)
in the reverse reaction, will accept a proton
Classifying Acids and Bases
Around the same time that Brønsted and Lowry were devising their acid/base scheme, our buddy Gilbert Lewis (yep, the same guy who did the dots) came up with yet another method of classifying them…it’s a broader method than Arrhenius, Brønsted, or Lowry ever postulated
Lewis Acid
A reactant that accepts an electron pair
Lewis Base
A reactant that donates an electron pair
Example #1
HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)
Or
HCl (aq) H+ (aq) + Cl- (aq)
Example #1
HCl is an acidIt dissociates to yield H3O+
(hydronium ion), which is really water with an extra H+. (Arrhenius)
It donates a proton (H+) to water in the first reaction written. (Brønsted-Lowry)
Example #1
H2O is an baseIt accepts a proton (H+) from
HCl in the first reaction written. (Brønsted-Lowry)
Example #1
H3O+ is a conjugate acidIt is produced when the water
accepts a proton (H+) from HCl in the first reaction written. (Brønsted-Lowry)
In the reverse reaction, it will donate a proton (H+) to Cl- in the first reaction written. (Brønsted-Lowry)
Example #1
Cl- is a conjugate baseIt is produced when the HCl
donates a proton (H+) to water in the first reaction written. (Brønsted-Lowry)
In the reverse reaction, it will accept a proton (H+) from H3O+ in the first reaction written. (Brønsted-Lowry)
Example #2
NH3 (g) + H2O (l) NH4+ (aq) + OH-
(aq)
Example #2
NH3 is a baseIt accepts a proton (H+) from
water. (Brønsted-Lowry)
Example #2
H2O is an acidIt donates a proton (H+) to
ammonia. (Brønsted-Lowry)
Example #2
NH4+ is a conjugate acid
It is produced when the ammonia accepts a proton (H+) from water. (Brønsted-Lowry)
In the reverse reaction, it will donate a proton (H+) to OH-. (Brønsted-Lowry)
Example #2
OH- is a conjugate baseIt is produced when the water
donates a proton (H+) to ammonia. (Brønsted-Lowry)
In the reverse reaction, it will accept a proton (H+) from NH4
+. (Brønsted-Lowry)
Water—our special friend
Did you notice that it behaved as a base in the first example and as an acid in the second example?
A substance that can behave as either an acid or a base is called, amphoteric or amphiprotic.
Example #3
H
H—N—H + H+ H—N—H
H H
Example #3
NH3 is a baseIt donates a pair of electrons to
H+. (Lewis)
Example #3
H+ is an acidIt accepts a pair of electrons
from NH3. (Lewis)
Autoionization of Water
Every acid and base will dissociate in water…even water (since it’s amphoteric)!
2H2O (l) H3O+ (aq) + OH- (aq)
orH2O (l) H+ (aq) + OH- (aq)
Autoionization of Water
Usually, the 2nd reaction is the one we will use since H3O+ is just water with an extra H+.
Write the K expression for the 2nd reaction, keeping in mind that we only include gaseous and aqueous phases.
Autoionization of Water
K = [H+][OH-]note that water is not included
because it is a liquid
This expression is known as the Kw, or equilibrium constant for water, expression K w = [H+][OH-]
Autoionization of Water
The value of Kw is 1 x 10-14 M2 at 25°C.
This is a small K value.
If the temperature changes, so does the value of Kw
Autoionization of Water
Make an equilibrium chart for the dissociation, or autoionization, of water.
Autoionization of Water
[H2O] [H+] [OH-]
Initial -- 0 0
Change
Eq.
Autoionization of Water
[H2O] [H+] [OH-]
Initial -- 0 0
Change -- +x +x
Eq.
Autoionization of Water
[H2O] [H+] [OH-]
Initial -- 0 0
Change -- +x +x
Eq. -- x x
Autoionization of Water
Determine the equlibrium concentrations of both the hydrogen ion and the hydroxide ion by plugging into the Kw expression.
1 x 10-14 M2 = [x][x]
Autoionization of Water
1 x 10-14 M2 = x2
1 x 10-7 M = x
[H+] = 1 x 10-7 M
[OH-] = 1 x 10-7 M
Autoionization of Water
Because the H+ and OH- concentrations are equal, the solution is neutral.
If [H+] > [OH-], then the solution is an acid.
If [H+] < [OH-], then the solution is a base.
Autoionization of Water
Since every acid or base dissociation we will entertain occurs in water, then the Kw expression is applicable to any of these dissociations.
Autoionization of Water
Thus, if you know the [H+] concentration of a solution, you can determine the [OH-] concentration.
And, if you know the [OH-] concentration of a solution, you can determine the [H+] concentration.
pH
Represents the “power of hydrogen”
Calculated by taking the opposite of the logarithm of the hydrogen ion concentration…
pH = -log [H+]
pH
Calculate the pH of water at 25°C knowing that the [H+] is 1 x 10-7 M.
pH = -log [1 x 10-7]
pH = 7
pH
If you already know the pH of a solution, then you can find the [H+] using:
[H+] = 10-pH
So,[H+] = 10-7
[H+] = 1 x 10-7 M
pOH
Represents the “power of hydroxide”
Calculated by taking the opposite of the logarithm of the hydroxide ion concentration…
pOH = -log [OH-]
pOH
Calculate the pOH of water at 25°C knowing that the [OH-] is1 x 10-7 M.pOH = -log [1 x 10-7]
pOH = 7
Ooh, ah!
So, the sum of pH and pOH for all aqueous solutions will be 14.pH + pOH = 14
Example #4—show all work
Find the pH, pOH, and [OH-] and state whether the solution is acidic, neutral or basic if the hydrogen ion concentration is3.48 x 10-4 M.
pH = 3.46pOH = 10.54[OH-] = 2.87 x 10-11 Macidic
Example #5—show all work
Find the pOH, [H+], and [OH-] and state whether the solution is acidic, neutral or basic if the pH is 9.84.
pOH = 4.16[H+] = 1.45 x 10-10 M[OH-] = 6.91 x 10-5 Mbasic
Example #6—show all work
Find the pH, [H+], and [OH-] and state whether the solution is acidic, neutral or basic if the pOH is 12.7.
pH = 1.30[H+] = 5.01 x 10-2 M[OH-] = 1.99 x 10-13 Macidic
Example #7—show all work
Find the pH, pOH, and [H+], and state whether the solution is acidic, neutral or basic if the[OH-] is 5.26 x 10-2 M.
pH = 12.7pOH = 1.28[H+] = 1.90 x 10-13 Mbasic