acids and bases this is our second conceptual definition. acids – substances that donate protons...
TRANSCRIPT
This is our second conceptual definition.
• Acids – Substances that donate protons (H+)• Bases – Substances that accept proton(s)
Compare with Arrhenius
HCl(aq) H+(aq) + Cl-(aq) (Arrhenius)
HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq) (B-L)
Acid
Gives up protons
Base
Accepts protons
Other Examples
NH3(aq) + H2O(l) NH4+
(aq) + OH-(aq)
Base:
Accepts protons
Acid: gives up protons
Note: Protons always move from the acid to the base.
Water is AMPHIPROTIC ( or AMPHOTERIC) meaning that it can function as either an acid or a base.
Check the strength of acid chart in your data booklet to see which other substances are amphiprotic
Conjugate Acid / Base Pairs
In any Bronsted-Lowry equation, there are 2 acids and 2 bases.
HSO4-(aq) + HPO4
2-(aq) SO4
2-(aq) + H2PO4
-(aq)
Acid
(Gives up H+)
Base
(accepts H+)
Acid
(gives up H+)
Base
(accepts H+)
A conjugate acid/base pair consists of 2 substances that differ by the gain or loss of a single proton (H+ )
Therefore the conjugate pair will always be the acid from one side of the equation and the base from the other side.
Each equation will have 2 conjugate acid/base pairs.
In the above equation, the conjugate pairs are:
HSO4-(aq) with SO4
2-(aq) and…
HPO42- (aq) with H2PO4
- (aq)
Check your data booklet. Notice that the formula for each conjugate is the result of removing a single H+ from the formula for the acid.
The strongest acid is the one that gives up protons the easiest. That is: it functions very well as an acid.
Important notes
The strongest base is the one that has the greatest attraction for protons. That is: it functions very well as a base.
How to represent substances in Bronsted-Lowry reactions
These are rules that you must know in order to be
able to write these reactions. Learn them
well.
I. Strong Electrolytes: Ionic saltsStrong AcidsStrong Bases
These are written in dissociated (ion) form. That is:
Ionic salts are written as individual ions.
Example: NaCl(s) → written as Na+(aq) and Cl-(aq)
Strong Acids are written as the hydronium ion.
Example: HCl (aq) → written as H3O+(aq)
ALL STRONG ACIDS ARE WRITTEN THE SAME WAY
Strong Bases are written as hydroxide ions.
Example: NaOH(s) → written as OH-(aq)
ALL STRONG BASES (HYDROXY BASES) ARE WRITTEN THE SAME WAY
II. Weak Electrolytes: Weak Acids
Weak Bases
These are written in associated form. That is: they are written “as is”.
Examples: CH3COOH(aq)
NH3(aq)
→ written as CH3COOH (aq)
→ written as NH3 (aq)
Brønsted-Lowry Acids and BasesBrønsted-Lowry Acids and Bases
Proton Transfer ReactionsProton Transfer Reactions
• Brønsted-Lowry: (conceptual definition)
–acid donates proton (H+)
–base accepts proton (H+)
• Brønsted-Lowry base does not need to contain OH-.
Brønsted-Lowry Acids and BasesBrønsted-Lowry Acids and BasesProton Transfer ReactionsProton Transfer Reactions
acid base
Brønsted-Lowry Acids and BasesBrønsted-Lowry Acids and Bases
Conjugate Acid-Base PairsConjugate Acid-Base Pairs
• Whatever is left of the acid after the proton is donated is called its conjugate base.
• Similarly, whatever remains of the base after it accepts a proton is called a conjugate acid.
Brønsted-Lowry Acids and BasesBrønsted-Lowry Acids and BasesConjugate Acid-Base PairsConjugate Acid-Base Pairs• Consider:
– After HA (acid) loses its proton it is converted into A- (base). Therefore HA and A- are conjugate acid-base pairs.
– After H2O (base) gains a proton it is converted into H3O+ (acid). Therefore, H2O and H3O+ are conjugate acid-base pairs.
• Conjugate acid-base pairs differ by only one proton.
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Brønsted-Lowry Acids and BasesBrønsted-Lowry Acids and BasesProton Transfer ReactionsProton Transfer Reactions
acid base
conjugate base conjugate acid
Strong Acids and BasesStrong Acids and BasesStrong AcidsStrong Acids
• The strongest common acids are HCl, HBr, HI, HNO3, HClO3, HClO4, and H2SO4.
• Strong acids are strong electrolytes.
• All strong acids ionize completely in solution:
Bronsted Lowry
Here is the procedure for
writing my reactions. It will
always work.
He always takes the credit for my
work! They should be Lowry-Bronsted reactions.
Procedure for writing Bronsted-Lowry reactions
1. List all species present. (don’t forget the H2O (l) )
2. Locate the strongest acid present using your data book. (start at the top left and work down.
The reaction will always occur between the strongest acid present and the strongest base present.
4. Transfer a single proton from the acid to the base.
5. Fill in the products. (these will be the conjugate acids and bases of your reactants)
3. Locate the strongest base present using your data book. (start at the bottom right and work up.)
Examples.1. HBr(aq) is mixed with NH3(aq)Species list
S.A. written as H3O+ (aq)
W.B. written “as is”H3O+(aq)
NH3(aq)
H2O(l)
H3O+ (aq) + NH3 (aq) NH4+ (aq) + H20(l)
Acid Base Con - Acid Con - Base
Example 2CaSO3(aq) mixed with HF(aq)
Species List
Ionic salt → Ca2+(aq) + SO3
2-(aq)
Ca2+(aq)
SO32-
(aq)
W.A.→ “as is” HF(aq)
HF(aq)
H2O(l)
Note: All metallic ions are spectator ions and can be omitted. They cannot function as an acid or a base.
HF(aq) + SO32-
(aq) HSO3-(aq) + F- (aq)
Acid Base Con - Acid Con - Base
Which side is favored?
• Strongest acid in the whole equation will dissociate the most. Therefore the side opposite the strongest acid is favored
Example 3KHCO3(aq) mixed with Na2HPO4(aq)
Species List
K+ + HCO3- Na+ + HPO4
2-HCO3-(aq)
HPO42-
(aq)
H2O(l)
HCO3- (aq) +
acid
HPO42-
(aq)
base
H2PO4-(aq) +
acid
CO32-
(aq)
base
Example 4.NaOH(aq) is mixed with HCl(aq)
Species List
S.B. → OH-(aq) S.A. → H3O+
(aq)OH-
(aq)
H3O+(aq)
H2O(l)
H3O+(aq) +
acid
OH-(aq)
base
H2O(l) + H2O(l)
acid base
Neutralization
Example 5.KCl(aq) is mixed with H2O(l)
Species List
K+(aq) + Cl-(aq)
Cl-(aq) H2O(l)
H2O(l)
H2O(l) + H2O(l) No Reaction
Or: H2O(l) + H2O(l) H3O+(aq) + OH-
(aq)
Same as in water →like we said…no reaction
Where do we find Kb?
• There are no Kb values in the data book.
• We need to use the Ka values on the strength of acid chart in our data book to calculate Kb.
Relationship Between KRelationship Between Kaa and K and Kbb
• For a conjugate acid-base pair
Ka Kb = Kw
•Therefore, the larger the Ka, the smaller the Kb. That is, the stronger the acid, the weaker the conjugate base.
Example 1
• Find Kb for CH3COO-.CH3COO-
(aq)+ H2O(l) ↔ CH3COOH(aq) + OH-(aq)
Look up Ka for conjugate acid (CH3COOH)
Ka = 1.8 x 10-5
Ka x Kb = Kw
1.8 x 10-5 x Kb = 1.0 x 10-14
Kb = 5.6 x 10-10
• Find Kb for Na2CO3.Na2CO3(aq) ↔ CO3
-2(aq) + 2Na+
(aq)
CO3-2
(aq) + H2O(l) ↔ HCO3-(aq) + OH-
(aq)
Look up Ka for conjugate acid (HCO3-)
Ka = 4.7 x 10-11
Ka x Kb = Kw
4.7 x 10-11 x Kb = 1.0 10-14
Kb = 2.1 x 10-4
Base
Example 2Ionic salt
Example 3A 0.100 mol/L propanoic acid solution (C2H5COOH(aq)), has a pH of 2.95. From these data, the Kb for the propanoate ion, C2H5COO-(aq) is________.
C2H5COOH(aq) + H2O (l) C2H5COO–(aq) + H3O+ (aq)
Ka = (1.122 x 10 -3) (1.122 x 10-3)
(0.100 – 1.122 x 10-3)
= 1.273 x 10-5
Kb = Kw (1.0 x10-14 )
Ka (1.273 x 10-5 )
= 7.9 x 10-10
Example 4Household ammonia is a cleaning product in which the concentration of ammonia is about 2 mol/L. The pOH of a 2 mol/L ammonia solution is ____________.
NH3(aq) + H2O(l) NH4+
(aq) + OH-(aq)
Kb = [NH4+] [OH-]
[NH3]Kb(NH3) = Kw
Ka (NH4+)
= 1.0 x 10-14
5.6 x 10-10
= 1.785 x 10-5
Now solve :
Kb = x2
2
1.785 x 10-5 = x2
2
X = 5.976 x 10-3
pOH = 2.2
Neutralization
• A reaction that involves equal moles of H+ ions and OH-
ions.
moles H+ = moles OH-
Basic Principle
Neutralization
• Strong or weak acids/bases are of no concern in neutralization reactions since weak acids will totally dissociate in the presence of a base.
• Recall Le Chatelier’s Principle:CH3COOH(aq) ↔ CH3COO-
(aq) + H+ (aq)
Addition of base will remove H+ (H+ + OH- → H2O). A decrease in [H+] will cause a shift to the right to replace the H+. i.e. dissociation increases
WS 18-2
ACIDS• Monoprotic – an acid that has one ionizeable proton.
• (egs…HCl, HI, HF, CH3COOH)
• Diprotic (egs…H2SO4, H2S, H2CO3, H2PO4-)
• Triprotic (egs…H3PO4)
• BASES• Monobasic species – a base that donates one OH-.
(or a base that accepts one proton)
(egs…KOH, LiOH, HCO3-)
• Dibasic species (egs…Ca(OH)2, Mg(OH)2, CO32-)
• Tribasic species (egs…Al(OH)3, PO43-)
Sample Calculations
1. What volume of 0.800 mol/L HCl(aq) is required to neutralize 50.0 mL of 0.500 mol/L NaOH(aq)?
2. 0.800 L of 0.300 mol/L Ca(OH)2 (aq) requires what
volume of 0.500 mol/L H3PO4 (aq) to neutralize it?
3. 150 mL of Ba(OH)2 (aq) is used to neutralize 50.0 mL of 0.040 mol/L of H3PO4 (aq) . What is the concentration of the base?
4. A 4.00 g sample of acetic acid requires 50.0 mL of NaOH(aq) for a neutralization. What is the concentration of the base?
A short cut formula
a CAVA = b CBVB
[Acid]Volume
of Acid[Base]
Volume
of Base
Number of OH- donated per mole of base (1,2 or 3)
Number of ionizeable protons (1,2 or 3)
• Titration is a physical process that is used to find the concentration of an unknown solution. Titration involves adding one solution (titrant) from a buret to another solution (sample) in an Erlenmeyer flask. (titration flask)
Qualitative Titration LabWe will titrate HCl(aq) with NaOH(aq)
CBL Settings: mode: time graph
interval: 0.5s – 1.0s
total time: 120s
Procedure
• Place 10.0 mL of HCl(aq) in a 250 mL beaker.
• Add about 85.0 mL of distilled H2O(l).
• Fill the right hand side buret with the NaOH (aq).
• Set up the apparatus as per the diagram.• Start the magnetic stirrer.• Push ‘start’ to begin collecting data.• Open the valve on the buret and allow the base to drain
into the beaker. (keep adding base until the pH is above 12 for several readings)
• Use graph link to print your graph.
Notes • Watch speed of the magnetic stirrer• Be careful when handling the pH probe. Make sure that
the probe is submerged in the buffer solution when you put it away.
• You must wear your and your
• Or you will be out!
• Have your bench checked before you leave the lab.
Student
Qualitative Titration # 2
• Follow the same procedure as lab 1 with the following changes.
• Substitute CH3COOH(aq) for the HCl(aq)
• That is CH3COOH (aq) with NaOH (aq)
Qualitative Titration # 3
• Follow the same procedure as lab 1 with the following changes.
• Substitute NaHCO3(aq) for the NaOH(aq)
•That is: HCl(aq) with NaHCO3(aq)
Strong acid/strong base
• Large vertical portion
• Wide range of indicators would be useful
• Equivalence point is 7.00
Weak acid/strong base
• Equivalence point is above 7.00
• Vertical segment of the graph is smaller (about 4 pH units)
• Fewer indicators will work. Select an indicator that changes color in a pH range above 7.00
Strong acid/weak base
• pH is less than 7.00 at equivalence point
• Select an indicator that changes color at a pH value less than 7.00
• Vertical segment of the graph covers about 4 pH units
• Fewer indicators will work
Weak acid/weak base
• Poor results
• No vertical segment on the graph
• No indicator will work
• pH at equivalence point is 7.00
• Equivalence point could be found with a pH meter.
WS-18-29
Titration procedure(quantitative)
1) Measured volume of unknown acid [ ] is added to a flask.
2) An appropriate indicator is added.
3) Measured amount of base of known [ ] is
added using a buret.4) Continue until solution changes color. This is
called the end point.
Procedure• Fill the left side buret with acid. (HCl(aq))
• Fill the right side buret with base.(NaOH(aq))
• Record the initial levels on each buret.• Add about 15-20 mL of acid to an erlenmeyer flask.• Add 1 drop of indicator.• Add base from the second buret until the expected color
change occurs.
• Record the final buret levels on each buret.
Notes
• Always keep the acid on the left and the base on the right.
• Stop at the first permanent pink.• Don’t forget to add the indicator.• Back titration is o.k.• You may wash the sides of the flask down with distilled
water.• Make sure that all glassware is very clean. Remember
that you will be marked on accuracy as well.
• [NaOH(aq) ] = 0.120 mol/L
Salt HydrolysisThe ions of a dissolved (dissociated) salt may accept H+
(aq) from water molecules OR donate H+
(aq) to water molecules.
This causes the solution to be slightly acidic or slightly basic depending on the direction of H+
(aq) transfer.
Salt Hydrolysis explains why some neutralization reactions do not result in a pH of 7.00 at the equivalence point.
Example # 1NaCH3COO(s) dissolved in H2O(l)
NaCH3COO(s) → Na+(aq) + CH3COO-
(aq)
Ion will interact with water
CH3COO-(aq) + H2O(l)
Base Acid
CH3COOH(aq) + OH- (aq)
Causes the solution to be slightly basic
Example # 2NH4Cl(s) dissolved in H2O(l)
NH4Cl(s) → NH4+
(aq) + Cl-(aq)
Ion will interact with H2O(l)
NH4+
(aq) + H2O(l)
Acid Base
H3O+(aq) + NH3(aq)
Graph Shapes
• Monoprotic → one vertical segment
• Diprotic → two vertical segments
• Triprotic → three vertical segments
Usually only the first two endpoints graph well
Measuring pHMeasuring pH• Most accurate method to measure pH is to
use a pH meter.
• However, certain chemicals change color as pH changes. These are called indicators.
• Indicators are less precise than pH meters.
• Many indicators do not have a sharp color change as a function of pH.
Indicators• All indicators are weak acids.• Exist in two forms:
– Associated (molecular)– Dissociated (ionized)
• Since they are weak acids, the associated form is favored in water.
An examplePhenolphthalein (HPh)
HPh(aq) ↔ H+(aq) + Ph-
(aq)
- associated form - dissociated form
- favored in acid/H2O - favored in base
- clear - pink
If a base is added to the above reaction, H+(aq) will be
removed. The equilibrium will shift to the right. The molecular HPh(aq) dissociates → solution turns pink.
If an acid is added to the above reaction, the build up of H+
(aq) causes an equilibrium shift to the left. The ions bond back together to form molecular HPh(aq) and the solution turns colorless.
Draw the reaction for what occurs to an indicator in an acid when a strong base is added as a titrant.
HIn(aq) + OH-(aq) ↔ H2O(l) + In-
(aq)
Show the reaction for what occurs to an indicator in a base when a strong acid is added.
In-(aq) + H3O+
(aq) ↔ HIn(aq) + H2O (l)
Buffers• A buffer consists of a mixture of a weak acid
and the salt of its conjugate base.• The pH of a buffer solution remains fairly
constant when the solution is diluted or when a small amount of strong acid or base is added.
• These are required in living systems and many industrial processes. (Any place where a constant pH is required.)
Buffers in the human body
• Blood and interstitial fluid are buffer solutions of carbonic acid and bicarbonate ions. If the blood were not buffered, the acid from a single glass of orange juice could be fatal.
ExampleThink of a buffer as a storage tank for Hydrogen ions. If we have too many, we can place some into storage and if we
don’t have enough, we can take some out of storage.
H2CO3 (aq) ↔ H+(aq) + HCO3
-(aq)
• Add base: H+ removed → equilibrium shifts right and replaces the H+ removed.
• Add acid: Equilibrium shifts left and removes the Excess H+ added. (stores them as H2CO3 molecules)
Storage tank→
H+ Into storage
Remove H+ from storage
Other Human Buffers
H2PO4-(aq) / HPO4 2-
(aq)
An example of buffering
initial pH (1.0L) pH after adding 1 mL
of 10.0 mol/L HCl (aq)
Neutral saline 7.0 2.0
Blood plasma 7.4 7.2