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11The Chemistry The Chemistry of Acids and of Acids and BasesBases
The Chemistry The Chemistry of Acids and of Acids and BasesBases SAVE PAPER AND INK!!! When you
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55Acids
Have a sour taste. Vinegar is a solution of acetic acid. CitrusHave a sour taste. Vinegar is a solution of acetic acid. Citrusfruits contain citric acid.fruits contain citric acid.
React with certain metals to produce hydrogen gasReact with certain metals to produce hydrogen gas..
React with carbonates and bicarbonates to produce carbon React with carbonates and bicarbonates to produce carbon dioxide gasdioxide gas
Have a bitter taste.Have a bitter taste.
Feel slippery. Many soaps contain bases.Feel slippery. Many soaps contain bases.
Bases
66
Some Properties of Acids
Produce H+ (as H3O+) ions in water (the hydronium ion is a
hydrogen ion attached to a water molecule)
Taste sour
Corrode metals
Electrolytes
React with bases to form a salt and water
pH is less than 7
Turns blue litmus paper to red “Blue to Red A-CID”
77
Anion Ending Acid Name
-ide hydro-(stem)-ic acid
-ate (stem)-ic acid
-ite (stem)-ous acid
Acid Nomenclature Review
No OxygenNo Oxygen
w/Oxygen w/Oxygen
An easy way to remember which goes with which…An easy way to remember which goes with which…
““In the cafeteria, you In the cafeteria, you ATEATE something something ICICky”ky”
88Acid Nomenclature Flowchart
h yd ro - p re fix-ic en d in g
2 e lem en ts
-a te en d in gb ecom es-ic en d in g
-ite en d in gb ecom es
-o u s en d in g
n o h yd ro - p re fix
3 e lem en ts
AC ID Ss ta rt w ith 'H '
99
• HBr HBr (aq)(aq)
• HH22COCO33
• HH22SOSO33
hydrohydrobromicbromic acidacid
carboncarbonicic acidacid
sulfursulfurousous acidacid
Acid Nomenclature Review
1111
Some Properties of Bases
Produce OHProduce OH-- ions in water ions in water
Taste bitter, chalkyTaste bitter, chalky
Are electrolytesAre electrolytes
Feel soapy, slipperyFeel soapy, slippery
React with acids to form salts and waterReact with acids to form salts and water
pH greater than 7pH greater than 7
Turns red litmus paper to blue “Turns red litmus paper to blue “BBasic asic BBlue”lue”
1212
Some Common Bases
NaOHNaOH sodium hydroxidesodium hydroxide lyelye
KOHKOH potassium hydroxidepotassium hydroxide liquid soapliquid soap
Ba(OH)Ba(OH)22 barium hydroxidebarium hydroxide stabilizer for plasticsstabilizer for plastics
Mg(OH)Mg(OH)22 magnesium hydroxidemagnesium hydroxide “MOM” Milk of magnesia“MOM” Milk of magnesia
Al(OH)Al(OH)33 aluminum hydroxidealuminum hydroxide Maalox (antacid)Maalox (antacid)
1313
Acid/Base definitions
• Definition #1: Arrhenius (traditional)
Acids – produce H+ ions (or hydronium ions H3O+)
Bases – produce OH- ions
(problem: some bases don’t have hydroxide ions!)
1414Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
1515
Acid/Base Definitions
• Definition #2: Brønsted – Lowry
Acids – proton donor
Bases – proton acceptor
A “proton” is really just a hydrogen atom that has lost it’s electron!
1616
A Brønsted-Lowry acid is a proton donorA Brønsted-Lowry base is a proton acceptor
acidconjugate
basebase conjugate
acid
1717
ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES
The Brønsted definition means NHThe Brønsted definition means NH33 is is aa BASEBASE in water — and water is in water — and water is itself anitself an ACIDACID
BaseAcidAcidBaseNH4
+ + OH-NH3 + H2OBaseAcidAcidBase
NH4+ + OH-NH3 + H2O
1919
Learning Check!
Label the acid, base, conjugate acid, and Label the acid, base, conjugate acid, and conjugate base in each reaction:conjugate base in each reaction:
HCl + OHHCl + OH-- Cl Cl-- + H + H22OO HCl + OHHCl + OH-- Cl Cl-- + H + H22OO
HH22O + HO + H22SOSO44 HSO HSO44-- + H + H33OO
++ HH22O + HO + H22SOSO44 HSO HSO44-- + H + H33OO
++
2020Acids & Base Acids & Base DefinitionsDefinitions
Lewis acid - a Lewis acid - a substance that substance that accepts an electron accepts an electron pairpair
Lewis base - a Lewis base - a substance that substance that donates an electron donates an electron pairpair
Definition #3 – Lewis Definition #3 – Lewis
2121
Formation ofFormation of hydronium ion hydronium ion is also an is also an excellent example.excellent example.
Lewis Acids & BasesLewis Acids & Bases
•Electron pair of the new O-H bond Electron pair of the new O-H bond originates on the Lewis base.originates on the Lewis base.
HH
H
BASE
••••••
O—HO—H
H+
ACID
2323
Lewis Acid-Base Lewis Acid-Base Interactions in BiologyInteractions in Biology
• The heme group The heme group in hemoglobin in hemoglobin can interact with can interact with OO22 and CO. and CO.
• The Fe ion in The Fe ion in hemoglobin is a hemoglobin is a Lewis acidLewis acid
• OO22 and CO can act and CO can act as Lewis basesas Lewis bases
Heme group
2424The The pH scalepH scale is a way of is a way of expressing the strength expressing the strength of acids and bases. of acids and bases. Instead of using very Instead of using very small numbers, we just small numbers, we just use the NEGATIVE use the NEGATIVE power of 10 on the power of 10 on the Molarity of the HMolarity of the H++ (or (or OHOH--) ion.) ion.
Under 7 = acidUnder 7 = acid 7 = neutral 7 = neutral
Over 7 = baseOver 7 = base
2626Calculating the pH
pH = - log [H+](Remember that the [ ] mean Molarity)
Example: If [H+] = 1 X 10-10
pH = - log 1 X 10-10
pH = - (- 10)
pH = 10
Example: If [H+] = 1.8 X 10-5
pH = - log 1.8 X 10-5
pH = - (- 4.74)
pH = 4.74
2727
Try These!Try These!
Find the pH of Find the pH of these:these:
1) A 0.15 M solution 1) A 0.15 M solution of Hydrochloric of Hydrochloric acidacid
2) A 3.00 X 102) A 3.00 X 10-7-7 M M solution of Nitric solution of Nitric acidacid
2828pH calculations – Solving pH calculations – Solving for H+for H+pH calculations – Solving pH calculations – Solving for H+for H+
If the pH of Coke is 3.12, [HIf the pH of Coke is 3.12, [H++] = ???] = ???
Because pH = - log [HBecause pH = - log [H++] then] then
- pH = log [H- pH = log [H++]]
Take antilog (10Take antilog (10xx) of both) of both sides and get sides and get
1010-pH -pH == [H[H++]][H[H++] = 10] = 10-3.12-3.12 = 7.6 x 10 = 7.6 x 10-4-4 M M *** to find antilog on your calculator, look for “Shift” or “2*** to find antilog on your calculator, look for “Shift” or “2nd nd
function” and then the log buttonfunction” and then the log button
2929pH calculations – Solving for pH calculations – Solving for H+H+
• A solution has a pH of 8.5. What is the A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the Molarity of hydrogen ions in the solution?solution?
pH = - log [HpH = - log [H++]]
8.5 = - log [H8.5 = - log [H++]]
-8.5 = log [H-8.5 = log [H++]]
Antilog -8.5 = antilog (log [HAntilog -8.5 = antilog (log [H++])])
1010-8.5-8.5 = [H = [H++]]
3.16 X 103.16 X 10-9-9 = [H = [H++]]
pH = - log [HpH = - log [H++]]
8.5 = - log [H8.5 = - log [H++]]
-8.5 = log [H-8.5 = log [H++]]
Antilog -8.5 = antilog (log [HAntilog -8.5 = antilog (log [H++])])
1010-8.5-8.5 = [H = [H++]]
3.16 X 103.16 X 10-9-9 = [H = [H++]]
3030
More About WaterMore About Water
Amphoteric: HAmphoteric: H22O can function as both an ACID and O can function as both an ACID and
a BASE.a BASE.
In pure water there can beIn pure water there can be AUTOIONIZATIONAUTOIONIZATION
Equilibrium constant for water = KEquilibrium constant for water = Kww
KKww = [H = [H33OO++] [OH] [OH--] =] = 1.00 x 101.00 x 10-14-14 at 25 at 25 ooCC
3131
More About WaterMore About Water
KKww = [H = [H33OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC
In a In a neutral neutral solution [Hsolution [H33OO++] = [OH] = [OH--]]
so Kso Kww = [H = [H33OO++]]22 = [OH = [OH--]]22
and so [Hand so [H33OO++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 M M
OH-
H3O+
OH-
H3O+
AutoionizationAutoionization
3232pOH
• Since acids and bases are Since acids and bases are opposites, pH and pOH are opposites, pH and pOH are opposites!opposites!
• pOH does not really exist, but it is pOH does not really exist, but it is useful for changing bases to pH.useful for changing bases to pH.
• pOH looks at the perspective of a pOH looks at the perspective of a basebase
pOH = - log [OHpOH = - log [OH--]]Since pH and pOH are on opposite Since pH and pOH are on opposite
ends,ends,pH + pOH = 14pH + pOH = 14
3434
[H[H33OO++], [OH], [OH--] and pH] and pHWhat is the pH of the What is the pH of the
0.0010 M NaOH solution? 0.0010 M NaOH solution?
[OH-] = 0.0010 (or 1.0 X 10[OH-] = 0.0010 (or 1.0 X 10-3-3 M) M)
pOH = - log (0.0010)pOH = - log (0.0010)
pOH = 3pOH = 3
pH = 14 – 3 = 11pH = 14 – 3 = 11
OR KOR Kww = [H = [H33OO++] [OH] [OH--]]
[H[H3OO++] = 1.0 x 10] = 1.0 x 10-11-11 M M
pH = - log (1.0 x 10pH = - log (1.0 x 10-11-11) = 11.00) = 11.00
3535The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater?
The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood?
3636
[OH[OH--]]
[H[H++]] pOHpOH
pHpH
1010 -pOH
-pOH
1010 -pH-pH-Log[H
-Log[H++]]
-Log[OH
Log[OH
--]]
14 -
pOH
14 -
pOH
14 -
pH
14 -
pH
1.0
x 10
1.0
x 10-1
4-14
[OH[O
H-- ]]
1.0
x 10
1.0
x 10-1
4-14
[H[H
++ ]]
3737Calculating [H3O+], pH, [OH-], and pOH
Problem 1: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H3O+], pH, [OH-], and pOH of the two solutions at 25°C.
Problem 2: What is the [H3O+], [OH-], and pOH of a solution with pH = 3.67? Is this an acid, base, or neutral?
Problem 3: Problem #2 with pH = 8.05?
3838
HNO3, HCl, H2SO4 and HClO4 are among the only known strong acids.
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
The strength of an acid (or base) is determined by the amount of IONIZATION.
The strength of an acid (or base) is determined by the amount of IONIZATION.
3939
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
• Generally divide acids and bases into STRONG or Generally divide acids and bases into STRONG or WEAK ones.WEAK ones.
STRONG ACID:STRONG ACID: HNOHNO3 3 (aq) + H(aq) + H22O (l) --->O (l) --->
HH33OO+ + (aq) + NO(aq) + NO33- - (aq)(aq)
HNOHNO33 is about 100% dissociated in water. is about 100% dissociated in water.
This is the same for other oxyacids (acids in which This is the same for other oxyacids (acids in which the acidic hydrogen is attached to an oxygen the acidic hydrogen is attached to an oxygen atomatom
4040
• Weak acidsWeak acids are much less than 100% ionized in are much less than 100% ionized in
water.water.
One of the best known is acetic acid = CHOne of the best known is acetic acid = CH33COCO22HH
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
4141
Organic Acids
• Acetic acid is an organic acid
• Organic acids are acids with a carbon backbone and a carboxyl group, which is the acidic part
– The carboxyl group consists of a carbon atom that is doubly bound to oxygen and is also bound to a hydroxyl group
– The proton (H+) of the hydroxyl group is donated
• These are weak acids
4242
Polyprotic Acids
• Some acids have more than one proton to donate
• Example (H2SO4): – Donation of the 1st proton
» H2SO4 (aq) => H+ (aq) + HSO4-
– Donation of the 2nd proton
» HSO4- (aq) => H+ (aq) + SO4
2- (aq)
– The first proton dissociates completely, making H2SO4 a strong acid
– Most of HSO4- remains intact, making HSO4
- a weak acid
4343
• Strong Base:Strong Base: 100% dissociated in 100% dissociated in water.water.
NaOH (aq) ---> NaNaOH (aq) ---> Na+ + (aq) + OH(aq) + OH- - (aq)(aq)
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
Other common strong Other common strong bases include KOH andbases include KOH and Ca(OH)Ca(OH)22..
CaO (lime) + HCaO (lime) + H22O -->O -->
Ca(OH)Ca(OH)22 (slaked lime) (slaked lime)CaOCaO
4444
• Weak base:Weak base: less than 100% ionized less than 100% ionized in waterin water
One of the best known weak bases is One of the best known weak bases is ammoniaammonia
NHNH3 3 (aq) + H(aq) + H22O (l) O (l) NH NH44+ + (aq) + OH(aq) + OH- - (aq)(aq)
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
4646
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Consider acetic acid, HCConsider acetic acid, HC22HH33OO22 (HOAc) (HOAc)
HCHC22HH33OO22 + H + H22O O H H33OO++ + C + C22HH33OO22 --
AcidAcid Conj. base Conj. base
Ka [H3O+][OAc- ]
[HOAc] 1.8 x 10-5Ka
[H3O+][OAc- ][HOAc]
1.8 x 10-5
(K is designated K(K is designated Kaa for ACID) for ACID)
K gives the ratio of ions (split up) to molecules K gives the ratio of ions (split up) to molecules
(don’t split up)(don’t split up)
4747Ionization Constants for Ionization Constants for Acids/Bases Acids/Bases
AcidsAcids ConjugateConjugateBasesBases
Increase strength
Increase strength
4848
Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids
Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids
Weak acid has KWeak acid has Kaa < 1 < 1
Leads to small [HLeads to small [H33OO++] and a pH of 2 - 7] and a pH of 2 - 7
4949
Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Weak base has KWeak base has Kbb < 1 < 1
Leads to small [OHLeads to small [OH--] and a pH of 12 - 7] and a pH of 12 - 7
5151Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
You have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H33OO++, OAc, OAc--, ,
and the pH.and the pH.
Step 1.Step 1. Define equilibrium concs. in ICE Define equilibrium concs. in ICE
table.table.
[HOAc][HOAc] [H[H33OO++]] [OAc[OAc--]]
initialinitial
changechange
equilibequilib
1.001.00 00 001.001.00 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
1.00-x1.00-x xx xx1.00-x1.00-x xx xx
5252Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Step 2.Step 2. Write KWrite Kaa expression expression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - xKa 1.8 x 10-5 =
[H3O+][OAc- ][HOAc]
x2
1.00 - x
This is a quadratic. Solve using This is a quadratic. Solve using quadratic formula.quadratic formula.
or you can make an approximation if x is very or you can make an approximation if x is very small! (Rule of thumb: 10small! (Rule of thumb: 10-5-5 or smaller is ok) or smaller is ok)or you can make an approximation if x is very or you can make an approximation if x is very small! (Rule of thumb: 10small! (Rule of thumb: 10-5-5 or smaller is ok) or smaller is ok)
5353Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Step 3.Step 3. Solve KSolve Kaa expression expression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - xKa 1.8 x 10-5 =
[H3O+][OAc- ][HOAc]
x2
1.00 - x
First assume x is very small because First assume x is very small because KKaa is so small. is so small.
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00
Now we can more easily solve this Now we can more easily solve this approximate expression.approximate expression.
5454
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve KSolve Kaa approximateapproximate expressionexpression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00
x =x = [[HH33OO++] = [] = [OAcOAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M
pH = - log [pH = - log [HH33OO++] = -log (4.2 x 10] = -log (4.2 x 10-3-3) =) = 2.372.37
5555Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Calculate the pH of a 0.0010 M solution of Calculate the pH of a 0.0010 M solution of formic acid, HCOformic acid, HCO22H.H.
HCOHCO22H + HH + H22O O HCO HCO22-- + H + H33OO++
KKaa = 1.8 x 10 = 1.8 x 10-4-4
Approximate solutionApproximate solution
[H[H33OO++] = 4.2 x 10] = 4.2 x 10-4-4 M, M, pH = 3.37pH = 3.37
Exact SolutionExact Solution
[H[H33OO++] = [HCO] = [HCO22--] = 3.4 x 10] = 3.4 x 10-4-4 M M
[HCO[HCO22H] = 0.0010 - 3.4 x 10H] = 0.0010 - 3.4 x 10-4-4 = 0.0007 M = 0.0007 M
pH = 3.47 pH = 3.47
5656Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 1.Step 1. Define equilibrium concs. in ICE tableDefine equilibrium concs. in ICE table
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial
changechange
equilibequilib
0.0100.010 00 000.0100.010 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx
5757Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 1.Step 1. Define equilibrium concs. in ICE tableDefine equilibrium concs. in ICE table
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial
changechange
equilibequilib
0.0100.010 00 000.0100.010 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx
5858Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression
Kb 1.8 x 10-5 = [NH4
+][OH- ][NH3 ]
= x2
0.010 - xKb 1.8 x 10-5 =
[NH4+][OH- ]
[NH3 ] =
x2
0.010 - x
Assume x is small, soAssume x is small, so x = [OHx = [OH--] = [NH] = [NH44
++] = 4.2 x 10] = 4.2 x 10-4-4 M M
and [NHand [NH33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10-4-4 ≈ 0.010 M ≈ 0.010 M
The approximation is validThe approximation is valid !!
5959Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 3.Step 3. Calculate pHCalculate pH
[OH[OH--] = 4.2 x 10] = 4.2 x 10-4-4 M M
so pOH = - log [OHso pOH = - log [OH--] = 3.37] = 3.37
Because pH + pOH = 14,Because pH + pOH = 14,
pH = 10.63pH = 10.63
6161pH testing
• There are several ways to test pHThere are several ways to test pH
–Blue litmus paper (red = acid)Blue litmus paper (red = acid)
–Red litmus paper (blue = basic)Red litmus paper (blue = basic)
–pH paper (multi-colored)pH paper (multi-colored)
–pH meter (7 is neutral, <7 acid, >7 pH meter (7 is neutral, <7 acid, >7 base)base)
–Universal indicator (multi-colored)Universal indicator (multi-colored)
– Indicators like phenolphthaleinIndicators like phenolphthalein
–Natural indicators like red cabbage, Natural indicators like red cabbage, radishesradishes
6262Paper testing
• Paper tests like litmus paper and pH Paper tests like litmus paper and pH paperpaper
– Put a stirring rod into the solution Put a stirring rod into the solution and stir.and stir.
– Take the stirring rod out, and Take the stirring rod out, and place a drop of the solution from place a drop of the solution from the end of the stirring rod onto a the end of the stirring rod onto a piece of the paperpiece of the paper
– Read and record the color change. Read and record the color change. Note what the color indicates. Note what the color indicates.
– You should only use a small You should only use a small portion of the paper. You can use portion of the paper. You can use one piece of paper for several one piece of paper for several tests.tests.
6464
pH meter
• Tests the voltage of the Tests the voltage of the electrolyteelectrolyte
• Converts the voltage to Converts the voltage to pHpH
• Very cheap, accurateVery cheap, accurate
• Must be calibrated with Must be calibrated with a buffer solutiona buffer solution
6565pH indicators
• Indicators are dyes that can be added that will change color in the presence of an acid or base.
• Some indicators only work in a specific range of pH
• Once the drops are added, the sample is ruined
• Some dyes are natural, like radish skin or red cabbage
6666
ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations
ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations
HH22CC22OO44(aq) + 2 NaOH(aq) --->(aq) + 2 NaOH(aq) --->
acidacid basebase
NaNa22CC22OO44(aq) + 2 H(aq) + 2 H22O(liq)O(liq)
Carry out this reaction using aCarry out this reaction using a TITRATIONTITRATION..
Oxalic acid,Oxalic acid,
HH22CC22OO44
6868
TitrationTitrationTitrationTitration1. Add solution from the buret.1. Add solution from the buret.2. Reagent (base) reacts with 2. Reagent (base) reacts with
compound (acid) in solution in compound (acid) in solution in the flask.the flask.
3.3. Indicator shows when exact Indicator shows when exact stoichiometric reaction has stoichiometric reaction has occurred.occurred.
pH= 7pH= 7[H[H++] = [OH] = [OH--]]=> Equivalence/stoichiometric point=> Equivalence/stoichiometric point
This is called NEUTRALIZATION.This is called NEUTRALIZATION.
7070
35.62 mL of NaOH is 35.62 mL of NaOH is
neutralized with 25.2 mL of neutralized with 25.2 mL of
0.0998 M HCl by titration to 0.0998 M HCl by titration to
an equivalence point. What an equivalence point. What
is the concentration of the is the concentration of the
NaOH?NaOH?
LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration.determine its concentration.
LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration.determine its concentration.
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Buffered Solutions
• A buffered solution resists changes in its pH
• A solution is buffered by the presence of a weak acid and its conjugate base
– Ex. Acetic acid, HC2H3O2 and a compound that contains its conjugate base (C2H3O2
-), sodium acetate
» In solution, the Na+ dissociates from the C2H3O2-
» Since acetic acid is a weak acid, we know acetate is a strong base (has a high affinity for the H=)
• Acetate will accept H+ that dissociates from a strong acid
» If a strong base is added to the solution, it will pull the H+ away from the acetic acid (OH- has a much stronger affinity for H+ than acetate)
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Buffer Characteristics
1. The solution contains a weak acid, HA and its conjugate base, A-
2. The buffer resists any changes in pH by reacting with any added H+ or OH- so that these ions do not accumulate
3. Any added H+ reacts with the base A-
• H+ (aq) + A- (aq) => HA (aq)
4. Any added OH- reacts with the weak acid HA1. OH- (aq) + HA (aq) => H2O (l) + A- (aq)