Chapter 14
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Transcript of Chapter 14
Chapter 14
Acids and Bases
Naming Acids
2 types of acids Binary ternary (sometimes called oxy)
binary -H and one other type of atom name them hydro _________ ic acid
Naming Acids
Ex1 HCl Hydrochloric Acid
Ex2 HBr Hydrobromic Acid
Ex3 H3P Hydrophosphoric Acid
Writing formulas from names for Acids
Criss Cross charges Ex4 Hydronitric Acid
H3N Ex5 Hydrosulfuric Acid
H2S
Naming Acids
ternary (oxy) acids H with a polyatomic ion Do not start with Hydro-
Change the –ate ending to –ic Change the – ite ending to –ous
Naming Acids
Ex6 H2SO4 Sulfuric Acid
Ex7 H2SO3 Sulfurous Acid
Ex8 HClO4 Perchloric Acid
Ex9 HClO Hypochlorous Acid
Writing Formulas From Names
Ex10 Nitric Acid HNO3
Ex11 Phosphorous Acid H3PO3
Some common acids:
Sulfuric – used for fertilizer, petroleum, production of metal, paper, paint
HCl – stomach acid, food processing, iron, steel
Acetic acid – vinegar, fungicide, produced by fermentation
Nitric acid – explosives, rubber, plastics, dyes, drugs
Phosphoric acid – beverage flavoring, animal feed, detergents
Properties of Acids:
Acid comes from Latin meaning acidus, or sour tasting.
Affect the colors of indicators. An indicator is a chemical that shows one color in an acid and another in a base. Acids turn blue litmus red.
Properties of Acids:
Acids react with bases to produce salt and water. This is called neutralization.
HCl(aq) + NaOH(aq) NaCl(aq) + HOH (l) 3H2SO4(aq)+ 2Al(OH)3(aq) Al2(SO4)3(aq)
+ 6 HOH (aq)
Properties of Acids:
Acids ionize in water. So, they conduct electricity (electrolytes).
Acids react with active metals to produce salts and hydrogen.
Mg+ + 2 HCl (aq) MgCl 2(aq) + H 2(g) Cu(s) + HCl(aq) NR
Arrhenius Acids
Substances that produces H+ ions when mixed with water. HCl(g) + H2O(l) H+1
(aq) + Cl-1 (aq)
It is now found that: H+1
(aq) + H2O(l) H3O+1 (aq)
so it is really… HCl(g) + H2O(l) H3O+1
(aq) + Cl-1 (aq)
Definitions of Acids
Bronsted-Lowery Acids – proton donors
Show HCl + water and HCl + ammonia
HCl + H2O H3O+(aq) + Cl-1(aq)
HCl + NH3 NH4+
(aq) + Cl-1(aq) (*not Cl2!!)
Types of Acids
Strong - HI, HBr, HCl, HNO3, H2SO4, HClO4 (one way arrows always!) HBr + H2O H3O+
(aq)+ Br-1 (aq)
Weak – HF, H2PO4, H2CO3, H2PO4
(double arrows always!) HF (aq) + H2O (l) H3O+1
(aq) + F-1 (aq)
Molecular, Total Ionic, and Net Ionic Equations for acids:
Molecular Equation: Zn(s) + 2 HCl(aq) ZnCl2(aq) + H 2(g)
Total Ionic Equation: Zn(s) + 2H+1
(aq) + 2Cl-1 (aq)
Zn+2(aq) + 2Cl-1(aq) + H2(g)
Net Ionic Equation:
Zn(s) + 2H+1(aq) Zn+2
(aq) + H2(g)
Some acids donate more than 1 proton….
Monoprotic (HF) - an acid that donates one proton (one hydrogen) Ex1: Write the reaction(s) showing
the complete ionization of HF. HF (aq) + H2O (l) H3O+1
(aq) + F-1 (aq)
Some acids donate more than 1 proton….
Diprotic (H2SO4) - an acid that donates two protons (two hydrogens) Ex2: Write the reaction(s) showing the complete
ionization of H2SO4. H2SO4 (aq) + H2O (l) H3O+1 (aq) + HSO4
-1 (aq)
HSO4-1
(aq) + H2O(l) H3O+1(aq) + SO4
-2 (aq)
__________________________________H2SO4 (aq) + 2 H2O(l) 2 H3O +1
(aq)+SO4-2
(aq)
*Note: When you lose a H+1, you gain a negative.
Some acids donate more than 1 proton….
Triprotic Acid: an acid that donates three protons (three hydrogens).
Ex3: Write the reaction(s) showing the complete ionization of H3PO4.
H3PO 4(aq) + H2O (l) H3O +1 (aq) + H2PO4 -1 (aq)
H2PO4 -1 (aq) + H2O (l) H3O +1 (aq) + HPO4
-2 (aq)
HPO4 -2
(aq) + H2O (l) H3O +1 (aq) + PO4 -3 (aq)
________________________________________H3PO 4(aq) + 3 H2O (l) 3 H3O +1 (aq) + PO4
-3 (aq)
Some acids donate more than 1 proton….
Diprotic and Triprotic can also be referred to as polyprotic.
2nd and 3rd ionizations are always weak (so, ).
Bases
Bases are used in cleaners (floors, drains, ovens), react with fats and oils so they become water soluble, used to neutralize stomach acid (antacids), used as laxatives
Properties of Bases
Bases are electrolytes. They dissociate in water. NaOH and KOH are strong electrolytes because they are both highly soluble.
Affect the colors of indicators. An indicator is a chemical
that shows one color in an acid and another in a base. Bases turn red litmus blue.
Bases react with acids to produce salt and water. This is called neutralization.
Bases taste bitter and feel slippery. Soap is an example
of a base.
Definition of Bases
A substance that has OH- ions. Bases dissociate in water to give OH- & positive metal ions.
Types of Bases
1. Traditional Bases (Arrhenius) – a substance that contains hydroxide ions and dissociates to give hydroxide ions in water.
NaOH(s) + H2O Na+(aq) + OH-
(aq)
Mg(OH)2(s) + H2O Mg+2(aq) + 2OH-
(aq)
Types of Bases
2. Bronsted- Lowry bases – proton acceptors
NH3(g) + H2O(l) NH4+1
(aq) + OH-1(aq)
* Water is amphoteric. It can act as an acid or base.
Types of Bases
List of strong bases: NaOH, KOH, CsOH, Ca(OH)2
List of weak bases: many organic compounds with N…NH3,
C6H5NH2, C2H3O2-
Hydroxides of Column I and II are strong bases
Neutralization reactions – hydronium + hydroxide yields water
It is a type of double replacement reaction. Note: H2O = HOH Acid + Base → Salt and Water
General Formula: HX + MOH MX + H2O
Neutralization Reaction
Example: hydrochloric acid + barium hydroxide ( molecular, total ionic, net ionic)
2HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2HOH(l)
2H+1(aq) + 2Cl-1(aq) + Ba+2
(aq) + 2OH-1(aq)
Ba+2(aq) + 2Cl-1(aq) + 2HOH(l)
2H+1(aq) + 2OH-1
(aq) 2HOH(l)
Relative Strengths of Acids and Bases
Some acids and bases are stronger than others. Bronsted (Danish) and Lowry (English)
independently discovered that acids are proton donors and bases are proton acceptors. A proton is a hydrogen ion.
Relative Strengths of Acids and Bases
Strong Acid Example: HCl(g) + H2O(l) H3O+1
(aq) + Cl-1(aq)
(Acid) (Base)
Weak Base Example: NH3(g)+ H2O(l) NH4
+1(aq) + OH-1
(aq)
(Base) (Acid)
Remember – water is amphoteric!
Conjugate Acids & Bases
Conjugate Acid: the substance that was the base and now acts as an acid.
Conjugate Base: the substance that was the acid and now acts as a base.
HCl(g) + H2O(l) H3O+1(aq) + Cl-1(aq)
(Acid) (Base) (Conjugate Acid) (Conjugate Base)
Conjugate Acids & Bases
NH3(g) + H2O(l) NH4+1
(aq) + OH-1(aq)
HF(l) + H2O(l) H3O+1
(aq) + F-1(aq)
Conjugate Acids and bases
H2CO3(aq) + H2O(l) H3O+1(aq) + HCO3
-1(aq)
The stronger the acid, the weaker the conjugate base.
Proton transfer reactions favor the production of the weaker acid and the
weaker base.
Conj. Acid/Base Practice
Complete the equation and label acid base pairs HSO4
-1(aq) + HCO3
-1(aq)
Write an equation showing how NH2-1 is a
stronger base than HSO4-1
Conj. Acid/Base Practice
Which one is correct? HSO4
-1(aq) + H3O+1
(aq) H2SO4(aq) + H2O(l)
(Base) (Acid) (Conj. Acid) (Conj. Base)
or HSO4
-1(aq) + OH-1
(aq) SO4-2
(aq) + H2O(l)
(Acid) (Base) (Conj. Base) (Conj. Acid)
The second reaction is favored because a weaker conjugate acid/base is produced.
Stuff to know for Acids and Bases
2nd and 3rd ionizations are always weak. This means a double yield sign ().
Memorize these strong acids. Strong means a single yield sign ().
HI, HBr, HCl, HNO3, H2SO4, HClO4 All other acids get double yield signs. Strong bases include metals from
column #1 and column #2 (below magnesium).
Proton reactions favor the formation of the weaker acid and base.
Reactions of Acids and Bases
Neutralization (double replacement): Acid + Base Salt + Water HX + MOH MX + H2O
Ex1: HCl(aq) + NaOH(aq) NaCl(aq) + HOH(l)
Reactions of Acids and Baes
Acid + Metal (single replacement): Metal + Acid Salt + Hydrogen
Ex2: Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g)
Reactions of Acids and Bases
Acid in water: Acid + Water Hydronium Ion + Negative Ion
Ex3: HCl(g) + H2O(l) H3O+1(aq) + Cl-1(aq)
Reactions of Acids and Bases
Traditional Base (ends with OH) in water (dissociation): Base + Water Positive Ion + Hydroxide
Ex4: Fe(OH)3(s) + H2O(l) Fe+3
(aq) + 3 OH-1(aq)
Reactions of Acids and Bases
Formation of acids and bases from anhydrides - synthesis (anhydride “without water”): Nonmetal oxide + water acid Ex5a:
CO2(g) + H2O(l) H2CO3(aq)
Ex5b: SO3(g) + H2O(l) H2SO4(aq)
Note: just add the nonmetal oxide to the water to determine the product.
Reactions of Acids and Bases
Metal oxide + water base Ex5c: Na2O(s) + H2O(l) 2 NaOH(aq)
Ex5d: MgO(s) + H2O(l) Mg(OH)2(aq)
Reactions of Acids and Bases
Acid and metal oxide (really just an acid and a base): Acid + Metal Oxide Salt + Water Ex6: H2SO4(aq) + CuO(s)
Turn CuO into Cu(OH)2 H2SO4(aq) + Cu(OH)2 (aq) CuSO4(aq) + HOH(l)
Now, re-write the original reactants, new products, and balance.
H2SO4(aq) + CuO(s) CuSO4(aq) + H2O(l)
Reactions of Acids and Bases
Base and nonmetal oxide (really just an acid and a base Base + Nonmetal Oxide Salt + Water
Ex7a: CO2(g) + NaOH(aq) NaHCO3(aq)
Reactions of Acids and Bases
This is a little confusing. So these reactions will be done like: CO2(g) + NaOH(aq)
Turn CO2 into H2CO3 H2CO3(aq) + NaOH(aq) Na2CO3(aq) + HOH(l)
Now, re-write the original reactants, new products, and balance. CO2(g) + 2 NaOH(aq) Na2CO3(aq) + H2O(l)
Reactions of Acids and Bases
Ex7b: 2 CO2(g) + Ca(OH)2(aq) Ca(HCO3)2(aq)
CO2(g) + Ca(OH)2(aq) Turn CO2 into H2CO3 H2CO3(aq) + Ca(OH)2(aq) CaCO3(aq) + HOH(l) Now, re-write the original reactants, new
products, and balance. CO2(g) + Ca(OH)2(aq) CaCO3(aq) + H2O(l)
Reactions of Acids and Bases
Metal oxide and nonmetal oxide It is like an acid base reaction they yield salt.
However, it does not produce water since no hydrogen is involved.
Ex8a: MgO(s) + CO2(g) MgCO3(s)
Note: just add the nonmetal oxide to the metal oxide to determine the product. Ex8b: CuO(s) + SO3(g) CuSO4(s)
Aqueous Solutions and the Concept of pH
Tap water conducts electricity – why? – many ions present: examples: Distilled water appears to not conduct
electricity, but it does – just a little, tiny bit
H2O + H2O H3O+1 + OH-1
Aqueous Solutions and the Concept of pH
The normal way to express the quantity of hydronium and hydroxide ions is in moles/L (M)
At 25 C0, [H3O+1] = 1 x 10-7 so [OH-1] = 1 x 10-7
These numbers are constant in neutral solution, so we can multiply them to get a constant
We call this constant Kw - ionization constant for water
Kw = [H3O+1][OH-1]
Aqueous Solutions and the Concept of pH
At 25 C0, [H3O+1] = 1 x 10-7
so [OH-1] = 1 x 10-7
so Kw = 1 x 10-14
Example: If the [H3O+1] is 1 x 10-3M, then what is the [OH-1]?
The solution is acidic because the hydronium ion concentration is greater than the hydroxide concentration.
Kw Practice
Beaker # [H3O+1] [OH-1] Acid or Base
1 1 x 10-5 1 x 10-9 Acid
2 1 x 10-12 1 x 10-2 Base
3 2 x 10-4 5 x 10-11 Acid
4 2.40 x 10-9 4.16 x 10-6 Base
Fill in the table below:
Aqueous Solutions and the Concept of pH
pH stands for parts per million of
Hydrogen ion.
How to calculate strengths of Acids and Bases
pH = - log 10 [H3O+]
pOH = - log 10 [OH-]
[H3O+] = 10 –pH
[OH-] = 10-pOH
pH + pOH = 14
How to calculate strengths of Acids and Bases
Log Review Logs are functions of exponents ex. log of 1000 = ex. log of .01 =
How to calculate strengths of Acids and Bases
To convert [H3O+1] to pH pH = - log 10 [H3O+] log, #, enter, then make it positive –
change the sign) [H3O+1] = pH =
1.0 x10 –1
1.0 x 10 –2
3.0 x 10 –4
How to calculate strengths of Acids and Bases
To convert pH to [H3O+1] [H3O+] = 10 –pH
2nd, log, (-), #, enter[H3O+1] pH Acid or Base
2
11
5.22
How to calculate strengths of Acids and Bases
[H+] [OH-] pH pOH Acid or Base
2 x 10-5
3.5 x 10-5
3.25
8.12
8.20
0.0016
2.8 x 10-11
Concentration units for Acids and Bases Chemical Equivalents: quantities of solutes that
have equivalent combining capactieies. Ex1: HCl + NaOH NaCl + H2O
To achieve a balance, 1 mole H+ needs to cancel out 1 mole OH-1.
So, for the above reaction: 1 mole of HCl is necessary to balance 1 mole of NaOH.
1 mole HCl = 1 mole NaOH
Concentration units for Acids and Bases Ex2: H2SO4 + 2 NaOH Na2SO4 + 2 H2O
To achieve a balance, 1 mole H+ needs to cancel out 1 mole OH-1.
So, for the above reaction: 1/2 mole of H2SO4 is necessary to balance 1 mole of NaOH.
½ mole H2SO4 = 1 mole NaOH
Concentration units for Acids and Bases
Ex3: To make H3PO4 chemically equivalent to NaOH, 1/3 mole of H3PO4 balances 1 mole NaOH
Equivalent Weight
the # of grams of acid or base that will provide 1 mole of protons or hydroxide ions.
HCl H2SO4 H3PO4
Moles of Acid 1 ½ 1/3
Moles of Hydrogen
1 1 1
Equivalent Weights
36.5 49.05 32.7
Equivalent Weight
Formula for calculating equivalent weight Eq. wt. = MW / equivalents
MW = molecular weight
Formula for calculating equivalents: # equivalents = (moles)(n)
n = # of H or OH in the chemical formula
Calculations
Example 1: How many equivalents in 9.30 g of H2CO3?
Step 1: Calculate the molecular weight of H2CO3:H = (2)1.0 = 2.0C = (1)12.0 = 12.0O = (3)16.0 = 48.0mw = 62.0 g/mole
Calculations
Step 2: Calculate the # of moles in 9.30 g of H2CO3: 9.30 g H2CO3 x 1 mole H2CO3 = .150 mole
62.0 g H2CO3 H2CO3
Step 3: Calculate the number of equivalents
# equivalents = (moles)(n) n = # of H or OH in the chemical formula
eq = (.150 moles)(2) eq = .300 eq
Calculations
Ex 2: Calculate the equivalent weight of 9.30 g H2CO3 (Use 00.300 equivalents calculated
above)eq. wt. = mw/eq
eq. wt. = 62.0 g / .300 equivalents eq. wt. = 206.7 g/eq
Calculations
Ex 3: How many grams of H2CO3 would equal .290 equivalents?
Step 1: Convert to moles # equivalents = (moles)(n)
moles = .29 / 2 moles = .145 moles
Step 2: convert moles to grams: .145 moles H2CO3 x 62.0 g H2CO3 = 8.99 g
1 mole H2CO3 H2CO3
Normality
In the past, we have used M for concentration. M = moles / L A more useful form of concentration for
acid/base reactions is Normality. N = # eq / L
# equivalents = (moles)(n) n = # of H or OH in the chemical formula
Normality
Also, in calculating pH, normality is used over molarity.
Normality is related to Molarity: N = (M)(n)
M = moles/liters (total) n = # of H or OH in the chemical formula
Normality
Ex 1: Find the normality of a solution that contains 1 mole H2SO4 in 1 L solution
Step 1: Calculate the molarity:1 mole H2SO4 = 1 M
1 L Step 2: Calculate the normality: N = (M)(n)
N = 1 M x 2N = 2 N
Normality
Ex2: Calculate the Normality if 1.80 g of H2C2O4 is dissolved in 150 mL of solution.
Step 1: convert grams to moles: 1.80 g H2C2O4 x 1 mole H2C2O4 = .0200 moles
90.0 g H2C2O4 H2C2O4
Step 2: Calculate the molarity: M = .0200 moles / .150 L .133 M Step 3: Calculate the normality:
N = (M)(n) N = (.133)(2) .267 N
Problems involving mixing unequal amounts of acid and base
Ex1: Find the pH of a solution made by mixing 50.0 mL of .100 M HCl with 49.0 mL of .100 M NaOH.
Step 1: Find the moles of HCl and moles of NaOH:HCl .100 M = x / .0500 L x
= .00500 moles HCl
NaOH .100 M = x / .0490 L x =.00490 moles NaOH
Find the pH of a solution made by mixing 50.0 mL of .100 M HCl with 49.0 mL of .100 M NaOH.
Step 2: Find the equivalents of H+ in HCl and the equivalents of OH- in NaOH:
eq = (moles)(n)
H+ (.00500 moles)(1) = .00500 eq of H+
OH- (.00490 moles)(1) = .00490 eq of OH-
Step 3: Find the equivalents of H+ or OH- left over by subtracting the equivalents of H+ and equivalents of OH- from each other (absolute value):
.00500 eq H+ - .00490 eq OH- = .00010 eq H+
Find the pH of a solution made by mixing 50.0 mL of .100 M HCl with 49.0 mL of .100 M NaOH.
Step 4:Calculate the normality of the resulting solution: N = eq / L
Total liters 50.0 mL + 49.0 mL = 99.0 mL .099 L
N = .00010 eq / .0990 L .00101 N
Step 5: Calculate the pH: pH = - log [H+] pH = - log (.00101) pH = 3.00
Example 2:
Find the pH when mixing 50.0 mL of .100 M sulfuric acid with 50.0 mL of 1.00 M NaOH
Step 1: Find the moles of H2SO4 and moles of NaOH Step 2: Find the equivalents of H+ in H2SO4 and the
equivalents of OH- in NaOH Step 3: Find the equivalents of H+ or OH- left over by
subtracting the equivalents of H+ and equivalents of OH- from each other
Step 4: Calculate the normality of the resulting solution
Step 5: Calculate the pH
Find the pH when mixing 50.0 mL of .100 M sulfuric acid with 50.0 mL of 1.00 M NaOH
Find the moles of H2SO4 and moles of NaOH:
H2SO4 .100 M = x / .0500 L x = .00500
moles H2SO4
NaOH 1.00 M = x / .0500 L x = .0500 moles NaOH
Find the pH when mixing 50.0 mL of .100 M sulfuric acid with 50.0 mL of 1.00 M NaOH
Find the equivalents of H+ in H2SO4 and the equivalents of OH- in NaOH:
eq = (moles)(n)H+ (.00500 moles)(2) = .0100 eq
OH- (.0500 moles)(1) = .0500 eq
Find the equivalents of H+ or OH- left over by subtracting the equivalents of H+ and equivalents of OH- from each other (absolute value):
.0500 eq OH-1 - .0100 eq H+ = .0400 eq OH-
Find the pH when mixing 50.0 mL of .100 M sulfuric acid with 50.0 mL of 1.00 M NaOH
Calculate the normality of the resulting solution:N = eq / L
total L 50.0 mL + 50.0 mL = 100.0 mL .1000L
N = .0400 eq / .1000 L .400 N
Calculate the pH: 1st calculate the pOH: pOH = -log[OH-] pOH = -log(.400) pOH =.399 Calculate the pH from the pOH pH = 14 – pOH pH = 14 - .399 pH = 13.6
Titrations
A controlled addition and measurement of the amount of a solution of a known concentration that is required to react completely with a measured amount of a solution of unknown concentration.
Titrations
Titrations Equivalence point –
In a neutralization reaction, the point @ which there are equivalent quantities of H3O+ and OH-
End Point – point in a titration where the indicator changes color
Titration of 20 mL 0.1 M HCL
mL of 0.1 M NaOH Titrant
Titration of HCl with NaOH
High amount of H+1
H+1= OH-
High amount of H+1
End point
Normality
Normality and Titration: Equation: NaVa =NbVb
N = normality V = volume in Liters
Ex1: A 15.5 mL sample of .215 M KOH requires 21.2 mL of acetic acid to titrate to the end point. What is the Molarity of the acid?
Step 1: Convert molarity to normality: N = (M)(n)
n = # of H or OH in the chemical formula N = (.215)(1) .215 N KOH
Step 2: Use the titration formula to find the normality of the acid:
NaVa =NbVb
(x)(21.2 mL) = (.215 N)(15.5 mL) x = .157 N
Step 3: Convert normality to molarity:M = N / n x = .157 N / 1 x = .157 M
Ex2: If 15.7 mL of sulfuric acid is titrated to the end point by 17.4 mL of .0150 M NaOH, what is the Molarity of the acid?
Convert molarity to normality: N = (M)(n)
N = (.015)(1) .0150 N KOH
Use the titration formula to find the normality of the acid:
NaVa =NbVb
(x)(15.7 mL) = (.0150 N)(17.4 mL) x = .0166 N
Convert normality to molarity: M = N / n x = .0166 N / 2 x = .00831 M
Titration Problems Ex 3: In a titration of 27.4 mL of
0.0154 M Ba(OH)2 solution is added to a 20 mL sample of an HCl solution. What is the Molarity of the HCl solution?
Percent Problems
Long way……. If 18.75 mL of .750 N NaOH is required to titrate
20.30 mL of acetic acid, calculate the % acetic acid in solution.
Step 1: Use the titration formula to find the normality of the acid:
NaVa =NbVb
(x)(20.30 mL) = (.750 N)(18.75 mL) x = .693 N
If 18.75 mL of .750 N NaOH is required to titrate 20.30 mL of acetic acid, calculate the % acetic acid in solution.
Step 2: Multiply the normality by the equivalent weight of the acid:
.693 N .693 eq/L
(.693 eq/L)(60.0g/eq) = 41.6 g/L
If 18.75 mL of .750 N NaOH is required to titrate 20.30 mL of acetic acid, calculate the % acetic acid in solution.
Step 3: Convert the liters to grams:
41.6 g/L 41.6 g/1000mL 41.6 g/1000g Step 4: Calculate the % by mass: (41.6 g / 1000g) 100 = 4.16 %
If 18.75 mL of .750 N NaOH is required to titrate 20.30 mL of acetic acid, calculate the % acetic acid in solution.
The short way…… Use the titration formula to find the normality of the acid:
NaVa =NbVb
(x)(20.30 mL) = (.750 N)(18.75 mL) = x = .693 N
Use: (N)(eq wt)/10 = %
(.693)(60.0) / 10 = x
x = 4.16 %