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Acids, Bases and Salts
Naming Acids
Naming Acids
Names of binary acids (two atoms) have the form: hydro-…ic acid. Ex: HF is hydrofluoric acid
HCl is hydrochloric acid Try:
H2S HI HBr
Hydrosulfuric Acid
Hydroiodic Acid
Hydrobromic Acid
Naming Acids When naming ternary acids (H and a poly-
ion). identify polyatomic ion in the formula. For polyions that end in “ate” change the
ending to “ic”. (Something I “ate” made me feel “ic”.)
For polyions that end in “ite” change the ending to “ous”.(A snake b“ite” is poison“ous”)
EX: HNO3 nitric acid
HNO2 nitrous acid
H2SO4
H2SO3
Sulfuric Acid
Sulfurous Acid
Name the following acids:
HCl
H2CO3
H3PO4
HF
HNO3
Hydrochloric Acid
Carbonic Acid
Phosphoric Acid
Hydrofluoric Acid
Nitric Acid
Properties of Acids
Acid Property #1. The word acid comes from the Latin word acere. which means "sour“. All acids taste sour.
Acid Property #2. In 1663. Robert Boyle wrote that acids would make a blue vegetable dye called "litmus" turn red.
Acid Property #3. Acids destroy the chemical properties of bases.
Properties of Acids Cont.
Acid Property #4. Acids conduct an electric current. (electrolytes)
Acid Property #5. Upon chemically reacting with an active metal. acids will evolve hydrogen gas (H2).
Properties of Bases
Base Property #1. The word "base" has a more complex history and its name is not related to taste. All bases taste bitter.
Base Property #2. Bases are substances which will restore the original blue color of litmus after having been reddened by an acid.
Base Property #3. Bases destroy the chemical properties of acids.
Properties of Bases Cont.
Base Property #4. Bases conduct an electric current. (electrolytes)
Base Property #5. Bases feel slippery. sometimes people say soapy. This is because they dissolve the fatty acids and oils from your skin and this cuts down on the friction between your fingers as you rub them together.
#6. Bases have a
pH greater than 7
VIDEO
Properties of A Salt
A salt is the combination of a cation(+ metal ion) and an anion (- nonmetal ion).
Salts are products of the reaction between acids and bases.
Solid salts usually make crystals. If a salt dissolves in water solution. it
usually dissociates into the anions and cations that make up the salt.
Theories of Acid/Base
Acids – produce H+
Bases - produce OH-
Acids – donate proton (H+)
Bases – accept protons (H+)
Acids – accept e- pair Bases – donate e- pair
1. Arrehenius
2. Bronsted-Lowry
3. Lewis
only in water
any solvent
used in organic chemistry.wider range of substances
The Acid Base TheoryBrønsted-Lowry
Two chemists. independent of one
another. proposed a new definition of an acid and a base.
An acid is a substance which can donate a proton (H+) to a base.
A base is a substance that can accept a proton (H+) from an acid.
Reactions Based on Brønsted-Lowry
HCl + H2O H3O+ + Cl¯
HCl - this is the acid, because it has a proton (H+) available to donate.
H2O - this is the base, since it accepts the proton (H+) that the acid lost.
Now, here comes an interesting idea: H3O+ - this is an acid, because it can
donate a proton. Cl¯ - this is a base, since it has the
capacity to accept a proton.
HCl + H2O H3O+ + Cl¯
Notice that each pair (HCl and Cl¯ as well as H2O and H3O+ differ by one proton (H+). These pairs are called conjugate pairs. HNO3 + H2O H3O+ + NO3¯
The acids are HNO3 and H3O+ and the bases are H2O and NO3¯.
The Bronsted-Lowry Concept
• Acids and bases are identified based on whether they donate or accept H+.
• “Conjugate” acids and bases differ only with the addition or elimination of one H+.
Bases and Conjugate Acid
Base NameConjugate acid Name
CH3COO-
Acetate ion CH3COOH Acetic acid
NH3 Ammonia NH4+ Ammonium
H2PO4- Dihydrogen
phosphate ionH3PO4 Phosphoric
acid
HSO4- Hydrogen sulfate
ionH2SO4 Sulfuric acid
OH- Hydroxide ion H20 water
NO3- Nitrate ion HNO3 Nitric acid
H2O water H30+ Hydronium ion
When a substance can donate or accept a proton (H+) it is an amphoteric substance:
HCO3-
H2CO3 CO3-2
+ H+ - H+
Acting like a base
Acting like an acid
accepts H+ donates H+
Water is considered an amphoteric substance because it can act as either an acid or a base.
Last slide
Strong Acids and Bases
Strong acids are those that ionize completely in water.
The dissociation of a strong acid looks like the diagram at the right in that it dissociates completely into positive and negative ions.
HA
Let’s examine the behavior of an acid. HA. in aqueous solution.
What happens to the HA molecules in solution?
HA
H+
A-
Strong Acid
100% dissociation of HA
Would the solution conduct electricity?
Weak Acids and Bases
Some acids and bases ionize only slightly in water.
These are considered weak.
The most important weak base is ammonia (NH3).
HA
H+
A-
Weak Acid
Partial dissociation of HA
Would the solution conduct electricity?
HA
H+
A-
Weak Acid
HA H+ + A-
At any one time. only a fraction of
the molecules
are dissociated.
Strong Acids
*HNO3 - nitric acid *HCl - hydrochloric acid *H2SO4- sulfuric acid *HClO4 - perchloric acid *HBr - hydrobromic acid *HI - hydroiodic acid*HClO3- chloric acid
*Refer to solubility rules.
Strong Bases
*LiOH - lithium hydroxide*NaOH - sodium hydroxide*KOH - potassium hydroxide*RbOH - rubidium hydroxide*CsOH - cesium hydroxide*Ca(OH)2 - calcium hydroxide*Sr(OH)2 - strontium hydroxide*Ba(OH)2 - barium hydroxide
*Refer to solubility rules for soluble hydroxides
Neutralization Reactions
The word "neutralization" is used to describe the double replacement reaction of an acid plus a base because the acid and base properties of H+ and OH- are destroyed or neutralized.
In the reaction. H+ and OH- combine to form water and a salt.
Bases Neutralize Acids
Milk of Magnesia contains magnesium hydroxide. Mg(OH)2. which neutralizes stomach acid. HCl.
2 HCl + Mg(OH)2 MgCl2 + 2 H2O
In general: Acid + Base Water + Salt
HCl + NaOH NaCl + HOH H2O
HCl + Mg(OH)2 H2O + ______
H2SO4 + NaOH H2O + ______
Bases Neutralize Acids
MgCl2
Na2SO4
• When acids and bases are equal in strength and concentration. a neutral (pH = 7) solution is formed.
• EX: HCl + NaOH --> H2O + NaCl (Strong acid) (Strong Base)
HF + Mg(OH)2 --> H2O + MgF2 (Weak acid) (Weak Base)
Neutral solutions have an equal number of hydroxide & hydronium
ions
Acidic NeutralSolution Solution
Calculations of Neutralization Reactions
Utilize equation: MAVA = MBVB
MA and VA= Molarity and Volume of Acid
MB and VB= Molarity and Volume of Base
*Must adjust the Molarity of strong acids or strong bases based on the number of moles of H+ or OH- they contribute.
Strong Acid: H2SO4 contributes 2 moles of H+ ions. Multiply the molarity (M) of H2SO4 by 2.
Strong Bases: There are three strong soluble bases that contribute 2 moles of OH- ions. Ca(OH)2. Sr(OH)2. and Ba(OH)2 molarities (M) must be multiplied by 2.
Adjust the molarity (M)
Sample Problems
1. How many mL of a 1.5 M HCl acid is needed to neutralize a 500. mL 1.5 M NaOH solution?
2. How many mL of a 2.0 M H2SO4 acid is needed to neutralize a 500. mL 1.5 M KOH solution?
3. How many mL of a 0.750 M HNO3 acid is needed to neutralize a 275 mL 1.5 M Ca(OH)2 solution?
4. Calculate concentration of acid if 300. mL of HBr is used to neutralize a 500. mL 2.50 M NaOH solution.
pH and pOH
pH and pOH
The scale is measured on a log scale of 0 to 14. with each unit representing a ten-fold change.
pH Scale
The pH scale is a measure of hydrogen ion [H+] concentration (acid molarity) as well as the hydroxide [OH-] concentration (base molarity).
Hydrogen ion concentration indicates: Acids will have a pH of 0-6
Hydroxide ion concentration indicates: Bases will have a pH of 8-14
The higher the [H+]. the higher the acidity. the higher the [OH-]. the higher the basicity.
Some common pH values
Calculating pH
The concentration (M or mol/L) of H+ is expressed in powers of 10. from 10-14 to 100.
Scientists use the following formula to calculate pH using acid molarity.
pH = -log [H+] (acid Molarity)(Remember that the [ ] mean Molarity)
Example pH CalculationspH = -log [H+]
EX: 0.50M HCl is added to water to make a final volume of 1 liter. What is the pH of this solution?
Step 1: Identify that you have an acid (starts with an H)
Step 2: Identify [H+] = 0.50 M Step 3: Place value in equation and solve. pH = -log[0.50] = 0.30 pH (Acidic) (pH less
than 7)
Practice pH CalculationspH = -log [H+]
Find pH of the following solutions if [H+] is:
1. 1.00 x 10-3 M = 3.00 pH2. 6.59 x 10-6 M = 5.18 pH3. 9.47 x 10-10 M = 9.02 pH
I-phones & Calculators: -log(6.59ee-6)enter = answer (2 decimal places)
Calculate pH of Strong Acids
pH = -log [H+] If you have the strong polyprotic acid,
H2SO4, you must adjust the molarity by the number of moles of H+ contributed.
EX: 0.250M H2SO4 is added to water to make a final volume of 1 liter. What is the pH of this solution?
Step 1: Identify [H+] = 0.250 M x 2 = 0.5 MStep 2: Place values in equation and solve.
pH = -log[0.50] = 0.30 pH (Acidic)
Calculate pH of Weak Acids
pH = -log [H+]
Weak acids will not dissociate 100%. A dissociation factor will be included in these problems.
EX: Calculate pH of a 0.150 M HNO2 solution (dissociation is 5.00%).
Step 1: Identify [H+] = 0.150 M x 5% = 0.0075 MStep 2: pH = -log[0.0075] = 2.13 pH
(Acidic)
Calculate H+ Concentration from pH
If given pH you can calculate the hydrogen ion concentration by performing the anti-log function. Calculator (10^)… I-Phone (alog)
[H+] = alog (-pH)
Calculator (10^)… I-Phone (alog)
Example [H+] Calculations[H+] = alog (-pH)
EX: Find the [H+] if the pH is 2.00. Alog (-2.00) = 1.00 x 10-2M 10^(-2.00) = 1.00 x 10-2M
Find [H+] if the pH is:1. 6.678 pH = 2.1 x 10-7 M2. 2.533 pH = 2.9 x 10-3 M3. 10.0 pH = 1.0 x 10-10 M4. 2.56 pH - = 2.75 x 10-3 M
Remember the unit for concentration is M.
pOH
Just like pH can be calculated from the [H+] concentration…
pH = -log [H+]
pOH can be calculated from the [OH-] concentration…
pOH = -log [OH-]
pOH
Just like [H+] concentration can be calculated from pH…
[H+] = alog (-pH)
[OH-] concentration can be calculated from pOH…
[OH-] = alog (-pOH)
Connecting pH to pOH
You can calculate the pH of a solution if you know the concentration of hydroxide ion. [OH-]
If we use the ion product constant of water we can derive this equation:
[pH] x [pOH] = 1.00 x 10-14
Working with this equation leads to:pH + pOH = 14
Practice pH Calculations Using pOH
EX: Find the pH of a solution with an [OH-] of 1.0 x 10-8 M.
Step 1: Calculate pOH by using equation: pOH = -log[OH-] = -log(1.0 x 10-8 )= 8
pH
Step 2: Subtract the pOH from 14 to find pH:
pH = 14 - pOH = 14 - 8 = 6 pH
Practice pH Calculations Using pOH
Find the pH of the following solutions with [OH-] of:
1. 1.00 x 10-4 M -log[1.00x10-4] = 4 pOH, 14-4=10 pH2. 2.64 x 10-13 M -log[2.64x10-13] = 12 pOH, 14-12=2 pH3. 5.67 x 10-2 M -log[5.67x10-2] = 1.25 pOH, 14-1.25=12.75 pH4. 3.45 x 10-11 M -log[3.45x10-11] = 10.46 pOH, 14-10.46=3.54 pH
Calculate [OH-] from pH or pOH
If given pH:14 – pH = pOH[OH-] = alog(-pOH) or 10^(-
pOH) If given pOH:
[OH-] = alog(-pOH) or 10^(-pOH)
Remember unit for concentration is M.
Summary of pH and pOH
pH = -log[H+] (acid concentration) pOH = -log[OH-] (base concentration) pH + pOH = 14 [H+] = alog(-ph) or 10^(-pH) [OH-] = alog(-pOH) or 10^(-pOH)
*Hints:1. Identify initial substance as acid or base to
determine if you have H+ or OH-.2. Label all concentrations with M unit.
Titrations
Titrations
Titration is a standard laboratory method of quantitative/chemical analysis which can be used to determine the concentration of a unknown reactant (acid or base).
An acid or base of known concentration (a standard solution) and volume is used to react with a measured volume of an unknown concentration of an acid or base.
Titrations
Using a buret to add the [unknown]. it is possible to determine the exact amount (V) that has been consumed when the endpoint is reached.
The endpoint is the point at which the titration is stopped.
This is classically a point at which the number of moles of [unknown acid or base] is equal to the number of moles of [known acid or base].
Titrations
Titrations
Due to the logarithmic nature of the pH curve. the transitions are generally extremely sharp. and thus a single drop of unknown just before the endpoint can change the pH by several points - leading to an immediate color change in a chosen indicator.
Titrations
Titrations of Strong Acids and Strong Bases
Titration of a strong acid and a strong base will result in an equivalence point of 7 because a neutral salt water solution is formed.
Solubility Rules Table The classification of ionic substances as “soluble” or “insoluble” in water is difficult. Nothing is completely “insoluble” in water. This table does not tell you how soluble a substance is, just that it either barely dissolves (“insoluble”) or that it dissolves to a greater extent.
Mainly Water Soluble (will dissociate and create ions) NO3
- All nitrates are soluble
C2H3O2- All acetates are soluble except AgC2H3O2
ClO3- All chlorates are soluble
Cl- All chlorides are soluble except AgCl, Hg2Cl2, PbCl2
Br- All bromides are soluble except AgBr, PbBr2, Hg2Br2, HgBr2
I- All iodides are soluble except AgI, Hg2I2, HgI2, PbI2
SO4-2 All sulfates are soluble except BaSO4, PbSO4, Hg2SO4, CaSO4, Ag2SO4, SrSO4
F- All fluorides are soluble except SrF, BaF2, MgF2, MnF2, CaF2
H+ (acids) All strong acids are soluble = HCl, HBr, HI, HClO3, HClO4, H2SO4, HNO3, H2SeO4
Alkali Metals All are soluble Cations (Group 1) All are soluble NH4
+ All are soluble
Mainly Water Insoluble (will NOT dissociate and create ions)
CO3-2, All carbonates are insoluble except those of group 1 elements and NH4
+
CrO4-2 All chromates are insoluble except those of the group 1 elements, NH4
+, CaCrO4, SrCrO4
OH- (bases) All hydroxides are insoluble except those of the group 1 elements, NH4OH, Ba(OH)2, Sr(OH)2, Ca(OH)2
PO43- All phosphates are insoluble except those of group 1 elements and NH4
+
SO32- All sulfites are insoluble except those of the group 1 elements and NH4
+
S2- All sulfides are insoluble except those of the group 1 and 2 elements and NH4+
Gases and Oxides
Keep all gases in molecular form (don’t dissociate) – examples- O2 (g), NH3 (g), CO2 (g)
Keep oxides in molecular form (don’t dissociate) except those containing group 1 elements
Acids & Bases Review
MAVA = MBVB (Don’t forget to adjust M x 2 if necessary…example H2SO4 or
Ba(OH)2)
pH = -log[H+] (acid concentration) pOH = -log[OH-] (base concentration) pH + pOH = 14 [H+] = alog(-ph) or 10^(-pH) [OH-] = alog(-pOH) or 10^(-pOH)
(Remember Acids (begin with H) & Bases (end with OH)
