Acid/base properties Definitions › Arrhenius › Bronsted-Lowry › Lewis Acid-Base Reactions...

Acids & BasesUnit 13

Overview

Acid/base properties Definitions

› Arrhenius› Bronsted-Lowry› Lewis

Acid-Base Reactions› Neutralization› Sulfides› Carbonates

pH/pOH Scale Acid/base strength

› Factor affecting› Ka, Kb, Kw

› Percent ionization

Vocabulary› Polyprotic Acids› Amphoteric › Anhydrides

Acids/Bases & Salts› Determine acidity› Calculations

Common Ion Effect Buffers

› Henderson-Hasselbalch Titration

› Indicators› 4 types of curves

Acids Sour taste React with active metals to produce hydrogen

gas Change the color of acid-base indicators React with bases to produce salt and water Conduct an electric current (electrolytes) Turn litmus paper red

Common Acids

Sulfuric Acid› Car batteries; production of metals, paints, dyes, detergents

Nitric Acid› Explosives, pharmaceuticals, rubber, plastics, dyes

Phosphoric acid› Soda, fertilizers, animal feed, detergents

Hydrochloric Acid› Stomach acid, cleaning metals, found in hardware stores

(muriatic acid) Acetic Acid

› Vinegar, food supplements, fungicide Citric Acid

› Fruit juices

Acids

Binary acids› Contain only two different elements› Name as “hydro - ic acid”› Example: HCl (hydrochloric acid)

Oxyacids› Acid consisting of hydrogen and a polyatomic

anion that contains oxygen (oxyanion)› To name, drop ending of polyatomic ion and ad

“- ic acid”› Example: HNO3 (nitric acid)

Bases Taste bitter Feel slippery Change the color of

acid-base indicators React with bases to

produce salt and water

Conduct an electric current (electrolytes)

Turn litmus paper blue

Common Bases

Ammonium hydroxide› Household cleaners, window cleaner

Ammonia› (Gas) inhalant to revive unconscious person

Sodium bicarbonate (baking soda)› Acid neutralizers in acid spills› Antacids for upset stomachs

Sodium hydroxide› Drain cleaner (drano), oven cleaner, production of

soap Magnesium hydroxide

› Antacids, milk of magnesia, laxatives

Definition: Arrhenius

Acid› Substance that ionizes in water and

produces H+ ions› Example: HCl H+ + Cl-

Base› Substance that ionizes in water and

produces OH- ions› Example: NaOH Na+ + OH-

Definition: Bronsted-Lowry

Acid› Substance that is capable of donating a

proton (H+ ion)

Base› Substance that is capable of accepting a

proton (H+ ion)

Examples: Bronsted-Lowry

HC2H3O2 + H2O ↔ C2H3O2- + H3O+

Acids: HC2H3O2 and H3O+

Bases: H2O and C2H3O2-

NH3 + H2O ↔ NH4+ + OH-

Acids: H2O and NH4+

Bases: NH3 and OH-

Notice that water can act as an acid or a base

Bronsted-Lowry

Conjugate Pair – a BL acid/base pair(one with H+ and one without H+)

› Examples: HC2H3O2 and C2H3O2

-

H3O+ and H2O

H2O and OH-

NH4+ and NH3

Bronsted-Lowry

The more easily a substance gives up a proton, the less easily the conjugate base accepts a proton (and vice versa)

› The stronger the acid, the weaker the conjugate base

› The stronger the base, the weaker the conjugate acid

Acid Base Reactions

Neutralization › Salt + water

Sulfides › Salt + sulfide gas

Carbonates › Salt + CO2 + H2O

Neutralization

Solution of an acid and solution of a base are mixed Products have no characteristics of either the acid or

the base

Acid + Base (metal hydroxide) salt + water› Salt comes from cation of base and anion of acid

HY + XOH XY + H2O

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Sulfides

Acid reacts with a sulfide Gaseous product (H2S) has a foul odor (rotten

eggs)

Acid + metal sulfide salt + hydrogen sulfide› Salt comes from cation of sulfide and anion of acid

HY + XS XY + H2S

HCl(aq) + Na2S(aq) NaCl(aq) + H2S(g)

Carbonates Carbonates and bicarbonates react with

acids

HY + XHCO3 XY + H2CO3

H2CO3 is not stable so breaks into H2O and CO2

Then HY + XHCO3 XY + H2O + CO2

HCl(aq) + NaHCO3(aq) NaCl(aq) + H2CO3 (aq)

HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O(l) + CO2(g)

The pH Scale

pH scale: measures concentration of hydrogen ions in solution

pH = -log[H+] therefore [H+] = 10-pH

Example: What is the pH of a solution with a [H+] of 1.4×10-5?

pH = -log[1.4×10-5] = 4.9

pH Scale

Acids: pH < 7

Neutral: pH = 7

Bases: pH > 7

Increasing [H+] means decreasing pH

Increasing pH means decreasing [H+]

The pOH Scale

pOH scale: measures concentration of hydroxide ions in solution

pOH = -log[OH-] therefore [OH-] = 10-pOH

Example: What is the [OH-] of a solution with a pOH of 6.2?

[OH-] = 10-6.2 = 6.3×10-7

Comparing pH and pOH

pH + pOH = 14

An acid has a pH of 4, what is the pOH?

4 + pOH = 14pOH = 10

Measuring pH

pH meter› Electrodes measure [H+]

Acid-base indicators› Change color in presence of acid or

base (or certain pH ranges)› Litmus paper, phenolphthalein,

cabbage juice, methyl orange, thymol blue…

Strong Acids and Bases

Completely ionize in solution (strong electrolytes)

Strong Acids Strong Bases

HCl Group 1 metals + OH

HBr (LiOH, NaOH, KOH,…)

HI

HClO3 Heavy group 2 metals + OH

HClO4 Ca(OH)2, Sr(OH)2, Ba(OH)2

HNO3

H2SO4If acid/base is not on this list, it is a weak acid/base

Weak Acids and Bases

Do not completely ionize in water (weak electrolytes)

Common weak acids:› HF, acids with -COOH group

Common weak bases:› NH3

Factors Affecting Acid Strength

1. Electronegativity of element bonded to H› Binary acids› More electronegative bond = stronger acid› Example: HCl stronger than HBr

2. Bond Strength› Stronger bonds do not allow hydrogen to dissociate as

easily› Reason why HF is not a strong acid (F is most

electronegative, but H-F bond is strongest bond)

3. Stability of Conjugate base› More stable the conjugate base, the stronger the acid

Factors Affecting Acid Strength

4. For polyatomic ions, the more electronegative the nonmetal, the stronger the acid (when comparing acids with same number of O atoms)› Example: HClO3 is stronger than HBrO3

5. For polyatomic ions, when nonmetal is the same, the more O atoms, the stronger the acid› Example: HClO3 is stronger than HClO2

Percent Ionization

Tells us what percent of an acid (or base) is ionized in water › Helps determine the strength of an acid (or

base)

Percent Ionization = ×100

[H+] at equilibrium

Initial Acid Concentration

Percent Ionization (Example)

A 0.035 M solution of HNO2 contains 3.7×10-3 M H+(aq). Calculate the percent ionization.

= = 11%

This means that 11% of the acid will dissociate in water.

3.7×10-3 M0.035 M

Strong Acids

HA(aq) + H2O(l) H+(aq) + A-(aq) acid water proton conjugate base

Strong acids dissociate completely The dissociation is not reversible

The acid is the only significant source of H+ ions, so pH can be calculated directly from the [H+]› Example: A 0.20 M solution of HNO3 has an [H+] of 0.20 M

› pH = -log[H+]

Strong Bases

Strong bases dissociate completely The dissociation is not reversible

The base is the only significant source of OH- ions, so pOH can be calculated directly from the [OH-]› Example: A 0.30 M solution of NaOH has a [OH-] of 0.30

M› pOH = -log[OH-]

Equilibrium Time!

Weak Acids and Bases

Do not dissociate completely Reversible reactions Need to use equilibrium to solve for

[H+]

K = [Products]

[Reactants]

Acid Dissociation

HA(aq) + H2O(l) H+(aq) + A-(aq)

acid water proton conjugate base

Write the equilibrium expression for the acid dissociation constant, Ka.

][

]][[

HA

AHKa

Base Dissociation

B(aq) + H2O(l) ↔ BH+(aq) + OH-(aq) base water conjugate hydroxide

acid ion

Write the equilibrium expression for the base dissociation constant, Kb.

[ ][ ]

[ ]b

BH OHK

B

Size of K

The greater the Ka, the stronger the acid

The smaller the Ka, the weaker the acid

The greater the Kb, the stronger the base

The smaller the Kb, the weaker the base

Example: Weak Acid Equilibrium Problem

What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?

Step #1: Write the dissociation equation

HC2H3O2 C2H3O2- + H+



Step #2: ICE

HC2H3O2 C2H3O2- + H+

I

C

E

0.50 0 0

- x +x +x

0.50 - x xx



Step #3: Set up the equilibrium expression

)50.0()50.0(

))((108.1

25 x

x

xxx

If percent ionization is less than 5%, you can ignore using the quadratic.

3100.3 xx



Step #5: Solve for pH

52.2)100.3log( 3 xpH

You can use the Kb expression to solve for pOH using the same method!

Example 2: Weak Acid Equilibrium - Solving for Ka

A student prepared a 0.10 M solution of formic acid (HCOOH) and measured its pH. The pH at 25°C was found to be 2.38. Calculate the Ka for formic acid at this temperature.

Step #1: Solve for [H+] from pH

[H+] = 10-2.38 = 4.2×10-3 M



Step #2: Set up ICE table

HCOOH(aq) HCOO- + H+

I

C

E

0.10 0 0

4.2×10-3



Step #3: Use stoichiometry to complete table HCOOH(aq) HCOO- + H+

I

C

E

0.10 0 0

4.2×10-34.2×10-3

4.2×10-34.2×10-34.2×10-3

0.0096



Step #4: Solve for Ka using equilibrium expression

Ka = 1.8×10-4

Rule of Thumb

The larger the value of Ka, the stronger the acid

(Self-) Auto-ionization of Water

According to Bronsted Lowry, H2O can act as either an acid or a base

Auto-ionization: One water molecule can donate a proton to another water molecule› Extremely rapid reaction and no molecule remains

ionized for long› At room temperature 1 out of every 109 molecule are

ionized at a given instant› Water is a nonelectrolyte and consists almost entirely of

H2O molecules

H2O(l) + H2O(l) ↔ H3O+ + OH-

Auto-ionization of Water

H2O(l) + H2O(l) ↔ H3O+ + OH-

Auto-ionization of water is an equilibrium process (use Kw - ion product constant)

Kw = [H3O+][OH-]

Also written as Kw = [H+][OH-]

At 25°C, Kw =1.4×10-14


1.4×10-14 = [H+][OH-]

In basic solutions, [OH-] > [H+]

In acidic solutions, [H+] > [OH-]

In neutral solutions, [H+] = [OH-]


1.4×10-14 = [H+][OH-]

If the concentration of one ion is known, you can solve for the concentration of the other ion

Example: Calculate the concentration of H+ in a solution in which the concentration of OH- is 0.010M.

1.4×10-14 = [H+][0.010][H+] = 1.0×10-12 M

Relating pKw to pKa and pKb

Acid or base dissociation constants are sometimes expressed as pKa and Kb. › pKa = –logKa

› pKb = -logKb

pKw = 14 = pKa+ pKb

Polyprotic Acids

Acids with more than one ionizable H+ ion

The acid-dissociation constants are Ka1, Ka2, etc…

The first proton is most easily removed› As protons are removed, it becomes more and

more difficult to remove protons

› Ka1>Ka2>Ka3….

H2SO3(aq) ↔ H+(aq) + HSO3-(aq) Ka1 = 1.7×10-2

HSO3-(aq) ↔ H+(aq) + SO3

-2(aq) Ka2 = 6.4×10-8

Polyprotic Acids

To calculate the overall K for the reaction, treat it as a multi-step equilibrium

Overall reaction…

H2SO3(aq) ↔ H+(aq) + HSO3-(aq) Ka1 = 1.7×10-

2

HSO3-(aq) ↔ H+(aq) + SO3

-2(aq) Ka2 = 6.4×10-

8

H2SO3(aq) ↔ 2H+(aq) + SO3-2(aq) Ka = Ka1 ×

Ka2

Ka = (1.7×10-2)(6.4×10-8) = 1.1×10-9

Polyprotic Acids

Strong acids (ex H2SO4) completely ionize with the first step› pH can be calculated by treating the acid as if

it were a monoprotic acid (one ionizable hydrogen)

If Ka values differ by a factor of 103 or more, acids can be treated as monoprotic

Polyprotic Acids

Monoprotic acid = 1 ionizable H+

Diprotic acid = 2 ionizable H +

Triprotic acid = 3 ionizable H +

Etc…

Amphoteric (amphiprotic) Substances

Substances that can act as either acids or bases › Example: H2O

Can give up H+ to become OH -

Can receive H + to become H3O +

› Example: H2PO4-

Can give up H + to become HPO4-2

Can receive H + to become H3PO4

Anhydrides

Acid Anhydride: combines with water to form an acid› CO2 + H2O H2CO3

› SO3 + H2O H + + HSO4-

Base Anhydride: combines with water to form a base› CaO + H2O Ca(OH)2

› Na2O + H2O 2Na+ + 2OH-

Acids Bases and Salts#1 If a salt is composed of the conjugates of a

strong acid and strong base, the solution will be neutral.

#2 If a salt is composed of the conjugates of a weak base and a strong acid, its solution will be acidic.

#3 If a salt is composed of the conjugates of a strong base and a weak acid, its solution will be basic.

#4 If a salt is composed of the conjugates of a weak base and a weak acid, the pH of its solution will depend on the relative strengths of the conjugate

acid and base of the specific ions in the salt.

Strong Acid + Strong Base

If a salt is composed of the conjugates of a strong acid and strong base, the solution will be neutral.

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

The Na+ and Cl- ions do not further ionize in water

Strong Acid + Weak Base

If a salt is composed of the conjugates of a strong acid and weak base, the solution will be acidic.

HCl(aq) + NH3(aq) NH4Cl(s)

The NH4Cl ionizes in water to produce some H+ ions (the Cl - ions do not react in water)

NH4Cl(s) NH4+(aq) + Cl-(aq)

NH4+ + H2O NH3

+ + H+Solution Is acidic

Weak Acid + Strong Base

If a salt is composed of the conjugates of a weak acid and strong base, the solution will be basic.

NaOH(aq) + HC2H3O2(aq) NaC2H3O2(s) + H2O(l)

The NaC2H3O2 ionizes in water

NaC2H3O2(s) Na+(aq) + C2H3O2-(aq)

The Na+ ions do not react at all, but the C2H3O2- ions

react to produce OH- in solution

C2H3O2-(aq) + H2O HC2H3O2 + OH-

Solution Is basic

Weak Acid + Weak Base

If a salt is composed of the conjugates of a weak acid and weak base, the pH of the solution will depend on the relative strengths of the conjugate acid and base of the specific ions in the salt.› The ion with the larger equilibrium constant

(Ka or Kb) will have the greater influence on pH)

Vocabulary

Hydrolysis› When the ions react with water to produce

hydrogen or hydroxide ions in solution Example: Strong Acid/Weak Base Example: Strong Acid/Strong Base

Acid Base Salt Calculations

Solve similar to weak acid/weak base problems

Solve for Ka or Kb depending on if the salt makes an acidic (Ka) or basic (Kb) solution› Use the ion of the strong acid or strong base

for equilibrium equation

Example (Acid Base Salt)

What is the pH of a 0.10 M solution of acetic acid, NaC2H3O2. The Ka of HC2H3O2 is 1.8 x 10-5 .

Step #1: We will use Kb because the solution will be basic. First find the Kb for HC2H3O2.

C2H3O2- + H2O HC2H3O2 + OH-



C2H3O2- + H2O HC2H3O2 +

OH-

Kw = KaKb

1.0 × 10-14 = (1.8x10-5)(Kb)

Kb = 5.6 × 10-10


Step #2: Write the equilibrium expression.


Kb = [HC2H3O2][OH-]

[C2H3O2-]



Step #3: ICEC2H3O2

- HC2H3O2 OH-

I

C

E

0.10 0 0

- x +x +x

0.10 - x xx


Step #4: Solve for OH-.


5.6×10-10 = = = 7.5×10-6

[x][x]

[0.10-x]

[x][x]

[0.10]


Step #4: Solve for OH-.


x = 7.5×10-6 = [OH-]


Step #5: Solve for pH.

[OH-] = 7.5×10-6

pOH = -log[OH-] = -log[7.5×10-6]pOH = 5.1

14 = pH + pOH14 = pH + 5.1

pH = 8.9

Common Ion Effect

Use LeChatlier’s Principle to determine how the addition of a substance will affect the equilibrium

The addition of the acetate ion (adding soluble salt) causes the equilibrium to shift to the LEFT

The hydrogen concentration will DECREASE

The acetic acid ionizes less with the addition of the acetate ion than it would alone in solution

HC2H3O2 C2H3O2 - + H+

Buffers

A solution with a very stable pH› You can add an acid or base to a buffer solution

without it greatly affecting the pH of the solution

Consists a weak acid-base conjugate pair› Usually a weak acid or base with the salt of that

acid or base› Resists changes in addition of strong acid or

base

Example: blood (pH of 7.4)

Buffers

Buffer Capacity – the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree

pH Range (of a buffer) – the range over which the buffer acts effectively

Henderson-Hasselbalch Equation

Reminder: pKa = –logKa and pKb = -logKb

][

][log

][

][log

acid

basepK

HA

ApKpH aa

][

][log

][

][log

base

acidpK

B

BHpKpOH bb

Given on AP Cheat Sheet!

Example (Henderson-Hasselbalch)

What is the pH of a buffer solution with concentrations of 0.20M HC2H3O2 and 0.50 M C2H3O2

-? The acid dissociation constant for HC2H3O2 is 1.8×10-5.

][

][log

][

][log

acid

basepK

HA

ApKpH aa

pH = -log(1.8×10-5) + log(0.50/0.20) pH = 4.7 + 0.4 = 5.1

Example 2(Henderson-Hasselbalch)


-? The acid dissociation constant for HC2H3O2 is 1.8×10-5.

][

][log

][

][log

acid

basepK

HA

ApKpH aa

pH = -log(1.8×10-5) + log(0.20/0.20) pH = 4.7 + 0 = 4.7

Choosing a Buffer

When concentrations of acid and conjugate base are the same, pH = pKa (and pOH = pKb)

When you want to prepare a buffer with a desired pH, choose an acid with a pKa close to the desired pH.

Weak AcidFormula

of the acidExample of a salt of the

weak acid Hydrofluoric HF KF – Potassium fluoride

Formic HCOOH KHCOO – Potassium formate

Benzoic C6H5COOH NaC6H5COO – Sodium benzoate

Acetic CH3COOH NaH3COO – Sodium acetate

Carbonic H2CO3 NaHCO3 - Sodium bicarbonate

Propanoic HC3H5O2 NaC3H5O2 - Sodium propanoate

Hydrocyanic HCN KCN - potassium cyanide

The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)

Acid/Salt Buffering Pairs

The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)

BaseFormula of the base

Example of a salt of the weak acid

Ammonia NH3 NH4Cl - ammonium chloride

Methylamine

CH3NH2 CH3NH3Cl – methylammonium

chloride

Ethylamine C2H5NH2 C2H5NH3NO3 - ethylammonium

nitrate

Aniline C6H5NH2 C6H5NH3Cl – aniline hydrochloride

Pyridine C5H5N C5H5NHCl – pyridine hydrochloride

Base/Salt Buffering Pairs

Buffer Solutions in ActionExample of a buffer of HC2H3O2 with NaC2H3O2

HC2H3O2 ↔ C2H3O2- + H+

When adding a strong acid to a buffer solution, the base acts to neutralize the acid (ex: add HCl)

NaC2H3O2 + HCl ↔ HC2H3O2 + NaCl

Na+ + C2H3O2- + H+ + Cl- ↔ HC2H3O2 + Na+ + Cl-

When adding a strong base to a buffer solution, the acid acts to neutralize the base (ex: add KOH)

HC2H3O2 + KOH ↔ KC2H3O2 + H2O

HC2H3O2 + K+ + OH- ↔ K+ + C2H3O2- + H2O

Buffer Solutions in Action (#1)


- after the addition of 0.10 M KOH? The acid dissociation constant for HC2H3O2 is 1.8×10-

5.

The strong base is neutralized by the acid:

HC2H3O2 + KOH C2H3O2 - + H+0.20 0.10 0.50

- 0.10

0.10 0.600

+ 0.10- 0.10 New conc. for acid and salt




5.

HC2H3O2 + KOH C2H3O2 - + H+

][

][log

HA

ApKpH a

pH = -log(1.8×10-5) + log(0.60/0.10) pH = 4.7 + 0.8 = 5.5



- after the addition of 0.05 M HBr? The acid dissociation constant for HC2H3O2 is 1.8×10-5.

The strong acid is neutralized by the base:

C2H3O2 - + HBr HC2H3O2 + Br-

0.50 0.05 0.20

- 0.05

0.45 0.250

+ 0.05- 0.05 New conc. for acid and salt




5.

HC2H3O2 + KOH C2H3O2 - + H+

][

][log

HA

ApKpH a

pH = -log(1.8×10-5) + log(0.45/0.25) pH = 4.7 + 0.3 = 5.0

Titration A base of known

concentration is slowly added to an acid of unknown concentration (or vice versa) to reach neutralization

Indicators are added to the solution that change color to signal the equivalence point› Equivalence point – point at

which stoichiometrically equivalent quantities of acid and base have been brought together - neutralization

Titration Curves

Titration Curve graph of the pH as acid or base is added to the solution

Indicators

Indicators are added to the solution that change color to signal the equivalence point› Equivalence point – point at which

stoichiometrically equivalent quantities of acid and base have been brought together – neutralization

Different indicators change at different pH levels

Indicator Low pH color Transition pH range High pH color

Gentian violet (Methyl violet 10B) yellow 0.0–2.0 blue-violet

Leucomalachite green (first transition) yellow 0.0–2.0 green

Leucomalachite green (second transition) green 11.6–14 colorless

Thymol blue (first transition) red 1.2–2.8 yellow

Thymol blue (second transition) yellow 8.0–9.6 blue

Methyl yellow red 2.9–4.0 yellow

Bromophenol blue yellow 3.0–4.6 purple

Congo red blue-violet 3.0–5.0 red

Methyl orange red 3.1–4.4 orange

Bromocresol green yellow 3.8–5.4 blue

Methyl red red 4.4–6.2 yellow

Methyl red red 4.5–5.2 green

Azolitmin red 4.5–8.3 blue

Bromocresol purple yellow 5.2–6.8 purple

Bromothymol blue yellow 6.0–7.6 blue

Phenol red yellow 6.8–8.4 red

Neutral red red 6.8–8.0 yellow

Naphtholphthalein colorless to reddish 7.3–8.7 greenish to blue

Cresol Red yellow 7.2–8.8 reddish-purple

Phenolphthalein colorless 8.3–10.0 fuchsia

Thymolphthalein colorless 9.3–10.5 blue

Alizarine Yellow R yellow 10.2–12.0 red

Litmus red 4.5-8.3 blue

Indicators

Selection of Indicators

Titration

1. Strong acid-strong base titration

2. Weak acid-strong base titration

3. Strong acid-weak base titration

4. Polyprotic acid-strong base titration

Strong Acid - Strong Base

1

2

3

4

5

6

7

8

9

10

11

12

13

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00

m illiliters NaOH (0.10 M)

pH

A solution that is 0.10 M HCl is titrated with 0.10 M NaOH

Endpoint is at pH 7

Weak Acid - Strong Base

1

2

3

4

5

6

7

8

9

10

11

12

13

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00m illiliters NaOH (0.10 M)

pH

A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH

Endpoint is above pH 7

Strong Acid – Weak Base

1

2

3

4

5

6

7

8

9

10

11

12

13

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00

m illiliters NH3 (0.10 M)

pH

A solution that is 0.10 M HCl is titrated with 0.10 M NH3

Endpoint is below pH 7

Polyprotic Acids

The titration curve will have as many bumps as there are hydrogen ions to give up

This curve has 2 bumps so it represents a diprotic acid

Titration Calculations

RememberM1V1 = M2V2

When finding the concentration or volume at which an acid and base neutralize each other, we use the same calculation

MaVa = MbVb

(a = acid and b=base)

Acid/base properties Definitions › Arrhenius › Bronsted-Lowry › Lewis Acid-Base Reactions...

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Transcript of Acid/base properties Definitions › Arrhenius › Bronsted-Lowry › Lewis Acid-Base Reactions...