Unit 4 systems and equilibrium acid base equilibrium
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Unit 4-Systems and Equilibrium Acid-Base Equilibrium. Properties of Acids and Bases. Which one will destroy the coke can faster, the acid or the base? http://youtu.be/WnPrtYUKke8 Acids and Bases have many similar properties, including the fact that they can both be dangerous.

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Unit 4-Systems and Equilibrium Acid-Base Equilibrium

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Unit 4 systems and equilibrium acid base equilibrium

Unit 4-Systems and EquilibriumAcid-Base Equilibrium


Properties of acids and bases

Properties of Acids and Bases

  • Which one will destroy the coke can faster, the acid or the base?

    • http://youtu.be/WnPrtYUKke8

  • Acids and Bases have many similar properties, including the fact that they can both be dangerous


Properties of acids and bases1

Properties of Acids and Bases

Produce H3O+ in solution

Produce OH- in solution

React with bases to form a salt and water

React with acids to form a salt and a water

Have a pH less than 7

Have a pH greater than 7

Are Electrolytes

Are Electrolytes

O-----H

Hydroxide Ion

OH-


Acid base definitions

Acid-Base Definitions

Definition 1- Arrhenius Theory of Acids and Bases

“An acid is a substance that produces H3O+ in water, a base is a substance that produces OH- in water”

Problem: there are compounds that are neither oxides nor hydroxides but are still bases

e.g. Na2CO3(aq) and NH3(aq)


Acid base definitions1

Acid-Base Definitions

Definition 2- Bronsted-Lowry Theory of Acids and Bases

“An acid is a proton donor, a base is a proton acceptor”

“Proton” = hydrogen without its electron


Acid base definitions2

Acid-Base Definitions

Definition 2- Bronsted-Lowry (B-L) Theory of Acids and Bases

“An acid is a proton donor, a base is a proton acceptor”

  • B-L Theory defines acids and bases in terms of the effect the substance has on the chemical reaction

    • NOT defined in terms of general properties of the substance itself

  • This means that a compound can be either an acid or a base depending on how it affects the chemical system it is in

    • Compound that can act as either acids or bases are called amphoteric (amphiprotic) compounds


Acid base definitions3

Acid-Base Definitions

  • Amphoteric Example: H2O

AcidBaseConjugateConjugate

Acid Base

Base Acid Conjugate Conjugate

Acid Base


Conjugate acids and bases

Conjugate Acids and Bases

Conjugate Pairs differ by one hydrogen!


Conjugate acids and bases1

Conjugate Acids and Bases

Label each conjugate acid/base pair

  • HClO4(aq) + H2O(l)⇄ H3O+(aq) + ClO4–(aq)

AcidBaseConjugateConjugate

Acid Base

  • HF(aq) + HSO3–(aq) ⇄ F–(aq) + H2SO3(aq)

AcidBase Conjugate Conjugate

Base Acid


Strength of conjugate acids and bases

Strength of Conjugate Acids and Bases

  • A strong acid has a weak conjugate base

  • A strong base has a weak conjugate acid


Acid base definitions4

Acid-Base Definitions

  • B-L Theory does not account for all substances, including those that form complexes in water

    • Examples: Al+3 and BF3

      Definition 3- Lewis Theory of Acids and Bases

      “An acid is an electron-pair acceptor, a base is an electron-pair donor”

  • Note: In order for a substance to be a Lewis base, it must have a lone electron pair to donate!


Lewis theory of acids and bases

Lewis Theory of Acids and Bases

  • Example 1:

    Lewis Acid Lewis Base Adduct

  • A bond is formed between Nitrogen and Boron, but it is the Nitrogen atom originating on the Lewis base that contributes (donates) to the bond


Lewis theory of acids and bases1

Lewis Theory of Acids and Bases

  • Example 2: Autoionization of water

    Lewis Lewis Hydroxide Hydronium Acid Base Ion Ion

  • A bond is formed between Oxygen and Hydrogen, but it is the Oxygen atom originating on the Lewis base that contributes (donates) to the bond

  • H2O can act as both and acid and a base!


More about water

[C]c[D]d

[A]a[B]b

More About Water

  • In a solution of pure water, the forward reaction, autoionization of water, is favoured.

H20(l) H20(l) OH-(aq) H30+ (aq)

Memorize this!

Kw= 1 x 10-14 @ 250C

[OH-][H3O+]

Keq=

Kw=

Equilibrium Constant for water


Water and ph

Water and pH

Memorize this!

Taking the –log of both sides gives us

14 = -log[OH-] + -log[H3O+]

14 = pOH + pH

Kw= 1 x 10-14 @ 250C

∴ @ 250C

1 x 10-14=

[OH-][H3O+]

Memorize this!

[OH-][H3O+]

Kw=


Ph and the environment

pH and the Environment


Acids and bases ph

Acids and Bases-pH

  • The strength of acids and bases is measured by how much they ionize to produce H+ or OH-

  • That means the pH is just a way of reporting the strength of an acid or base

  • pH = 7 neutral

  • pH < 7 Acidic

  • pH > 7 Basic

    ….why?


Acids and bases ph1

Acids and Bases-pH

  • pH = 7 neutral pH =-log [H30+]

  • pH < 7 Acidic

  • pH > 7 Basic

    If pH = 7 then

    14 = pH + pOH

    14 = 7 + pOH

    7 = pOH

    pH = pOH = 7, which means that

    [H30+] = [OH-]  neutral


Acids and bases ph2

Acids and Bases-pH

Example:

If pH = 5.5 then what is the concentration of H30+? OH-?

pH =-log [H30+]

-pH = log[H30+]

10-pH = [H30+] (antilog = 10x)

10-5.5 = [H30+]

[H30+] = 3.16 * 10-6 mol/L

pOH = 14 - 5.5 = 8.5

8.5 = -log[OH-]

-8.5 = log[OH-]

10-8.5 = [OH-]

[OH-] = 3.16 * 10-9 mol/L ……. <[H3O+] ∴ acidic


Acids and bases ph3

pH =-log [H30+]

Acids and Bases-pH

Your turn:

If pH = 9.3 then what is the concentration of H30+? OH-?

pH =-log [H30+]

-pH = log[H30+] (antilog = 10x)

10-pH = [H30+]

10-9.3 = [H30+]

[H30+] = 5.01 * 10-10 mol/L

pOH = 14 – 9.3 = 4.7

4.7 = -log[OH-]

-4.7 = log[OH-]

10-4.7 = [OH-]

[OH-] = 2.00 * 10-5 mol/L ……. > [H3O+] ∴ basic


Strong acids and bases

Strong Acids and Bases

  • The strength of an acid or base is measured by how many ions it forms in solution, NOT BY ITS CONCENTRATION!

  • A strong acid is defined as “an acid that undergoes complete dissociation to form ions in solution”

    • It has nothing to do with the concentration of the acid, only how much of the acid reacts to form H3O+ in solution

  • A strong base is defined as “a base that undergoes complete dissociation to form ions in solution”

    • It has nothing to do with the concentration of the base, only how much of the base reacts to form OH- in solution


Strong acids and bases1

Strong Acids and Bases

  • Strong acids and bases are 100% ionized

    HA(aq) +H2O(l)  H3O+(aq)+ A-(aq)

    B(aq) + H2O(l)  BH+(aq) + OH-(aq)

  • There is no equilibrium setup because all strong acids and bases becomes ions

  • The initial [HA] = [H3O+] (for a 1:1 ratio)

  • The initial [B] = [OH-]

  • The pH (and pOH) of strong acids and bases is easy to determine

100%

0%


Strong acids and bases finding ph

Strong Acids and BasesFinding pH

pH =-log [H30+]

  • Determine the pH for a 0.15 mol/L solution of HCl

pH =-log [H30+]

pH =-log [0.15M]

pH =-log [-0.82]

pH =-(-0.82)

pH = 0.82


Strong acids and bases finding ph1

Strong Acids and BasesFinding pH

pH =-log [H30+]

  • Determine the pH for a 0.0010 mol/L solution of NaOH

pOH =-log [OH-]

Kw=

[OH-][H3O+]

[0.0010M][H3O+]

1 x 10-11=

1 x 10-14 =

1 x 10-14=

[OH-][H3O+]

[H3O+]

pOH =-log [0.0010M]

OR

pOH = 3

14 = pOH + pH

pH = 14 - pOH

pH = -log [1 x 10-11]

pH = 11

pH = 14 - 3

pH = 11


Strong acids and bases finding ph2

Strong Acids and BasesFinding pH

pH =-log [H30+]

  • What is the pH of a 2.0 x 10-3 M solution of HBr?

    • HBr is a strong acid100% dissociation

pH =-log [H+]

pH =-log [0.0020M]

pH = 2.7

pH = 14 - pOH


Strong acids and bases versus weak acids and bases

Strong Acids and Bases Versus Weak Acids and Bases

  • Strong Acids and Bases

    • Are 100% ionized

    • There is no equilibrium, only ions

    • HNO3, HCl, HBr, HI, HClO4, H2SO4strong acids

    • Hydroxides with groups 1 and 2strong bases

      • E.g. Ca(OH)2, KOH

  • Weak Acids and Bases

    • Do not dissociate completely

    • There is an equilibrium formed between the ions in solution and molecules of acid/base

    • The equilibrium constant for acids and bases tells us the ratio of ions to molecules (acid/base)


Weak acids and bases

[A-][H3O+]

[HA]

Weak Acids and Bases

HA(aq) + H2O (l) ⇄ A-(aq)+ H3O+(aq)

  • A weak acid means that the [H3O+] and [A-] is small compared to the [HA]

    • Ka < 1

acid

Ka=

NO H2O(l)


Weak acids and bases1

[C2H3O2 -][H3O+]

[HC2H3O2]

Weak Acids and Bases

Example 1: Acetic Acid in Water

HC2H3O2(aq) + H2O (l) ⇄C2H3O2-(aq)+ H3O+(aq)

Ka=

NO H2O(l)


Weak acids and bases2

[BH+][OH-]

[B]

Weak Acids and Bases

B(aq) + H2O (l) ⇄ BH+(aq)+ OH-(aq)

  • A weak base means that the [OH-] and [BH+] is small compared to the [B]

    • Kb < 1

base

Kb=

NO H2O(l)


Weak acids and bases3

[NH4+][OH-]

[NH3]

Weak Acids and Bases

Example 1: Ammonia in Water

NH3(aq) + H2O (l) ⇄ NH4+(aq)+ OH-(aq)

Kb=

NO H2O(l)


Setting up the equilibrium for weak acids and bases weak acid

[CH3COO- ][H3O+]

[CH3COOH ]

Setting Up the Equilibrium for Weak Acids and Bases-Weak Acid

Example 1: Acetic Acid in Water

What is the pH of a solution of 0.100 mol/L of acetic acid in water? What are the equilibrium concentration of all species in equilibrium in this system? Ka = 1.8 *10-5

Step 1: Balanced Equation

CH3COOH(aq) + H2O(l)⇄ H3O+(aq) + CH3COO- (aq)

Step 2: Equilibrium Expression

Ka=

NO H2O(l)


Setting up the equilibrium for weak acids and bases weak acid1

x*x

[0.100-x ]

Setting Up the Equilibrium for Weak Acids and Bases-Weak Acid

Example 1: Acetic Acid in Water

CH3COOH(aq) + H2O(l)⇄ H3O+(aq) + CH3COO- (aq)

Step 3: I.C.E. Table

Step 4: Equilibrium Expression

Ka=


Setting up the equilibrium for weak acids and bases weak acid2

x2

[0.100 ]

Setting Up the Equilibrium for Weak Acids and Bases-Weak Acid

Example 1: Acetic Acid in Water

CH3COOH(aq) + H2O(l)⇄ H3O+(aq) + CH3COO- (aq)

Step 5: Can we approximate?

Since Ka is so small, we can assume that x will also be very small and that 0.100-x ≈ 0.100

Step 6: Solve with approximation and then check to make sure the approximation is valid!

1.8 * 10-5=


Setting up the equilibrium for weak acids and bases weak acid3

x2

[0.100 ]

Setting Up the Equilibrium for Weak Acids and Bases-Weak Acid

Example 1: Acetic Acid in Water

CH3COOH(aq) + H2O(l)⇄ H3O+(aq) + CH3COO- (aq)

X = 1.3 * 10-3 mol/L

0.100 – 0.0013 ≈ 0.100

The approximation is valid!

[CH3COOH] = 9.9 * 10-2 mol/L

[H3O+] = 1.3 *10-3 mol/L

[CH3COO-] = 1.3 * 10-3 mol/L

1.8 * 10-5=


Setting up the equilibrium for weak acids and bases weak acid4

Setting Up the Equilibrium for Weak Acids and Bases-Weak Acid

Example 1: Acetic Acid in Water

CH3COOH(aq) + H2O(l)⇄ H3O+(aq) + CH3COO- (aq)

[CH3COOH] = 9.9 * 10-2 mol/L

[H3O+] = 1.3 *10-3 mol/L

[CH3COO-] = 1.3 * 10-3 mol/L

pH = -log[H3O+]

= -log[1.3 *10-3 ]

= 2.88


Setting up the equilibrium for weak acids and bases weak base

[NH4+ ][OH-]

[NH3]

Setting Up the Equilibrium for Weak Acids and Bases-Weak Base

Example 2: Ammonia in Water

What is the pH of a solution of 0.100 mol/L of ammonia in water? What are the equilibrium concentration of all species in equilibrium in this system? Kb = 1.8 *10-5

Step 1: Balanced Equation

NH3 (aq) + H2O(l)⇄ NH4+(aq) + OH- (aq)

Step 2: Equilibrium Expression

Kb=

NO H2O(l)


Setting up the equilibrium for weak acids and bases weak base1

x*x

[0.100-x ]

Setting Up the Equilibrium for Weak Acids and Bases-Weak Base

Example 2: Ammonia in Water

NH3 (aq) + H2O(l)⇄ NH4+(aq) + OH- (aq)

Step 3: I.C.E. Table

Step 4: Equilibrium Expression

Kb=


Setting up the equilibrium for weak acids and bases weak base2

x2

[0.100 ]

Setting Up the Equilibrium for Weak Acids and Bases-Weak Base

Example 2: Ammonia in Water

NH3 (aq) + H2O(l)⇄ NH4+(aq) + OH- (aq)

Step 5: Can we approximate?

Since Kb is so small, we can assume that x will also be very small and that 0.100-x ≈ 0.100

Step 6: Solve with approximation and then check to make sure the approximation is valid!

1.8 * 10-5=


Setting up the equilibrium for weak acids and bases weak base3

x2

[0.100 ]

Setting Up the Equilibrium for Weak Acids and Bases-Weak Base

Example 2: Ammonia in Water

NH3 (aq) + H2O(l)⇄ NH4+(aq) + OH- (aq)

X = 1.3 * 10-3 mol/L

0.100 – 0.0013 ≈ 0.100

The approximation is valid!

[NH3] = 9.9 * 10-2 mol/L

[NH4+] = 1.3 *10-3 mol/L

[OH-] = 1.3 * 10-3 mol/L

1.8 * 10-5=


Setting up the equilibrium for weak acids and bases weak base4

Setting Up the Equilibrium for Weak Acids and Bases-Weak Base

Example 2: Ammonia in Water

NH3 (aq) + H2O(l)⇄ NH4+(aq) + OH- (aq)

[NH3] = 9.9 * 10-2 mol/L

[NH4+] = 1.3 *10-3 mol/L

[OH-] = 1.3 * 10-3 mol/L

pOH = -log[OH-]

= -log[1.3 *10-3 ]

= 2.88

pH + pOH = 14

14- pOH = pH

pH = 11.12


Equilibrium for weak acids and bases practice practice practice

Equilibrium for Weak Acids and BasesPractice, Practice, Practice!

Weak Acid Problems

What is the pH of the resulting solution when 0.500 moles of acetic acid are dissolved in water and diluted to 1.00 L?Unbalanced reaction: CH3COOH(aq)  H+(aq) + CH3COO-(aq)     Ka = 1.8x10-5 at 25oC.

What is the pH of the resulting solution when 100.0 mL of 0.500 M of hypochlorous acid is mixed with 100.0 mL water?Unbalanced reaction: HOCl(aq)  H+(aq) + OCl-(aq)     Ka = 3.0x10-8 at 25oC.

What is the pH of the resulting solution when 0.500 moles of trichloroacetic acid are dissolved in water and diluted to 1.00 L?Unbalanced reaction: CCl3COOH(aq)  H+(aq) + CCl3COO-(aq)     Ka = 1.3x10-1 at 25oC.

Weak Base Problems

What is the pH of the resulting solution when 250.0 mL of a 0.300 M solution of ammonia added to 750.0 mL of water?Unbalanced reaction: NH3(aq) + H2O  NH4+(aq) + OH-(aq)     Kb = 1.8x10-5 at 25oC.

Calculate Kb for methylamine if a solution prepared by dissolving 0.100 moles of methylamine in 1.00 L of water has a measured pH of 11.80.Unbalanced reaction: CH3NH2(aq) + H2O  CH3NH3+(aq) + OH-(aq)


Relationship between k a and k b

Relationship Between Ka and Kb

  • The bigger the Ka of an acid, the smaller the Kb of its conjugate base

  • The bigger the Kb of a base, the smaller the Ka of its conjugate acid


How can we relate k a and k b for conjugate acids and bases

[A-][H3O+]

[HA][OH-]

[HA]

[A-]

How Can We Relate Ka and Kb For Conjugate Acids and Bases?

  • Acid Reaction

    HA (aq) + H2O (l)  H3O+ (aq) + A- (aq)

  • Conjugate Base Reaction

    A- (aq) + H2O (l)  HA (aq) + OH- (aq)

Ka=

Kb=


How can we relate k a and k b for conjugate acids and bases1

[A-][H3O+]

[HA][OH-]

[HA]

[A-]

How Can We Relate Ka and Kb For Conjugate Acids and Bases?

  • Recall: When we add the 2 reactions together, we multiply their K’s

  • HA (aq) + H2O (l)  H3O+ (aq) + A- (aq) Ka

  • A- (aq) + H2O (l)  HA (aq) + OH- (aq) Kb

  • 2H2O (l)  H3O+(aq) + OH-(aq) Ka*Kb

    Ka*Kb = x

    = [H3O+]x[OH-]

    = Kw


How can we relate k a and k b for conjugate acids and bases2

How Can We Relate Ka and Kb For Conjugate Acids and Bases?

  • Therefore, we can say the following relationship between Ka and Kb exists for CONJUGATE PAIRS

  • Kw=Ka*Kb

  • Ka=Kw/Kb

  • Kb=Kw/Ka

  • So if only Ka is given, we can find the Kb of its conjugate base when needed

  • This formula shows that the stronger the weak acid/base, the weaker it’s conjugate base/acid is

    • Stronger weak acid = weaker conjugate base

    • Stronger weak base = weaker conjugate acid


How can we relate k a and k b for conjugate acids and bases example

How Can We Relate Ka and Kb For Conjugate Acids and Bases?Example

  • Example 1:

    The Kb for the rocket fuel hydrazine, N2H4(g) is 1.7 x 10-6. What is the Ka of its conjugate acid N2H5+?

    • 5.9 x 10-9


Percent ionization for weak acids another way of determining reporting h 3 o

Mark received

[Ion in Solution]

Total Marks Possible

[Total Weak Acid]

Percent Ionization for Weak AcidsAnother Way of Determining/Reporting [H3O+]

  • Percent Ionization is

    • The amount of acid that has dissociated or ionized in solution

  • We can calculate the percent of weak acid that has ionized in solution in the same manner we calculate the percent we receive on a test…

  • Percent on test = x 100%

  • Percent Ionization = x 100%


Percent ionization

[Ion in Solution]

[H3O+(aq)]

[OH-(aq)]

[Total Weak Acid/Weak Base in Solution]

[HA(aq)]

[B(aq)]

Percent Ionization

Percent Ionization = x 100%

Percent Ionization Acid = x 100%

Percent Ionization Base= x 100%


Percent ionization for weak acids examples

Percent Ionization for Weak AcidsExamples

  • Example 1:

    The pH of a 0.10mol/L methanoic acid (HCOOH) solution is 2.38. Calculate the percent ionization of methanoic acid

    • 4.2%

  • Example 2:

    Calculate the acid ionization constant, Ka, of acetic acid (CH3COOH) if a 0.1000mol/L solution at equilibrium has a percent ionization of 1.3%

    • 1.7 x 10-5


  • Salt solutions

    Salt Solutions

    • Salts are formed when an acid reacts with a base

      • The result is a salt and water

        HCl + NaOH NaCl + H2O

        acid base salt water

    • Salts may either form neutral, acidic or basic solutions depending on where the ions came from

    • This means that when added to a solution, salts may or may not affect the pH of that solution


    Salt solutions1

    Salt Solutions

    • How do we tell if a salt added to water will have an effect on the pH of a solution?

      • We must look at the reaction of the dissolved ions from the salt with water

      • If a reaction with an ion from a salt with water produces an acidic or basic solution, it is called hydrolysis

    • There are 3 possible scenarios:

      • The salt is a neutral salt and does not hydrolyze (react with water)

      • The salt is an acidic salt and hydrolyzes (reacts with water) to form H3O+ ions in solution

      • The salt is a basic salt and hydrolyzes (reacts with water) to form OH- ions in solution


    Scenario 1 nuetral salts

    Scenario 1: NUETRAL SALTS

    • In general, salts that are composed from strong acids and bases will be too weak to react with water

      • Because they come from the conjugate base or acid

      • Strong acid/base = weak conjugate base/acid

        NaOH+ HCl→NaCl +H2O

        Strong base Strong Acid Neutral Salt Water


    Scenario 2 acidic salts

    Scenario 2: Acidic Salts

    • In general, acidic salts come from the combination of a weak base with a strong acid

      NH3+ HCl→NH4Cl

      Weak Base Strong Acid Acidic Salt

      Cl-+ H2O  Nothing….but

      NH4++ H2O  NH3 + H3O+

      Weak Weak Hydronium

      Conjugate Base Ion

      Acid Ion


    Scenario 3 basic salts

    Scenario 3: Basic Salts

    • In general, basic salts come from the combination of a weak acid with a strong base

      HCN+NaOH→ NaCN +H2O

      Weak Acid Strong Base Basic Salt Water

      Na++ H2O  Nothing….but

      CN-+ H2O  HCN + OH-

      Weak Weak Hydroxide

      Conjugate Base Ion

      Base Ion


    Salt solutions how to predict acid base behavior of a salt

    Salt SolutionsHow to Predict Acid/Base Behavior of a Salt

    • Step 1: Determine if the cation is the conjugate acid of a weak base. If so, it will make the solution more acidic, if not, it will NOT affect the pH of a solution

    • Step 2: Determine if the anion is the conjugate base of a weak acid. If so, it will make the solution more basic, if not, it will NOT affect the pH of a solution

    • Step 3: If the salt has both an anion and a cation that can react with water (hydrolyze) then compare their Ka and Kb values.

      • If Ka>Kb, the solution becomes acidic

      • If Kb> Ka, the solution becomes basic


    Salt solutions examples

    Salt SolutionsExamples

    • Example 1:

      Predict whether a 0.10M solution of NaNO2(aq) will be acidic, basic, or neutral. Then calculate its pH. Ka = 7.2 x 10-4

      NaNO2(s)  Na+(aq) + NO2-(aq)

    From NaOH

    Strong

    Base

    From

    HNO2

    Weak

    Acid


    Salt solutions examples1

    [HNO2][OH-]

    [NO2-]

    Salt SolutionsExamples

    NO2-(aq) + H2O(l) HNO2(aq) + OH-(aq)

    Kb =

    Kb=Kw/Ka

    =1.0 x 10-14/7.2 x 10-4

    = 1.4 x 10-11


    Salt solutions examples2

    x2

    x2

    [0.10-x]

    [0.10]

    Salt SolutionsExamples

    NO2-(aq) + H2O(l) HNO2(aq) + OH-(aq)

    Kb = 1.4 x 10-11 = Approximate to get:

    1.4 x 10-11 =

    x = [OH-] = 1.2 x 10-6 mol/L


    Salt solutions examples3

    Salt SolutionsExamples

    NO2-(aq) + H2O(l) HNO2(aq) + OH-(aq)

    x = [OH-] = 1.2 x 10-6 mol/L

    pOH = -log[OH-]

    = 5.92

    pH = 14 - pOH

    = 14 – 5.92

    = 8.08


    Acid base titrations

    Acid-Base Titrations

    • What is a titration?

      • A technique used in chemistry used to find unknown concentrations of a sample

      • Commonly used in industry:

        • To determine fatty acid content in foods

        • To test water in aquariums and marine environments

        • To determine proper concentrations of chemicals used in pharmaceuticals

        • To determine proper concentrations in biology when using anesthetic or in euthanizing animals


    Acid base titrations1

    Acid-Base Titrations

    • How does it work?

      • We use our knowledge of stoichiometry and acids and bases to determine unknowns from what we do know

        Acid + Base ⇄ Salt + Water

    • If we know the concentration of one and have a way to figure out when the reaction finishes, then we can use stoichiometry to figure out the concentration of the other


    Acid base titrations2

    Acid-Base Titrations

    • When we titrate, there are 3 different scenarios we look at:

      • A strong acid and a strong base

      • A strong acid with a weak base

      • A strong base with a weak acid

      • The general process is the same for each:

      • We have a solution who’s concentration is known

      • We react it with a solution who’s concentration is unknown

      • We use an indicator or pH probe to tell us when the reaction is finished

      • We figure out the concentration of the unknown using our knowledge of acid-base reactions and stoichiometry


    Acid base titrations3

    Acid-Base Titrations

    • So how do we actually do it?

      • We put our “known solution”-our “titrant” into a buret

        • We know this solution’s concentration

      • We put our solution with unknown concentration-our analyte-in a flask

        • With an “indicator” that tells us when the reaction is done by changing colour

        • We could also use a pH probe to tell us when the reaction is done


    Acid base titrations4

    Acid-Base Titrations

    • We record how much (the volume) of the titrant we put in to make the reaction complete

      • From there we can figure out the number of moles of titrate we used

    • We use our knowledge of acids and base reactions to relate that information to the number of moles in our unknown

      • We use that plus the volume of unknown we put in to calculate the concentration of the unknown sample


    Acid base titrations5

    Acid-Base Titrations

    McGraw-Hill Animation


    Acid base titrations 1 strong acid with strong base

    Acid-Base Titrations1. Strong Acid with Strong Base

    • When we titrate a strong acid with a strong base, we end up with a neutral solution:

      NaOH+ HCl→ NaCl +H2O

      Strong base Strong Acid Neutral Salt Water

    • If we have NaOH in the buret and HCl in the flask, then as we add NaOH, the OH- reacts with the H+ from HCl to form water-to “neutralize” the solution

    • With every drop of NaOH added, we end up with less H+.

      • Which causes our pH to go up (become more basic)


    Acid base titrations titration curves

    Acid-Base TitrationsTitration Curves

    Titration Curves

    A plot of pH versus volume of titrant added (solution in buretwith known concentration)

    As the volume of base added increases, the pH increases slightly UNTIL…

    We hit the point where there is no more H+ from HCl left to react with = neutralization

    Then as more NaOH is added, the pH rises dramatically as [OH-] dominates the solution


    Acid base titrations titration curves1

    Acid-Base TitrationsTitration Curves

    Titration Curves

    The “Equivalence Point”-the point at which exactly enough titrant has been added to react with all analyte

    All titrant added after the equivalence point has no analyte left to react with

    In this case, with NaOH being our titrant, the pH rises quickly in response to the addition of more and more base


    Acid base titrations titration curves2

    Acid-Base TitrationsTitration Curves

    Because we are titrating a strong acid with a strong base, our pH at the equivalence point is “neutral pH” = 7

    This is NOT the case when we mix a weak acid/base with a strong base/acid


    Acid base titrations calculating unknown concentrations with strong acid and strong base titrations

    Acid-Base TitrationsCalculating Unknown Concentrations with Strong Acid and Strong Base Titrations

    NaOH+ HCl→ NaCl +H2O

    Strong base Strong Acid Neutral Salt Water

    If the concentration of NaOH is 0.100mol/L and the volume of HCl in the flask is 200mL, what is the molarity of HCl?

    mL of NaOH added


    Acid base titrations calculating unknown concentrations with strong acid and strong base titrations1

    Acid-Base TitrationsCalculating Unknown Concentrations with Strong Acid and Strong Base Titrations


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