TOPIC 3: MORE ABOUT ACIDS AND BASES

Unit 12.2: acids, bases and salts

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Learning outcomes

•Recall the definitions of acids and bases

•Describe the reactions of acids and bases

•Define strong acids

•Calculate the pH of strong acids and weak acids

•Calculate the pH of strong bases and weak bases

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Instructions

• This power point concentration contains definitions on acids and bases, reactions of acids and bases, and activities on how to calculate the pH of strong acids, weak acids, strong bases and weak bases.

• It should take you at least 2 weeks (if you allocate an hour per day for Chemistry) to do the examples and complete all activities.

• You need your calculator for this topic.

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Acidity and basicity

DEFINITIONS OF ACIDITY DEFINITIONS OF BASICITY

The acidity of an acid refers to the relative number of protons it donates per mole of acid (in aqueous solution)

Basicity refers to the relative number of protons accepted per mole of base (in aqueous solutions)

Monoprotic acids release one mole of hydrogen ions per mole of acid. Hydrochloric acid, HCL, is a Monoprotic acid.

HCl H+ + Cl-

Monobasic substances, such as potassium hydroxide, KOH, accept one mole of protons per mole of substance.

KOH + H+ K+ + H2O

Diprotic acids release two moles of hydrogen ions per mole of acid. Sulphuric acid, H2SO4, is a diprotic acid.

H2SO4 2H+ + SO42-

Dibasic substances accept two moles of protons per mole of a substance. Calcium hydroxide is dibasic.Ca (OH)2 + 2H+ Ca2+ + + 2H2O.

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ACID-BASE REACTIONS

• The reaction between aqueous acids and aqueous bases involves proton transfer.

• The hydronium ions released by the acid donate protons to the hydroxide ions released by the base:

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• Example A: Reaction between HCl and NaOH

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

• Hydronium ions from HCl react with OH ions

from NaOH:

H3O+ (aq) + OH-(aq) 2H2O(l)

Na+ and Cl- are spectator ions

Example B: Acid-base reaction between hydrogen chloride and ammonia

•HCl(g) + NH3(g) NH4Cl(s)

•The hydrogen chloride donates protons to ammonia

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Strong acids

• A strong acid is virtually 100% dissociated in water to form hydronium ions and the conjugate base.

• Nitric acid, HNO3(aq) and Hydrochloric acid, HCl(aq), are strong acids.

• e.g. When hydrogen chloride gas dissolves in water, it fully dissociates to form H3O+(aq) and Cl-(aq).

• HCl(g) + H2O(l) H3O+(aq) + Cl-(aq)

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Calculating ph of strong acids

• pH = -log[H3O+]

• E.g. What is the pH of hydrochloric acid with a concentration of 0.00600 ml/L?

•On your calculator, press – sign followed by log, then enter the concentration of acid

• ph(HCl) = -log0.000600 Therefore pH(HCl) = 2.22

ACTIVITY: Calculate the pH of:

a. 0.015 mol/L of HCl

b. 0.500 mol/L of HCl

c. 0.00213 mol/L of HCl

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WEAK ACIDS • A weak acid only partially dissociates in water – often less than 5% of the acid

molecules present dissociate. This means that most of the acid molecules do not release their protons.

• The equilibrium position for the acid dissolving in water lies towards the left-hand side of the equilibrium equation.

• HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq) HA = acid, A- = conjugate base

• EXAMPLE: Ethanoic acid, CH3COOH, is a weak acid

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

0.1 mol L- ≈0.001mol L- ≈0.001mol L-

In this equilibrium:

• Concentration of ethanoic acid remains high

• Concentration of both hydronium ions and ethanoate ions are identical and small.

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The acidity constant, Ka

• Ka: the acid dissociation constant

• A measure of how much an acid breaks apart into its ions

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EXAMPLE: Give the Ka expression for CH3COOH (aq) + H2O(l) ⇌ CH3COO-(aq) + H3O+(aq)

Ka = 𝐻

3𝑂+

[𝐶𝐻3𝐶𝑂𝑂

−

][𝐶𝐻

3𝐶𝑂𝑂𝐻]

Calculating ph of weak acids• Finding the pH of weak acid is a bit more

complicated than finding the pH of strong acids because the acid does not fully dissociate into its ions.

• The equation is still the same (pH = -log[H+]), but you need to use the acid dissociation constant (Ka) to find [H+].

1. 𝐻3𝑂+= 𝐾𝑎 𝑥 𝑐(𝐻𝐴)

2. pH = -log[H3O+]

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EXAMPLE: Calculate the pH of 0.2 mol L- of CH3COOH (Ka= 1.74 x 10-5) at constant temperature.

Activity

•ANSWER Q3, Q5 and Q6 in the link below

•GR12 SCI CHEM SKAILOU4 STRONG AND WEAK ACIDS ACTIVITY.pdf

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Ka and pKa

• Ka is the acid dissociation constant and represents the strength of an acid.

• pKa is the –log of Ka, having a smaller comparable values for analysis.

• They have an inverse relationship. The larger the Ka, smaller the pKa and stronger the acid.

• Smaller pKa, stronger the acid.

Ka = 10-pKa

pKa = -logKa

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Calculating ka and pKa• EXAMPLE

1. Calculate the pKa of HCOOH (Ka = 1.82x10-4)

Solution: pKa = -logKa

= -log(1.82x10-4)

= 3.74

2. Calculate the Ka of NH4+ with pKa of 9.24.

Solution: Ka = 10-pKa

= 10-9.24

= 5.75 x 10-10

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ACTIVITY

•Answer Q2 and Q3 in the link below

•GR12 SCI CHEM SKAILOU3 Ka and pKa activity.pdf

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Ionic product, kw, FOR WATER

• Kw is the ion product constant for water.

• It is actually the equilibrium constant for the dissociation of water. H2O ⇌ H+ + OH-

• Kw = [H3O+] [OH-]

• The value of Kw varies with temperature.

• The value of Kw at 25°C is 1x10-14

• Higher the temperature, higher the Kw

• The ionic product, Kw = [H3O+] [OH-] can be used to determine the pH of strong bases.

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Calculating the ph of strong bases using kw

• A strong base completely dissociates in water.

• E.g. NaOH(aq) + H2O(l) Na+(aq) + OH-(aq)

• pH of a strong base is calculated using

Step 1: [H+] = 𝐾𝑤

[𝑂𝐻−]Step 2: pH = -log[H+]

Example

Calculate the pH of 0.0200 mol L- of NaOH at 25°C.

[H+] = 𝐾𝑤

[𝑂𝐻−]pH = -log[H+]

= 1 𝑥 10−14

0.0200= -log(5.00x10-13)

= 5.00x10-13 = 12.3

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ACTIVITYAnswer Q1 and Q2 in the link belowGR12 SCI CHEM SKAILOU2 Activity on pH of bases.pdf

Weak bases

• A weak base reacts only to a small extent with water.

• Example 1: Ammonia is a weak base, only a very small percentage reacts.

• NH3aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)

0.1 mol/L 0.001 mol/L 0.001 mol/L

• Example 2: Ethanoate ion, CH3COO-, is an even weaker base than ammonia

CH3COO- (aq) + H2O(l) ⇌ CH3COOH(aq) + OH-(aq)

0.1 mol/L 0.00001 mol/L 0.00001 mol/L

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Calculating ph of weak base solution

1. Using Ka to calculate pH of a solution of a weak base

[H3O+] = 𝐾𝑎 𝑥 𝐾𝑤

𝑐(𝐵)

pH = - log[H3O+]

2. Using Kb (base dissociation constant) to calculate pH of a solution of a weak base

STEP 1: [OH-] = 𝐾𝑏 𝑥 𝑐(𝐵) Kb = 𝐾𝑤

𝐾𝑎

STEP 2: [H3O+] = 𝐾𝑤

[𝑂𝐻−]

STEP 3: pH = - log[H3O+]

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Example: Calculate the pH of a weak base

Calculate the pH of a 0.400 mol L- solution of sodium ethanoate, CH3COOHNa.

Ka(CH3COOH) = 1.74 x 10-5 mol L-

CH3COOHNa(s) + H2O(l) CH3COO-(aq) + Na+(aq)

STEP 1

[H3O+] = 𝐾𝑤 𝑥 𝐾𝑎

[𝐶𝐻3𝐶𝑂𝑂−]

[H3O+] = 1𝑥10−14 𝑥 1.74 𝑥 10−5

[0.400]= 6.60 x 10-10

STEP 2

pH = -log[H3O+]

= -log(6.60 x 10-10)

= 9.18

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ACTIVITY

•ANSWER Q3 on the link below

•GR12 SCI CHEM SKAILOU2 Activity on pH of bases.pdf

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