Chapter 5 Equilibra

Question 1

A Brønsted acid

Answer 1

• is a species that can donate a proton
  
  • For example, hydrogen chloride (HCl) is a Brønsted acid as it can lose a proton to form a hydrogen (H⁺) and chloride (Cl^-) ion

HCl (aq) → H⁺ (aq) + Cl^- (aq)

Question 2

A Brønsted base

Answer 2

• is a species that can accept a proton
  
  • For example, a hydroxide (OH^-) ion is a Brønsted base as it can accept a proton to form water

OH^- (aq) + H⁺ (aq) → H₂O (l)

Question 3

In an equilibrium reaction, the products are formed at the

Answer 3

• same rate as the reactants are used
• This means that at equilibrium, both reactants and products are present in the solution

Question 4

A conjugate acid-base pair

Answer 4

is two species that are different from each other by an H⁺ ion

• Conjugate here means related
• In other words, the acid and base are related to each other by one proton difference

Question 5

Conjugate acid-base pairs are a pair of reactants and products that are linked to each other by the

Answer 5

transfer of a proton

Question 6

The pH indicates the

Answer 6

• acidity or basicity of an acid or alkali
• The pH scale goes from 0 to 14
  
  • Acids have pH between 0-7
  • Pure water is neutral and has a pH of 7
  • Bases and alkalis have pH between 7-14

Question 7

calculation of pH

Answer 7

• The pH can be calculated using: pH = -log₁₀ [H⁺]

where [H⁺] = concentration of H⁺ ions (mol dm^-3)

• The pH can also be used to calculate the concentration of H⁺ ions in solution by rearranging the equation to:

[H⁺] = 10^-pH

Question 8

The K_a is the

Answer 8

• acidic dissociation constant
  
  • It is the equilibrium constant for the dissociation of a weak acid at 298 K
• For the partial ionisation of a weak acid HA the equilibrium expression to find K_a is as follows:

HA (aq) ⇌ H⁺ (aq) + A^- (aq)

Question 9

When writing the equilibrium expression for weak acids, the following assumptions are made:

Answer 9

• The concentration of hydrogen ions due to the ionisation of water is negligible
• The dissociation of the weak acid is so small that the concentration of HA is approximately the same as the concentration of A^-

Question 10

The value of K_a indicates the extent of

Answer 10

dissociation

• A high value of K_a means that:
  
  • The equilibrium position lies to the right
  • The acid is almost completely ionised
  • The acid is strongly acidic
• A low value of K_a means that:
  
  • The equilibrium position lies to the left
  • The acid is only slightly ionised (there are mainly HA and only a few H⁺ and A^- ions)
  • The acid is weakly acidic

Question 11

Since K_a values of many weak acids are high/low what values are used to compare

Answer 11

• very low, pK_a values are used instead to compare the strengths of weak acids with each other

pK_a= -log₁₀K_a

• The less positive the pK_a value the more acidic the acid is

Question 12

The K_w is the

Answer 12

• ionic product of water
  
  • It is the equilibrium constant for the dissociation of water at 298 K
  • Its value is 1.00 x 10^-14 mol² dm^-6
• For the ionisation of water the equilibrium expression to find K_w is as follows:

H₂O (l) ⇌ H⁺ (aq) + OH^- (aq)

Question 13

K_w : As the extent of ionisation is

Answer 13

• is very low, only small amounts of H⁺ and OH^- ions are formed
• The concentration of H₂O can therefore be regarded as constant and removed from the K_w expression
• The equilibrium expression therefore becomes:
• K_w* = [H⁺] [OH^-]
• As the [H⁺] = [OH⁺] in pure water, the equilibrium expression can be further simplified to:
• K_w* = [H⁺]²

Question 14

Calculating [H+] & pH

Answer 14

• If the concentration of H⁺ of an acid or alkali is known, the pH can be calculated using the equation:

pH = -log [H⁺]

• Similarly, the concentration of H⁺ of a solution can be calculated if the pH is known by rearranging the above equation to:

[H⁺] = 10^-pH

Question 15

Strong acids are completely

Answer 15

• ionised in solution

HA (aq) → H⁺ (aq) + A^- (aq)

• Therefore, the concentration of hydrogen ions ([H⁺]) is equal to the concentration of acid ([HA])
• The number of hydrogen ions ([H⁺]) formed from the ionisation of water is very small relative to the [H⁺] due to ionisation of the strong acid and can therefore be neglected
• The total [H⁺] is therefore the same as the [HA]

Question 16

Strong alkalis are

Answer 16

• completely ionised in solution

BOH (aq) → B⁺ (aq) + OH^- (aq)

• Therefore, the concentration of hydroxide ions ([OH^-]) is equal to the concentration of base ([BOH])
  
  • Even strong alkalis have small amounts of H⁺ in solution which is due to the ionisation of water

Question 17

• The concentration of OH^- in solution can be used to calculate the pH using the ionic product of water

Answer 17

K_w = [H⁺] [OH^-]

• Since K_w is 1.00 x 10^-14 mol² dm^-6

• Once the [H⁺] has been determined, the pH of the strong alkali can be founding using pH = -log[H⁺]
• Similarly, the ionic product of water can be used to find the concentration of OH^- ions in solution if [H⁺] is known

Question 18

A buffer solution is a

Answer 18

solution in which the pH does not change a lot when small amounts of acids or alkalis are added

• A buffer solution is used to keep the pH almost constant
• A buffer can consists of weak acid - conjugate base or weak base - conjugate acid

Question 19

A common buffer solution is an

Answer 19

aqueous mixture of ethanoic acid and sodium ethanoate

Question 20

Ethanoic acid is a …acid

Answer 20

• weak acid and partially ionises in solution to form a relatively low concentration of ethanoate ions

Question 21

• When H⁺ ions are added:

Answer 21

• The equilibrium position shifts to the left as H⁺ ions react with CH₃COO^- ions to form more CH₃COOH until equilibrium is re-established
• As there is a large reserve supply of CH₃COO^- the concentration of CH₃COO^- in solution doesn’t change much as it reacts with the added H⁺ ions
• As there is a large reserve supply of CH₃COOH the concentration of CH₃COOH in solution doesn’t change much as CH₃COOH is formed from the reaction of CH₃COO^- with H⁺
• As a result, the pH remains reasonable constant

Question 22

• When OH^- ions are added:

Answer 22

• The OH^- reacts with H⁺ to form water

OH^- (aq) + H⁺ (aq) → H₂O (l)

• The H⁺ concentration decreases
• The equilibrium position shifts to the right and more CH₃COOH molecules ionise to form more H⁺ and CH₃COO^- until equilibrium is re-established

CH₃COOH (aq) → H⁺ (aq) + CH₃COO^- (aq)

• As there is a large reserve supply of CH₃COOH the concentration of CH₃COOH in solution doesn’t change much when CH₃COOH dissociates to form more H⁺ ions
• As there is a large reserve supply of CH₃COO^- the concentration of CH₃COO^- in solution doesn’t change much
• As a result, the pH remains reasonable constant

Question 23

When hydroxide ions are added to the solution, the hydrogen ions react with them to form water; The decrease in hydrogen ions would mean that the pH would increase however the equilibrium moves to the right to replace the removed hydrogen ions and keep the pH constant

Answer 23

Question 24

Uses of buffer solutions in controlling the pH of blood

Answer 24

• In humans, HCO₃^- ions act as a buffer to keep the blood pH between 7.35 and 7.45
• Body cells produce CO₂ during aerobic respiration
• This CO₂ will combine with water in blood to form a solution containing H⁺ ions

CO₂ (g) + H₂O (l) ⇌ H⁺ (aq) + HCO₃^- (aq)

Question 25

* This equilibrium between CO₂ and HCO₃^- is extremely important * If the concentration of H⁺ ions is not regulated, the blood pH would

Answer 25

drop and cause ‘**acidosis**’ * **Acidosis** refers to a condition in which there is too much acid in the body fluids such as blood * This could cause body malfunctioning and eventually lead to coma

Question 26

* **If there is an increase in H⁺ ions in the blood**

Answer 26

* The equilibrium position shifts to the **left** until equilibrium is restored **H⁺ (aq) + HCO₃^- (aq) → CO₂ (g) + H₂O (l)** * This reduces the concentration of H⁺ and keeps the pH of the blood **constant**

Question 27

**If there is a decrease in H⁺ ions in the blood**

Answer 27

* The equilibrium position shifts to the **right** until equilibrium is restored **CO₂ (g) + H₂O (l) → H⁺ (aq) + HCO₃^- (aq)** * This increases the concentration of H⁺ and keeps the pH of the blood **constant**

Question 28

The pH of a **buffer** **solution** can be calculated using:

Answer 28

* The *K_a* of the **weak acid** * The **equilibrium concentration** of the **weak acid** and its **conjugate base** (salt)

Question 29

To determine the pH, the concentration of **hydrogen** **ions** is needed which can be found using the equilibrium expression

Answer 29

* To simplify the calculations, **logarithms** are used such that the expression becomes: * Since -log₁₀ [H⁺] = pH, the expression can also be rewritten as:

Question 30

**Solubility** is defined as the

Answer 30

**number of grams** or **moles** of compound needed to **saturate** 100 g of **water**, or it can also be defined in terms of 1 kg of water, at a given temperature * For example, sodium chloride (NaCl) is considered to be a **soluble** salt as a saturated solution contains 36 g of NaCl per 100 g of water

Question 31

The **solubility product** (***K_sp***) is

Answer 31

* The product of the concentrations of each ion in a saturated solution of a relatively soluble salt * At 298 K * Raised to the power of their relative concentrations **C (s) ⇌ aA^x+ (aq) + bB^y- (aq)** ***K_sp*** **= [A^x+ (aq)]^a [B^y- (aq)]^b**

Question 32

When an **undissolved ionic compound** is in contact with a **saturated solution of its ions**, an (what?)

Answer 32

* equilibrium is established * The ions move from the solid to the saturated solution at the same rate as they move from the solution to the solid * For example, the undissolved magnesium chloride (MgCl₂) is in equilibrium with a saturated solution of its ions **MgCl₂ (s) ⇌ Mg²⁺ (aq) + 2Cl^- (aq)**

Question 33

The *K_sp* is only useful for

Answer 33

* **sparingly soluble salts** * The smaller the value of *K_sp*, the lower the solubility of the salt

Question 34

The general equilibrium expression for the **solubility product** (*K_sp*) is:

Answer 34

**C (s) ⇌ aA^x+ (aq) + bB^y- (aq)** ***K_sp*** **= [A^x+ (aq)]^a [B^y- (aq)]^b**

Question 35

Calculations involving the **solubility product** (*K_sp*) may include:

Answer 35

* Calculating the solubility product of a compound from its **solubility** * Calculating the solubility of a compound from the **solubility** **product**

Question 36

A **saturated solution** is a solution that contains the

Answer 36

* **maximum** amount of dissolved salt * If a second compound, which has an ion **in common** with the dissolved salt, is added to the saturated solution, the solubility of the salt reduces and a solid **precipitate** will be formed * This is also known as the **common ion effect**

Question 37

The **solubility product** can be used to predict whether a precipitate will

Answer 37

l actually form or not * A precipitate will form if the product of the ion concentrations is **greater than** the solubility product (*K_sp*)

Question 38

**Common ion effect in silver chloride**

Answer 38

* When a **KCl** solution is added to a saturated solution of **AgCl**, an AgCl precipitate forms * In a **saturated** AgCl solution, the silver chloride is in equilibrium with its ions **AgCl (s) ⇌ Ag⁺ (aq) + Cl^- (aq)** * When a solution of potassium chloride is added: * Both KCl and AgCl have the common Cl^- ion * There is an increased Cl^- concentration so the equilibrium position shifts to the left * The increase in Cl^- concentration also means that [Ag⁺ (aq)] [Cl^-(aq)] is **greater** than the *K_sp* for AgCl * As a result, the AgCl is **precipitated**

Question 39

The **partition coefficient (*****K_pc*****)** is the

Answer 39

ratio of the concentrations of a **solute** in two different immiscible solvents in contact with each other when equilibrium has been established (at a particular temperature)

Question 40

A **separating funnel**

Answer 40

* is shaken with the organic solvent and aqueous methylamine * The methylamine is **soluble** in both solvents, so when the mixture is left to settle an **equilibrium** is established * The **rate** of methylamine molecules moving from the organic layer into the aqueous layer is **equal** to the rate of molecules moving from the aqueous layer to the organic layer **CH₃NH₂**(aq) **⇌ CH₃NH₂**(organic solvent)

Question 41

The value of its **equilibrium constant** is also called the

Answer 41

**partition coefficient**

Question 42

***The partition coefficient is the ratio of methylamine molecules in the organic and aqueous layer once equilibrium has been established***

Answer 42

Question 43

The **partition** **coefficient** (*K_pc*) for a system

Answer 43

in which the solute is in the same physical state in the two solvents can be calculated using the equilibrium expression

Question 44

The partition coefficient (*K_pc)* depends on the

Answer 44

**solubilities** of the **solute** in the two **solvents**

Question 45

The degree of solubility of a solute is determined by how strong the

Answer 45

**intermolecular bonds** between **solute** and **solvent** are

Question 46

The strength of these intermolecular bonds, in turn, depends on the

Answer 46

**polarity** of the **solute** and **solvent molecules**

Question 47

When *K_pc* is \< 1

Answer 47

* the **solute** is more soluble in water than the organic solvent

Question 48

When *K_pc* is \> 1

Answer 48

the **solute** is more soluble in the organic solvent than the water