The Oceans Flashcards

Question

What will happen if Δ_tot*S* = 0 for a reversible process?

Answer 1

The process will reach equilibrium where neither forwards/backwards favoured

Answer 2

If the states of the products have a higher entropy than those of the reactants (e.g. are gaseous not solid) it will be feasible If there are more molecules of product than reaction it will be feasible as this means there are more ways of arranging the molecules + energy quanta

Answer 3

Because the E_a might be very high - not taken into account when calculating entropy changes

Answer 4

**Solar energy** reaches the Earth mainly as **visible + UV radiation** The **Earth** **absorbs** some of this radiation + **radiates** some in the form of **IR** **Water vapour** in the atmosphere absorb **some frequences of IR**, helping keep the **temp**. of the Earth **constant** However, **greenhouse gases** in the **troposphere** absorbs the **remaining freqs**. of IR (the **'IR window'**) This causes the **vibrational energy** of their **bonds** to **increase** + energy is **transfered** to **other molecules by collisions**, increasing their **kinetic energy** and thus **raising temp**. Greenhouse gagses also **re-emit** some of the absorbed IR in **all directions**, contributing to the heating of the Earth

Answer 5

The troposphere

Answer 6

Methan CO₂

Answer 7

Increased concentrations of greenhouse gases

Answer 8

The can dissolve/absorb some CO₂ preventing it building up in the atmosphere and acting as a greenhouse gas

Answer 9

It leads to ocean acidification

Answer 10

CO_2(aq) + H₂O_(l) ⇌ HCO₃^-_(aq) + H⁺_(aq) HCO₃^-_(aq) ⇌ CO₃^-_(aq) + H⁺_(aq)

Answer 11

A substance which donates H⁺ ions when in an aqueous solution It is a proton donor

Answer 12

A substance that accepts H⁺ ions Is a proton acceptor

Answer 13

Used to categorise acids + bases Based on ability to transfer/accept H⁺ ions

Answer 14

The substance that is formed from the dissociation of an acid but can act as a base to reverse the reaction/dissociation HA_(aq) ⇌ H⁺_(aq) + A^-_(aq) HA and A^- are the conjugate acid-base pair

Answer 15

A species that donates protons in the reverse reaction of a dissociation of an acid

Answer 16

An acid which fully dissociates into H⁺ and A^- ions

Answer 17

An acid which doesn't dissociate fully Undergoes an incomplete ionisation to H⁺ and A^- so reaches equilibrium

Answer 18

They are **equal** if the acid is **monoprotic**

Answer 19

It doesn't have any

Answer 20

pH = -log₁₀ [H⁺]

Answer 21

The acidity constant K_a = [A^-][H⁺] / [HA]

Answer 22

The stronger the acid, the greater K_a

Answer 23

Used when the value for K_a very small due to acid being very weak ## Footnote **pK_a = -log K_a**

Answer 24

That the acid is very weak The higher pK_a, the weaker the acid

Answer 25

In **equilibrium concentration [HA]** for a weak acid is the **same as the initial concentration of the acid** The **equilibrium concentration of [A^-]** is **equal** to the **equilibrium concentration of [H⁺]**. A few protons will be provided by water but these are insignificant compared to those donated by the acid

Answer 26

That an acid will donate 1 H⁺ (proton) when it dissociates as it contains 1 (acidic) H

Answer 27

Write an expression for K_a: K_a = [CH₃COO^-][H⁺] / [CH₃COOH] Assumption 1: Few molecules dissociate, so [CH₃COOH] = 0.01moldm^-3 Assumption 2: For every 1mol H⁺ there's 1mol CH₃COO^- so [H⁺] = [CH₃COO^-] So K_a = [H⁺]² / 0.01 = 1.7x10^-5 [H⁺] = √1.7x10^-7 [H⁺] = 4.12x10^-5 pH = log₁₀ (4.12x10^-5) = 3.4

Answer 28

The base fully dissociates so [OH^-] is equal to the initial concentration of the base That any dissociation from water molecules is insignificant so can be ignored

Answer 29

NaOH → Na⁺ + OH^- Strong bases fully dissociate so [OH^-] = 0.01moldm^-3 K_w = [H⁺] [OH^-] [H⁺] = K_w / [OH^-] [H⁺] = 1x10^-14mol²dm^-6 / 0.01moldm^-3 = 1x10^-12moldm^-3 pH = -log (1x10¹²0 = 12.0

Answer 30

The ionic product of water Formed due to the slight dissociation of water H₂O_(l) ⇌ H⁺_(aq) + OH^-_(aq) K_w = [H⁺] [OH^-] = 1x10^-1⁴mol²dm^-6 at 298K *Used instead of K_a in calculations involving strong bases*

Answer 31

A solution which has **little/no change in pH** when a **small amount of acid/alkali added** Made from a weak acid/base and one of its salts

Answer 32

Large amounts of a proton donor - either weak acid or conjugate acid Large amounts of proton acceptor - either weak base of conjugate base Means that any additions of acid/alkali react with these + keeps pH constant

Answer 33

Equilibrium shifts to the left to oppose the change in conc. of H⁺ This is done by the ethanoate ions reacting with the excess H⁺ to produce more ethanoic acid Hence the pH remains constant

Answer 34

Because the acid alone doesn't dissociate to produce enough conjugate acid ions to oppose a change in pH if more H⁺ ions were added

Answer 35

Shampoos In food + drink as 'acidity regulators' In the blood

Answer 36

To protect the body from changes in pH due to the formation of CO₂ + H⁺ in metabolic processes Otherwise these pH changes could affect the action of enzymes which is bad for our health

Answer 37

By mixing equal vols, each original conc halved, so: [CH₃COOH] = 0.10moldm^-3 [CH₃COO^-] = 0.05moldm^-3 K_a = [H⁺] [CH₃COO^-] / [CH₃COOH] [H⁺] = K_a x [CH₃COOH] / [CH₃COO^-] [H⁺] = 1.7x10^-5 x 0.1 / 0.05 = 3.5x10^-5 pH = -log (3.4x10^-5) = 4.5

Answer 38

K_a of the weak acid Conc. of the salt + weak acid Needed to calculate [H⁺] and therefore pH

Answer 39

The solubility product An equilibrium constant for the dissolution of a sparingly soluble salt

Answer 40

AB_(s) ⇌ A⁺_(aq) + B^-_(aq)

Answer 41

K_sp = [A⁺] [B^-]

Answer 42

Because the concentration of a solid cannot be measured

Answer 43

CaI₂ ⇌ Ca²⁺ + 2I^- x ⇌ x + 2x **K_sp = [Ca²⁺] [2I^-]²**

Answer 44

If the product of the conc. of the ions is smaller/equal to value for K_sp the ions will stay in solution and a precipitate will not form

Answer 45

In order to do so, the **concentrations of the ions in solution** needs to be **measured** e.g. with Ca(OH)₂ this can be done via **titration** to determine [OH^-] **Undissolved salt** should be **filtered** before titrating The conc. [Ca²⁺] can then be found once [OH^-] known using the **molar ratio** ([OH^-] will be x2 [Ca²⁺])

Answer 46

K_sp [AgCl] = 2.0x10^-10mol²dm^-6 at 298K [Ag⁺] and [Cl^-] = (5x10^-6) x (5x10^-6) = 2.5x10^-12mol²dm^-6 Answer smaller than K_sp so no precipitate will form