Equilibrium and Aqueous

Level 3 Chemistry Study!

strong electrolytes are _______ compounds _______ _______ in water to form _______

completely dissolve, ions

weak electrolytes are substances that _______ _______ in water to form _______ _______

partially ionise, some ions

acidic/basic salts fully _______ first in water, then the _______ ion will _______

dissociate, weak, hydrolyse

_______ of salts is made up of two reactions

hydrolysis

non-electrolytes will fully _______ in water, but not into _______

dissolve, ions

species that _______ dissociate/ionise will have a very high concentration of _______, and a very _______ concentration of reactants

completely, products, low

species that _______ ionise produce a _______ of reactants and _______, assuming more _______ are left using the 5% estimate

partially, mixture, products, reactants

electrical conductivity requires _______ _______ _______, or _______

mobile charged particles, ions

high [_______] = high conductivity

ions

it is important to look at _______ ratios when comparing [ions] for relative electrical _______

mole, conductivity

the _______ of a solution depends on the [H₃O⁺]

pH

if [H₃O⁺] > [OH^-], then the solution is _______ and pH __ 7

acidic, <

if [H₃O⁺] = [OH^-], then the solution is _______ and pH __ 7

neutral, =

if [H₃O⁺] < [OH^-], then the solution is _______ and pH __ 7

basic, >

the higher the [OH^-], the lower the [_____]

H₃O⁺

K_a is the _______ _______ _______

acid dissociation constant

pK_a = ____K_a

-log

K_a = ____pK_a

10^-

a low K_a means a _______ _______ _______

weaker weak acid

a higher K_a means a _______ _______ _______

stronger weak acid

low pK_a = _______ K_a

high

stronger weak acid = _______ _______

more dissociated

a stronger weak acid which is more dissociated = _______ [H₃O⁺]

higher

higher [H₃O⁺] = _______ pH

lower

higher pK_a = _______ K_a = _______ weak acid = _______ dissociated = _______ [H₃O⁺] = _______ pH

lower, weaker, less, lower, higher

for a titration of a weak acid with a strong base:

1. only _______ _______

2. __ _______ _______ - _______ of the weak acid has been reacted to form the _______

3. _______ _______ - _______ of the weak acid has been reacted, and no _______ are left

4. _______ - extra _______ _______ has been added, but there is no _______ _______ left to react with it so it _______ to form OH^-

weak acid, ½ equivalence point, half, salt, equivalence point, all, reactants, excess, strong base, weak acid, dissociates

at the ½ equivalence point and equivalence point, [H₃O⁺] is determined by the _______ of _______ acid to base

ratio, conjugate

to calculate [H₃O⁺] for a weak acid (HA): [H₃O⁺] = √(__ × [__])

Ka, HA

to calculate [H₃O⁺] for a weak base (A-): [H₃O⁺] = √(__ × __ / [__])

Ka, Kw, A-

K_w = __×10^-__

1, 14

the “henderson hassleback” equation calculates __ for a/an _______

pH, buffer

to calculate pH of a buffer, use pH = ___ + ___([__] / [__])

pKa, log, A-, HA

HA and A^- are known as a/an _______ pair

conjugate

pKa = -log__

Ka

for strong acids, [H₃O⁺] = _______

concentration

for strong bases, [OH^-] = _______, then [H₃O⁺] = ___ / [___]

concentration, Kw, OH-

when calculating [H₃O⁺] of strong acids and bases, be aware of dissociation mole _______!

ratios

a/an _______ is a solution that contains a/an _______ of _______ acid / conjugate _______

buffer, mixture, weak, base

buffers prevent changes in ___ when _______ quantities of _______ acid or alkali are added, by reacting with added ___ or ___

pH, small, strong, H3O+, OH-

do you need the volume of the solution to calculate the pH of a buffer?

no

at _______ point, __(HA) = __(A^-), so C₁V₁ = C₂V₂

equivalence, n, n

to make pH calculations at equivalence point, initial and total _______ can be read from the _______

volumes, graph

to calculate pH when _______ strong acid / base is added:

1. calculate number of _______ of excess strong acid / base, by subtracting n(added strong _______ / base) - n(_______ base / _______)

2. calculate [excess strong acid / base] using __ = __ / __ with _______ volume

3. calculate [H₃O⁺], using _______ acid / base equations

4. use pH = _____[H₃O⁺] to calculate __

excess, moles, acid, weak, acid, C, n, V, total, strong, -log, pH

½ equivalence point is in the mid-point of the _______ zone, so can use this equation to calculate pH

buffer

when using the _______ equation to calculate pH at ½ equivalence point, [A^-] : [HA] = __, so pH = ___

buffer, 1, pKa

to calculate pH somewhere in the buffer zone (when adding strong base to weak acid):

1. use n = ___ to calculate _______ moles of weak _______

2. use the same equation to calculate initial moles of strong _______ reactant, which is equal to the moles of _______ base _______

3. subtract n(_______ base) from n(_______ acid to find n(_______ _______)_final

4. use the mole _______ of [A^-] : [HA] to complete the _______ pH calculation

CV, initial, acid, base, conjugate, product, conjugate, weak, weak acid, ratio, buffer

an _______ buffer is made from equimolar weak acid + conjugate base

acidic

an _______ buffer is made from equimolar weak base + conjugate acid

alkaline

_______ means equal number of moles

equimolar

when a small amount of H₃O⁺ is added to an acidic buffer, it will react with _______ _______ ions to form _______ _______ molecules. the added H₃O⁺ is _______, and the __ of the solution changes very little

conjugate base, weak acid, consumed, pH

when a small amount of OH^- is added to an acidic buffer, it will react with _______ _______ ions to form _______ _______ molecules. the added OH^- is _______, and the __ of the solution changes very little

weak acid, conjugate base, consumed, pH

the _______ of a buffer depends on the _______ _______ of the weak _______ and _______ base

effectiveness, relative concentrations, acid, conjugate

if [weak acid] > [conjugate base], the buffer will be more effective in consuming a _______ _______

strong base

if [weak acid] < [conjugate base], the buffer will be more effective in consuming a _______ _______

strong acid

if [weak acid] > [conjugate base] or pH = pKa, the buffer will be _______ effective in consuming both strong acids and bases

equally

if pH > pKa, the buffer will be more effective in consuming a _______ _______

strong acid

if pH < pKa, the buffer will be more effective in consuming a _______ _______

strong base

an _______ changes colour within pH ±__ of its ___

indicator, 1, pKa

the pH of _______ point is measured from half-way up the vertical straight on a pH profile

equivalence

an indicator should be chosen with its ___ within ±_ of the ___ of equivalence point for that reaction

pKa, 1, pH

most _______ solids dissolve in water to a/an _______

ionic, extent

a _______ _______ is at the point where the [_______] remains _______ and no more solid will _______

saturated solution, ions, constant, dissolve

a saturated solution is depicted with a/an _______ equation

equilibrium

_______, __, is the concentration of a _______ solution, or the _______ dissolved solid per L

solubility, s, saturated, maximum

to calculate K_s from solubility for an AB type salt in a 1:1 ratio, use K_s = __s^__

1, 2

to calculate K_s from solubility for an A₂B type salt in a 2:1 ratio, use K_s = __s^__

4, 3

to calculate between concentration from gL^-1 to molL^-1, _______ the value by its molar mass, gmol^-1

divide

to calculate between concentration from molL^-1 to gL^-1, _______ the value by its molar mass, gmol^-1

multiply

to calculate solubility knowing K_s of a salt in a __:1 ratio, use s = √K_s

1

to calculate solubility knowing K_s of a salt in a __:1 ratio, use s = ³√(K_s / 4)

2

the concentration of each ion in a AB salt, with solubility s, is:

• [A⁺] = __

• [B^-] = __

s, s

the concentration of each ion in a A₂B salt, with solubility s, is:

• [A⁺] = __

• [B^-] = __

note that K_s = [A⁺]^__ × [B^-]

2s, s, 2

IP stands for _______ _______

ionic product

to calculate IP, [new] = [original] × V_{_____} / V_{_____} for each ion, then substitute into the ___ expression

note that if the volume doubles, then the concentration _______

original, total, Ks, halves

if IP < K_s, does a precipitate form?

no

if IP = K_s, does a precipitate form?

no

if IP > K_s, does a precipitate form?

yes

if IP = K_s, the solution is _______

saturated

if there is already a _______ _______ in solution, _______ (more/less) solid will be able to dissolve

common ion, less

to calculate s when a common ion is present, substitute Ks and [common ion] into equation, and the _______ of the other ion = _______ of solid

concentration, solubility

Le Chatelier’s Principle states that if two reactions both occur…

1. the _______ of a solid, according to its _______

2. the _______ of that solid/ions with another chemical (for example, when in acidic/basic conditions)

… then the _______ of the solid will _______

dissolving, solubility, reaction, solubility, increase

Le Chatelier’s Principle states that if a system at _______ is changed, the system will _______ to _______ that change

equilibrium, adjust, minimise

_______ is a further reaction with H₂O / H⁺

hydrolysis

if an excess of OH^-, NH₃ or CN^- is added, then a _______ _______ may form with some metals (given on resource booklet)

complex ion

if a complex ion is formed, it reduces [ions], which means the _______ will _______ to _______ this change by _______ the forwards/reverse reaction which will result in more/less (original solid) _______, leading to a change in __

equilibrium, adjust, minimise, favouring, dissolving, s