Unit 3 Flashcards
equilibrium reaction
the concentrations of all reactants and products remain constant with time. it is highly dynamic as the reaction never stops. there is no limiting reactants
stoichiometric reaction
proceed to completion therefore the amount of limiting reactant the remains is negligible
equilibrium lies to the right
it favours the products, concentrations of reactants is never 0
equilibrium lies to the left
it favours the reactants
how can you disturb the “system”
temperature, change volume, changing something will stress the system and change the equilibrium
when is equilibrium reached
when rate of forward reaction equals the rate of reverse reaction
the equilibrium expression (K or Kc)
is a thermodynamic quantity
eqn is products to power of coefficients/reactants to power of coefficients
it has no units and concentrations are always in Mol/L
must always be accompanied by a balanced them eqn
reverse K
always needs a new equation! it is written as K^-1
when the reaction is multiplied by a factor
multiply the exponents by the factor.
equilibrium position
K is always the same for a certain temperature but the concentrations might change.
each equilibrium concentration is called the position and there is an infinite number of positions at a certain temperature
when K>1
the system will favour the products
equilibrium lies to the right
K»10^3, reaction will essentially be products
when K<1
the system will favour the reactants
equilibrium lies to the left
K«10^3, reaction will essentially be reactants
this would be a non-reaction but that is a lie!
equilibrium and kinetics
- size of k and time are not directly related
- time to reach equilibrium depends on the rate of reaction which is determined by kinetics
- size of K is determined by thermodynamics factors such as difference in energy between products and reactants
heterogeneous equilibrium
not all reactants are in the same state
- pure solis and liquids have concentrations that don’t change and are constant so they can be ADDED to the constant equilibrium value (remove them from the eqn!)
decrease the temperature of equilibrium
that decreases the energy to be more endothermic, the equilibrium shifts right and makes the reaction more exothermic to make up for the energy lost
increase the temperature of equilibrium
that increases the energy to be more exothermic, the equilibrium shifts left and makes the reaction more endothermic to make up for the energy
increase the pressure
increase the pressure, the system respond by decreases the pressure by shifting to the side with less gas moles therefore shifting (left/right)
decrease the pressure
decrease the pressure, the system respond by increases the pressure by shifting to the side with more gas moles therefore shifting (left/right)
effect of concentration
has no effect on k,
if you increase the concentration, the system respond by decreasing the concentration of that reactant and shifting (left/right)
reaction quotient, Q
calculate the same way as K but it is used if you don’t know the concentrations at equilibrium
Q>K
too many products and shifts left to reach equilibrium
reaction will shift to reach K
Q
too many reactants
shifts right to reach equilibrium
Q=K
at equilibrium
ICE tables
initial
change
equilibrium
table can be made using moles, concentrations, pressures
le chatalier’s principle
if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change
how to change the pressure of equilibrium
- add or remove a gaseous reactant or product (addition of an inert gas has no effect on the concentrations or partial pressures of the reactants or products) so NO CHANGE
- add another inert gas that does not react
- change the volume of the container
when you change the volume of the container
- concentrations of both reactants and products are also changed
- when volume is reduced, pressure increases and system decreases its own pressure
- it decreases the total number of gas moles in the system
by changing the temperature, you affect…
the value of K, and you treat energy as a reactant or product
adding a catalyst
does not change the equilibrium in any way, it just changes the time it takes to get to equilibrium
when a salt dissolves
ions get surrounded by water. its never entirely soluble so an equilibrium occurs
more soluble ionic compounds
the equilibrium lies to the right
insoluble ionic compound
the equilibrium lies to the left (more solid)
when writing ionic compounds
always write the ionic compound (solid) on the left
Arrhenius acid def
anything that has an H+
arrhenius base
donates OH-
bronsted-lowry acid
anything that donates an H+
bronsted-lowry base
accepts an H+
how are acids and bases defined? (water)
defined by their reaction with each other. H2O can’t be an acid or a base as it depends on the reactant that is associated with it
bronzed lowry acid base equilibrium
the products from the reaction can act as acids and bases themselves.
conjugate
results from the transfer of H+
Acid to conjugate….
base
take away an H+ ion from acid to form the conjugate base
base to conjugate…
acid
add an H+ ion from base to form conjugate acid
remember… charges
loosing an H, leaves the thing with a -ve charge
amphiprotic/amphoteric
bronstead lowry term to mean it can act as an acid or a base
ex.
water and HCO3-
hydronium
H3O+
pH
measures the concentration of H3O+ in a solution
usually positive and unitless
change by one is a 10 fold change in H+
pH eqns
pH= -log[H3o+]
[H+]=10^-pH
pOH eqns
1.0x10^-14 = [H+][OH-] at 25 celsius
pOH= -log[OH-]
pH + pOH = 14
significant figures in log
only the stuff after the number (AKA decimals) are significant
strong acids
completely dissociate in water, not in equilibrium
strong acids you should know
HCL, HBr, HI, H2SO4, HNO3
HF has a small radius and too strong intermolecular force to ionize!
weak acid
partially dissociates, it is in equilibrium
K a
acid dissociation constant, a stronger acid will have a larger Ka
more H3O+ because H+ is dissociated
the equilibrium lies more towards the right compared to a weaker acid
ka eqn
= [H3O+][A-]/[HA]
HA is the acid
strong bases
completely ionize
strong bases to know
group 1 and 2 + OH-
Weak bases
partially ionize, in equilibrium
Kb
a stronger base will have a larger Kb
a stronger base will have an equilibrium which lies more towards the right
auto ionization of water
water can react with itself to make hydronium
kw at 25 degrees celsius= 1.0x10^-14
This is an endothermic reaction in the forward direction. increasing the temperature decreases the pH more H30+
kb eqn
[HA][OH-]/[A-]
KaKb relationship
Ka times Kb= [H3O+][OH-]= kw=1.0 x 10^-14
that means that if ka is large then kb is small
in a neutral solution
hydronium concentration equals OH- concentration
in an acidic solution
hydronium concentration is more than OH-
in a basic solution
hydronium concentration is less than OH-
-logkw
pH+pOH
in calculations, if the concentration of HA is much greater (>100) than Ka then
you can make the approximation that the concentration of HA initial is equal to the concentration of HA at equilibrium. must be stated each time you use this
if b initial> 100Kb
assume B initial equals B equilibrium
ionic compounds and acidic/ basic properties
determining whether acid/base is weak acid/natural salt/weak base
hydrolysis (salts)
certain ions react with water to change pH
salts that are made from the conjugate base of a strong acid and the cation of a strong base do NOT react with water (no change)
strong bases that don’t react with water
group 1 and 2 cations come from strong bases
conjugate bases that don’t react with water
BR-, I-, Cl-, NO3-, CLO4-, CLO3- come from strong acids
ions and pH
conjugate base of weak acid will react which adds OH+ and changes the pH (basic)
conjugate acids of a weak base will react with water to change its pH (acidic)
two ions that hydrolyse
a salt can be made from the anion of a weak acid and the cation of a weak acid
this is ammonium acetate and ammonium formate
compare relative strength of their ka and kb
buffer
resist pH after the addition of OH- or protons
buffered solutions contain
a weak acid and its salt (its conjugate) HF and NaF OR a weak base and its salt (its conjugate) NH3 and NH4CL
buffers in word problems
the concentration is on the other side of the equilibrium as well, not 0
buffer mechanism
buffer solutions needs a large quantity of weak acid and conjugate base (or other way) compared to the amount of strong acid or base added
buffer mechanism for hydroxide ions
hydroxide ions will not accumulate as they will react with the weak acid or conjugate base
OH- + HA –> A- + H2O
OH- +BH+ –> B + H20
*stoichiometry arrows
buffer mechanism for protons
protons will react with conjugate base or weak base
H+ + A—> HA
H+ + B –> BH+
pH and buffers
will be determined by the ratio of HA to A-
if the amounts of HA and A- originally present is large then any small changes shouldn’t affect the ratio by much therefore shouldn’t affect the pH by much
how to create a buffer
1) combine a weak acid and its salt (or base ad its salt)
- should be about equal moles
- ex. NH3 and NH4Cl (source of NH4+)
2) taking a weak acid and combing it with a lower amount of stronger base (or weak base and strong acid)
- this is more realistic
- take about half amount of strong acid to weak base
titration
used to determine the amount of acid or base in a solution of unknown concentration
end point
indicator changes colour
equivalence point
when the moles of acid = moles of base if monoprotic
titration curve
a graph of pH as a function of volume of titrant added
partly sigmoidal
equivalence point is the POI of the curve
how can end point show equivalence point
because the tangent of the equivalence point and end point are essentially the same as the tangent is vertical (same x)
drawing a titration curve of strong acid and strong base
strong acid starts at pH 1-2
weak acid is at 3-4 pH
weak base is 9-11 pH
strong base is 11-12
curve is symmetrical around equivalence point
pH rises sharply at equivalence point
equivalence point is 7 pH
weak acid and strong base half equivalence
ka= {h30+] at half equivalence (1/2 moles required for equivalence) and creates a buffer with two curves on graph
weak acid and strong base initial
HA
weak acid and strong base equivalence
A- so basic pH
weak acid and strong base after equivalence
OH- basic pH
titration of weak acid and strong base graph
initial pH is much higher (around 3)
pH equivalence point is greater than 7
curve is not symmetrical around equivalence point
after equivalence the curve flattens out at a high value (pH of strong base)
buffer region:
- will be flat until the equivalence point
half of the equivalence point volume (x six) is the half equivalence point
weak base and strong acid initial
B
weak base and strong acid half equivalence
B + BH+ –> buffer
weak base and strong acid equivalence
BH+ so acidic
weak base and strong acid after equivalence
H+
weak acid and weak acid curve
initial pH is high (3) from weak acid
addition of base makes pH rise gradually
rise around the equivalence point is less sharp
after equivalence point, the curve flattens out at relatively low pH (around 9ish)
hard to find the equivalence point (POI)
le chatelier statement
if I _(increase/decrease)___ the system responds by ___(doing opposite)__ by shifting __(left/right)___
indicator
a weak base that changes colour. they have a Ka
when its colour A, its an acid
Colour b is the dissociated side with the H+ ion and is the base
the Ka/H+ ratio will show the indicator colour at a pH
when to see a indicator colour change
tipping point of colour a and b is ka= [H+]
visible change is noticed when ratio of Ln-/ HLN = 0.1 to 10
colour change also is pKA = +/_ 1 because at half equivalence, Ka is equal to the [H+]???
indicator for strong acid and strong bas
pretty much anything
indicator for weak acid/ strong base
any indicator that is basic
indicator for weak base and strong acid
acid indicator
molar solubility
- is the amount of moles that dissolve
- it is the x value in the ICE table!
for le chatilier
think about what is in K, temp, volume moles and stuff
KSP is super small
so anything that has an initial volume the isn’t zero is insignificant in the table
acid to conjugate base
loose an H+
conjugate base to acid
add an H+
base to conjugate acid
add an H+
conjugate acid to base
loose an H+
weak acids to kno
CH3COOH, HF, HNO2, NH4+
calculating pH of acid/base strong
its just stoichiometry of the concentration as its monoprotic 1-1
Basic anions in water
CH3CO2-, F-, CO3 2-, S 2- (conjugate bases)
acidic cations in water
Al3+, NH4+ (conjugate acid)
salt can be made from anion of weak acid and cation of weak base
just determine the strength of their relative Ka and Kb
whichever is larger
titration of strong acid and strong base
1) initial is just -log of thing being titrated
2) eon with H+ and OH- makes water (stoichiometry) then divide by the full volume after subtracting the LR
3) equivalence point is pH 0
4) same as 2
titration of weak acid and strong base
1) weak acid so its in equilibrium with water, use Ka to get H3O+ concentration
2) half equivalence point is a buffer. first to the titrated thingy with strong base (OH-) in eon, then get concentration of ion, then do that ion concentration in equilibrium which is now a buffer and is Ka= [H3O+]
3) equivalence. first do the weak acid and OH- stoichiometry, then get ion concentration then do it in equilibrium, and get pH
4) just stoichiometry as OH- is in excess (subtract LR then do pH)
titration of a weak base and strong acid
1) equilibrium of weak base, then use Kb
2) half equivalence so stoichiometry, ion concentration, equilibrium to get OH- then pOH then pH as pOH= pKb at half equivalence
3) stoichiometry to get ion, the use equilibrium to get pH
4) stoichiometry as H+ is in excess (- Lr)
titration of weak base and strong acid at half equivalence
pOH= pKb
when doing acidic/basic salt calculations
decide first if anion is going to react, then cation, then do equilibrium with only that iON AND DO THE PH
when asked about rate and equilibrium constant
Equilibrium and rate are measures of two different things and are not related. The ____reaction____ occurs to a __large/small__ extent and so the reaction has a _same as above (large/small)__ equilibrium constant. It is quite a __slow/fast__ reaction under normal conditions of temperature and pressure and so rate of reaction is __same as above (slow/fast)__.
when you change the volume of vessel or increase/decrease temperature (something that changes the whole equilibrium)
all concentrations drop immediately, then proceed to increase/decrease as you would predict for the new equilibrium and then flatten out
when volume is suddenly decreased, it gives a greater concentration of the solution and temporarily darkens the colour. then, the solution lightens up in favour of equilibrium as concentrations level out
when asked about ions and equilibrium le chatelier
as you add water, it dilutes all species, and the reaction will proceed to side with the greater number of moles (as it will be ‘diluted’ to the greatest extent).
remove water will shift to more solid form
auto ionization of water
transfer of proton from one water molecule to another to produce hydroxide ion and hydronium ion
can happen in liquid ammonia as well, it turns to NH4+ and NH2-
water properties
water is a stronger base than the conjugate base of a strong acid but a weaker base than the conjugate base of a weak acid
to check if a reaction is endo or exo based on K value changes over temperature
see if K increases or decrease with increase or decrease of temperature. then think of le chatelier and which way equilibrium will shift when you change it and think of ratio of products to reactants to find if its endo or exo
when the concentration of strong acid is super super low (like 10^-10)
its just pH 7
when H+ is added to a buffered solution
reacts with the weak base
H+ + A- –> HA
or H+ + B –> BH+
when OH- is added to a buffered solution
reacts with the weak acid
OH- + HA –> A- + H20
OH- + BH+ –> B + H20
when there is a titration and indicator changes colour quickly
that means that titration solution (in burette) is very concentrated and no concentration can be determined
h2CO3 and sO3
dissociates!! to water and carbon dioxide!
when asked about the indicator and changing colour pH
look at the Ka of the indicator, the do the ratio of HIn/Ln- the top ratio should be 10 times greater than the bottom so multiply the Ka by 10 and find pH
