Topic 4b- Acids and Bases Flashcards
what is the point of a buffer
to maintain the pH when an acid or base is added
adding an acid ion into a solution will
increase the pH therefore will shift the opposite way to keep pH level
strong acid vs a weak acid
strong acid- fully dissociate
weak acid- partially dissociate
ka formula
prod/reac
can we ignore the x again with the 400 rule
yes
after solving x in ice table, what might you have to do
add the x to the original H3O concentration
what are buffers made of
solutions of a weak conjugate acid-base pair
weak acid and its conjugate base
eg HF (WA) / NaF- (F is CB)
resistant to pH changes, even when strong acid or base is added
for our buffer equations what is the A- and what is the HA
A- is base
HA is acid
with buffer equations , when the ratio is 1:1 and they have the same amount of mols what happens
you can cross out the base/acid and your left with pH= pKa
if your acid is more then the conj base then
if your conj base is greater then the acid then
the pka is greater then pH
your pH will be greater then pKa
when do we need to use before and after tables
with STRONG acids and bases added to a buffer
-NOT TITRATION (these would be ice tables)
-weak ones we can just use the HH equation
if were only given mols and no vol, what can we do when plugging it into HH equation
the mols will cancel out, we dont have volume so we dont need to find M
when we have two different volumes what do we need to do
add them together
what happens when strong acids or bases are added to a buffer
all strong acids or bases are consumed
how to determine pH range
-range of pH where buffer system works
-need to choose an acid with a pKa closr to desired pH
- or - of one pH unit
-the buffering capacity is proportional to the concentration or the conjugate acid/base in a solution
steps of adding a strong A or B to a buffer
- determine the neutralization and how its affects solution
- use HH equation to determine pH
-addition of acid ( dissolve completely) A- + HA= HA
-addition of base: HA + OH = A+ + H2O
when writing on our formulas with a strong acid or base being added, what can we leave out
you can take out the ion (Na) from all formulas. eg Na NaC2H3O2 and NaOH
if base= OH- + CH3COOH= H2O + CH3COO-
if 0.020 of a strong acid is added into a buffer solution and its fully consumed, what will happen to the WA and CB before and after the reaction
before: given mols
After: subtract or add the 0.020 mols depending on the side from the before and OH will be 0
titrations will add either a
base to an acid or an acid to a base
pH meter indicates
determines when a solution has reached equivalence point , when the stoich of A equal base
titration of strong acids with strong base will equal
pH 7
-H+ + OH- = H2O
-the pH slowly goes up at the start
-before and after the the equiv point, the pH increases rapidly
-at equiv point, moles acid= moles base and the solution contains only water and salt from the base and acid
-as more base is added, the pH levels off
-base will start higher and go down with the same steps
when working with a titration question that doesn’t say diluted, what cant we do
is C1V1
titrant means
-what is going in the titration
-the other will have a volume with the M if we asked to find volume of the titrant
titration of weak acid with strong base
- OH- + HA -> A - + H2O
-the CB of WA affects the pH when it is formed :
-initial pH of WA is the start
-before equivalence point is the buffer range of pH increasing
-equivalence point is weak CB (wont be 7.0 as it increases)
-after equiv point the OH in excess and level off
how to find pH from ice tables etc
-log [H3O]
-log [OH-]
if you have pH and need concentration you will 10-e
14-pOH= pH
half equiv point volume formula
equiv point/2
Titration of a weak acid with a strong base (titrant)
weaker acids will have a higher initial pH and the pH change is not as rapid near the equiv point
-greater then 7
titration of a weak base with strong acid
Base will start lower and pH change will not be as rapid when near equiv point
-less than 7
polyprotic acid titrations
-multiple equivalence points
solubility ksp formula
ksp= [prod] [prod]
-dont forget exponents
is ksp the same as solubility
no
-generally expressed as mass of solute dissolved in g/l g/ml M
since solubility will have solids as reactants we can
ignore the products but use its ksp for the formula
-remember the stoich ratio
remember x for
solubility questions with 2:1 ratio
Factors affecting solubility
-common ion affect: if ion is already dissolved then the reaction will shift left
pH: if substance has a basic anion, it will be more soluble in acidic solution, vis versa with basic, shift to left
complex ions: metals can act as lewis acids and form complex ions with lewis bases and increases solubility
-amphoterism: can dissolve in either A or B
will a precipitate form?
if q=ksp, then equilibrium
if Q < ksp, towards products , no
if Q > ksp, towards reactants,
strong AB titrations will not have
a buffer region (before, after, midpoint)
no pka
for titrations with weak acid or weak base, where do we do HH equations
before or after equiv point
when does the pH equal the pka in titrations
at midpoint
at equiv point for buffer titration the equation will be
opposite the the midpoint
calculating equiv point for WA or WB has to to do with the ____ WA or WB
-initial concentration of the base or acid at midpoint (in the before-after)
-initial volume of at the start of WA or WB
-mols of initial will equal the mols of other at equiv point
-calculate the new volume with that initial and concentration that was added
-add inital volume and volume needed to reach
when dealing with an after titration question, what’s the difference between having a Wa titrated by strong base, and WB titrated by SA
WA titrated by strong base- need to do before and after and kb
WB with SA- just find new mol and divide by concentration
steps from ksp to solubility vis versa
ksp- [prod] [prod]
-find concentration of indiv ion which is the [prod] in formula
-ratio to find concentration of compound
-if we need mass, multiply by molar mass
what is solubility
The maximum amount of a solute that can dissolve in a given amount of solvent at equilibrium.
for solubility (precipitate), When two solutions of equal volume are mixed, the concentrations of all species ______ because
are halved
-volume doubles while the number of moles remains the same.
Strong acid titrated by strong base
vs
strong base titrated by strong acid
-at the start of titration
-equation
-pH
SA:
-H only present
-HCl->H + Cl-
-[H3O]=[HA] pH=-log[H3O]
SB:
-OH only present
-NaOH-> Na + OH
-pOH= - log [OH], 14-pOH
Strong acid titrated by strong base
vs
strong base titrated by strong acid
-before equiv point
-equation
-pH
SA:
-some OH added to neutralize HA (HCl)
-HA + OH= A- + H20, Ha excess
-find mols of H3O from unreacted HA, use vol to get concentration, pH
SB:
-some H+ added to neutralize OH
-OH + H= H3O + H2O , OH excess
-find mols of unreacted OH, use vol to get concentration, pOH, PH
Strong acid titrated by strong base
vs
strong base titrated by strong acid
-equivalence point
-equation
-pH
SA:
-all mols of H+ used up
-moles acid= moles base, same equation as before equiv point
-7.0 pH neutral
SB:
all mols of OH used up
-moles acid= moles base, same formula as equiv point
-7.0 pH neutral
Strong acid titrated by strong base
vs
strong base titrated by strong acid
-after equiv point
-equation
-pH
SA:
-No HA remains, OH in excess
-no reaction, OH mol increasing
-find mols of OH added equiv point after equiv
SB:
-no OH remains, H in excess
-no reaction, mols of H increasing
-find mols of H added after equiv point (add H+ volume and divide by new mols)
weak acid titrated by strong base
vs
weak base titrated by strong acid
-start of titration
-equation
-pH
WA:
-only weak acid present
-HA + H2O = A + H3O
-ice table, initial conc not including the volume, calculate pH
WB:
-only weak base present
-A + H2O = HA + OH
-ice table, find kb, pOH, pH
weak acid titrated by strong base
vs
weak base titrated by strong acid
-before or after mid point
-equation
-pH
WA:
-some OH added to partially neutralize H
-H + OH = H2O + A, HA in excess
-find mols HA and OH before, subtract or add from OH to HA and A
-divide both mols by total volume for new concentration
-HH equation
WB:
-some H added to partially neutralize H
-A- + H3O= H2O + HA, A in excess
-find mols HA and A before, subtract or add from strong acid to CA and WB
-divide both mols by total volume for new concentration
-find ka
-HH equation
weak acid titrated by strong base
vs
weak base titrated by strong acid
-midpoint
-equation
-pH
1/2 volume of of titrant needed to reach equiv point
WA
-exactly half of titrant added
-same equation as B or A mid, concentration equal
-pH=pka
WB:
-exactly half of titrant added
-same equation as B or A mid, concentration equal
-pH=pka
weak acid titrated by strong base
vs
weak base titrated by strong acid
-equivalence point
-equation
-pH
WA:
-mols A= mol B
-reaction flipped from mid now base is first
-take initial mols of WA from beginning (M and vol in question) and divide by total volume to get flipped base concentration, find kb, pOh, pH
WB:
mols B= mol A
-reaction flipped from mid now acid is first
-take initial mols of WB from beginning (M and vol in question) and divide by total volume to get flipped acid concentration, find ka, pH
weak acid titrated by strong base
vs
weak base titrated by strong acid
-after equiv point
-equation
-pH
WA:
-no Ha remains, OH in excess
-OH mols increasing, no reaction
-find excess mols (find new mols and subtract the acid original from M and vol in equaition), new conc/ total volume, pOH, pH
-WB
-no base remains, H in excess
-H mols increase, no reaction
-find new concentration from taking HCL original concentration x new vol to get mols and divide total volume + added vol, pH
only time we use inital M of WA or WB that doesnt include the volume
-at the start as its not being titrated yet
to get mass from HH
ph given= pka we calc + log [solve for this]
10e
new conc= intial con/inital conc
