mcat chemistry chapter 10 acids and bases kaplan
Terms
undefined, object
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- litmus paper turns red in ?
- acid
- litmus paper turns blue in?
- bases
- bases turn litmus paper?
- blue
-
arrhenius def of acid and base
acid is?
base is? -
acid prod H+
base prod OH- - what is the shortcoming of arrhenius def?
- only desc a and b in aq media, not in non aqueous
-
bronst lowry a and b
a is
b is? -
acid donate h+
bases accepts H+ -
t or f
NH3 and CL- are bronst lowry bases but not arrhenius bases - t
-
t or f
a bronsted lowry defin of an a or a b is limited to aq solutions - f
- the ______________ defin of an a or a b is limited to aq solut
- arhhenius
- bronst lowry acid and b occur in pairs called?
- conjugate a b pairs
-
bronst lowry
2 members of a conj pair are related by a transf of? - a proton
- H3O+ is the conjugate acid of the base?
- H2O
- NO2 - is the conjugate b of ?
- HNO2
-
lewis def of a and b
acid
base? -
acid is electr pair accept
base is electr pair donator. - AlCl3 is a _______________ acid but not a ________________ acid
- lewis but not a bronst lowry
- AlCl3 can accept a?
- electron pair
- name of an acid is related to its?
- parent anion
- anion ending in -ide will become ______________ acid
- hydro ic
- flouride bec acid?
- hydrofluoric acid
- bromide becomes ________________ acid
- hydrobromic acid
- HF is?
- hydrofluoric acid
- HBr is ?
- hydrobromic acid
- hydrofluoric acid is?
- HF
- MnO4- is what anion?
- permanganate, even though there are no manganate or manganite ions
- acids from oxyanions are called?
- oxyacids
- anion ends in ite it will be __________ acid
- ous acid
- anion ends in ate, it will be ____________ acid
- ic acid
- CLO-
- hypochlorite
- ClO2-
- chlorite
- ClO3-
- chlorate
- ClO4-
- perchlorate
- NO2-
- nitrite
- NO3-
- nitrate
- HClO
- hypochlorous acid
- hypochlorite
- CLO-
- chlorate
- CLO3-
- perchlorate
- ClO4-
- nitrite
- NO2-
- nitrate
- NO3-
- HClO2
- chorous acid
- chlorous acid
- HClO2
- HClO3
- chloric acid
- perchloric acid
- HClO4
- HNO2
- nitrous acid
- HNO3
- nitric acid
- nitric acid
- HNO3
- ph=
- -log[H+]=log(1/[H+])
- pOH is=
- -log[OH-]= log (1/[OH-])
- in any aq solut the H2O does what?
- dissociates slightly
- Kw is the?
- water dissociation constant
- Kw= [H+][OH-]=
- 10-14
- Kw=
- [H+][OH-]=10-14
- ph + pOH=?
- 14
-
math reminder
log(xy)=? - logx + logy
- pure water _____________ is equal to ________________
- H+ is equal to OH-
- pH below 7 is
- acidic
- pH above 7 indicates an excess of?
- OH- ions
-
how can you estimate a log
n x 10-m = -
log(n x 10-m)= -m + logn
the negative log is
m-logn -
m-logn
since n is a number betw 1 and 10 its log will be a fractoin betw ? - 0 and 1
-
m-logn
since n will be a number betw 0 and 1.
m-logn will be between m-1 and ? - m-0
-
m-logn
the larger n is the larger ? - the fraction logn will be the answer will be closer to m-1
-
If Ka = 1 .8 x 10-5, then pKa =
estimate please -
5 - log 1.8. Since 1 .8 is small,
its log will be small, and the answer will be closer to 5 than
to 4. (The actual answer is 4.74.) - strong acids and bases are those that?
- completely dissociate into their component ions in aq solution
- when NaOH dissoc in water, why can you normally ignore the contribution of OH- from H2O?
- as long as the amount of OH- or H+ is greater than 10-7 then you can ignore contribution of water.
- when cant you ignore the contrib of water to acid and base calculations?
- when it is a very weak acid or base whose H+ or OH- concentr is close to 10-7
-
1 x 10-8 M HCL solution
Kw=(x+ 1 xlO-8)(x)=1.O x 1O-14 molecules).
what is X? -
where x=[H+]=[OH-] (bothfrom
the dissociation of water -
1 x 10-8 M HCL solution.
Kw=(x+ 1 x lO-8)(x)=1.O x 1O-14 molecules).
when do use this kind of equation. ? -
when you have to calcul the H+ concentr with a weak acid.
you could also use the equation for a weak base - name some strong acids
-
HClO4- perchloric acid
HNO3 nitric acid
H2SO4 sulfuric acid
HCl hydrochloric acid - sulfuric acid is?
- H2SO4
- HCl is?
- hydrochloric acid
- HNO3 is?
- nitric acid
- name some strong bases
-
NaOH sodium hydroxide
KOH potassium hydroxide
other soluble hydroxides of gr IA and IIA metals - weak acids and bases only partly?
- dissociate in aq sol
- Ka measures?
- the degree to which an a dissoc
- the degree to which an an a dissoc is measur by?
- the acid dissoc const Ka
-
t or f
Ka=[H3O][A-]/[HA] - t
- Ka=
- =[H3O][A-]/[HA]
-
Note that as a weak acid or
base, the effect on pH will
always be _______________ of a
strong acid or base of the same
concentration. - less than that
- weaker the acid, the smaller the?
- Ka
- the weaker the acid, the _____________ the Ka
- smaller
- Kb is the
- base dissociation constant
- the weaker the base the ____________ the Kb
- smaller
- a conjugate base is formed when?
- an a loses a proton
-
HCO3-/CO3-
which is the conjugate a and b? -
conjugate a HCO3-
conjugate b CO3- -
To find the Ka of the conjugate acid HC03, the reaction with __________
must be considered. - water
-
HC03- (aq) + H20 (I) ---> H3O+ (aq) + C03 2- (aq)
this reaction is used to calculate the ? - Ka of HCO3-
-
Ka X Kb=Kw=l x10-14
this is for an acid and its? - conjugate base
- what can we calculate for an acid and its conjugate base?
- Ka X Kb=Kw=l x10-14
- Ka and Kb are _____________ related
- inversely
- Ka and _______ are inversely related
- Kb
- if Ka is large then Kb?
- will be small
- if the conjugate a is strong then the conjugate base?
- will be weak
- to calculate the concentration of H+ in a 2.0 M aq sol of acetic acid CH3COOH (Ka=1.8 X 10-5) what should you do first?
- write the equilibrium reaction
-
CH3COOH (Ka=1.8 X 10-5)
write the equilibrium react - CH3COOH (aq)---> H+ (aq) + CH3COO- (aq)
- to calculate the concentration of H+ in a 2.0 M aq sol of acetic acid CH3COOH (Ka=1.8 X 10-5) what do you do after writing the equil react?
- write the express for the acid dissoc const
-
to calculate the concentration of H+ in a 2.0 M aq sol of acetic acid CH3COOH (Ka=1.8 X 10-5)
write the acid dissoc const? -
Ka=[H+][CH3COO-]
------------ = 1.8 X 10-5
[CH3COOH] -
to calculate the concentration of H+ in a 2.0 M aq sol of acetic acid CH3COOH (Ka=1.8 X 10-5)
what is the concentr of acetic acid at equil equal to? - its initial concentr 2.0 M- X, the amount dissociated
-
to calculate the concentration of H+ in a 2.0 M aq sol of acetic acid CH3COOH (Ka=1.8 X 10-5)
[H+] is=? - =[CH3COO-]=x
-
to calculate the concentration of H+ in a 2.0 M aq sol of acetic acid CH3COOH (Ka=1.8 X 10-5)
How can you rewrite the Ka expression? - Ka= [X][X]/[2.0-X]=1.8 X 10-5
-
to calculate the concentration of H+ in a 2.0 M aq sol of acetic acid CH3COOH (Ka=1.8 X 10-5)
what can you approximate? - that 2.0-x is approx 2.0 bec acetic acid is a weak acid
-
Ka= [X][X]/[2.0]=1.8 X 10-5
what did you approximate? - that 2.0-x is approx = to 2.0
-
Ka= [X][X]/[2.0-X]=1.8 X 10-5
if when you solve for X it is close the the original concentr of acetic acid (2.0 M) what do you need to do? -
you cant approx that 2.0-x is =2.0
instead you have to use the quadratic equation - a neutrialization react is when?
- an a and a base reat w each other, forming a salt and usually water
- equation for neutralization
- HA + BOH ---> BA + H2O
-
HA + BOH ---> BA + H2O
this is the equatrion for? - neutralization
-
t or f
neutralization reactions usually go to completion - t
- salt ions reacting with water to give back the acid or base is ?
- hydrolysis
- hydrolysis (in terms of a and b) is?
- reverse react of neutralization, when salt ions react w water to give back the a or b
- what combinations of a and b are possible?
-
1. str a and str b
2. str a and weak b
3. weak a and str b
4. weak a and weak b -
this is an ex of?
HClO + NaOH ---> NaClO + H2O - weak acid w a str base
- prod of a react betw equal concentr of a str acid and str base are?
- salt and water, a and b neutr each oth so ph is 7
- prod of a react betw str a and weak b are?
- SALT, but usually no water is formed bec weak b are usually not hydroxides
- weak bases are usually not?
- hydroxides
-
str a w a weak b
the cation of the salt formed will do what? - will react w the water solvent, reforming the weak b
- what are the react for HCL reacting with NH3?
-
HCl (aq) + NH3 (aq) --> NH4+ (aq) + Cl-(aq) Reaction I
NH4+ (aq) + H20 (aq) --> NH3 (aq) + H3O+(aq) Reaction II - str acid with a weak base the resulting solut will be?
- acidic
- weak a with a str b the resulting solut is?
- basic.
- weak a with a str b why is the result solut basic?
- bec of hydrolysis of the salt to reform the aci and format of OH- from hydrolyzed water molecules.
- ph of a weak a and a weak b depends on?
- relative strenght of the reactants
-
the acid HClO has a Ka = 3.2 X 10-8,
and the base NH3 has a Kb = 1.8 x 1O-5.
what will the resulting solut be? acidic, basic, or neutral? -
an aqueous solution of HCIO
and NH3 is basic since Ka for HCIO is less than Kb for NH3 - an a equiv is equal to ?
- one mole of H+
- a b equiv is equal to?
- one mole of OH- ions
-
H2SO4 is
a. monoprotic
b. diprotic
c. triprotic - b. diprotic
-
H2S04(aq) --> H+(aq) + HS04(aq)
HS04(aq) --> H+(aq) + S042-(aq)
what is this reaction showing? - that H2SO4 is diprotic
- what is the equation for the dissoc of H2SO4?
-
H2S04(aq) --> H+(aq) + HS04(aq)
HS04(aq) --> H+(aq) + S042-(aq) - one mole of H2SO4 can prod ______ acid equiv
- 2
- 2M H3PO4 would have be _________ N
- 6
- how do you calcul equival weight?
- divide the gram molecular weight by how many moles of H+ it liberates
- what is equival weight of H2SO4?
-
98 g/mol is molec weight
since each mole liberates 2 acid equiv, the gram equiv of H2SO4 is 98/2 or 49 g. - name 3 polyvalent acids?
-
H2SO4
H3PO4
H2CO3 - amphoteric means?
- can act as an acid or a base
- a substance that can act as an a or a b is?
- amphoteric
-
bronsted lowry
an amphoteric species can? - gain or lose a proton
-
t or f
water is amphoteric - t
- the partially dissoc conjug base of a polyprotic acid is?
- usually amphoteric
- the hydroxides of Al, Zn, Pb, and Cr are?
- amphoteric
- the hydroxides of which metals are usually amphoteric?
- Al, Zn, Pb, and Cr
-
t or f
species that can act as either oxidiz or reducing agents are considered to be? - amphoteric
-
t or f
spec that can act as red or oxidiz agents are consid to be amphoteric - t
- oxid or reduc agents act amphoteric when they ?
- accept or donate electr pairs, acting like Lewis a or bases
- titration is used to ?
- determ the molarit of an a or a b
-
in titration
you react a ______ volume of a solution of unknown concentrat with a _______ volume of a sol of known concentrat -
known vol
known vol o - equival point is when the?
- number of acid equival equals the number of base equival
-
t or f
the equival point is always at ph of 7 -
f
only for str a with a str base - when titrating polyprotic a or b there are several?
- equivalence points
- you estimate the equival point in titrat by what 2 methods?
-
1. graphical method- use ph meter
2. watching for a color ch - indicators are?
- weak organic a or bases that have differ colors in their undissoc or dissoc states
- why dont indicators change the equival point
- bec they are in low concetr
- point at which the indicat ch color is called the?
- end point
- the _______ of the indicator and the __________ of the a and b should be close.
-
end point
equivalence point - what formula do yo use to calculate the volume added to reach the endpoint?
- NaVa=NbVb
-
accord to this graph, has the best indicator been chosen? - no, bec the end point has a pretty different ph than the equival point of the titration
-
in the early part of the curve, the __________ species - acidic species
-
in the early part of the curve, the acidic spec dominates so small amounts of base will do what to the ph? - it will only change the ph a little
-
the addit of base most ch the concentr of H+ and OH- where on the graph? - near the equivalence point. this is where the ph the most
-
this is a titration of a ____________ with a strong base - weak acid
-
t or f
the titration of any monoprotic a with a str b will give a similar curve to this - t
-
the equival point is in the _______ range - basic
- a buffer consists of a weak acid or a weak base and?
- its salt
- give 2 examples of buffers?
-
examples of buffers are: a solution of acetic acid (CH3COOH) and its salt,
sodium acetate (CH3COO-Na+); and a solution of ammonia (NH3) and its
salt, ammonium chloride (NH4+Cl-). - why are buffers useful?
-
Buffer solutions have the useful
property of resisting changes in pH when small amounts of acid or base
are added. -
Consider a buffer solution of acetic acid and sodium acetate:
CH3COOH <---> H+ + CH3COO-
what happens when NaOH is added to the buffer? -
When a small amount of NaOH is added to the buffer, the OH- ions
from the NaOH react with the H+ ions present in the solution; subse-
quently, more acetic acid dissociates (equilibrium shifts to the right),
restoring the [H+]. Thus, an increase in [OH-] does not appreciably
change pH. -
Consider a buffer solution of acetic acid and sodium acetate:
what happens when a small amount of HCl is added? -
H+ ions from the HCl react with the acetate ions to form acetic acid. Thus
[H+] is kept relatively constant and the pH of the solution is relatively
unchanged. - the henderson-Hasselbalch equation is used for?
-
The Henderson-Hasselbaich equation is used to estimate the pH ot
a solution in the buffer region where the concentrations of the species
and its conjugate are present in approximately equal concentrations. - Henderson- Hasselbach equation is?
- pH = PKa + log [conjugate base]/[weak acid]
-
henderson-hasselbach
pH = PKa + log [conjugate base]/[weak acid]
when the concentr of conj base equals the conc of the weak acid then? - ph=pKa, bec log1=0
-
henderson-hasselbach
pH = PKa + log [conjugate base]/[weak acid]
when will the concentr of conj base equals the conc of the weak acid ? - in a titration, half-way to the equivalent point
- you can use the henderson-hasselbach equation to make a buffer at any?
- pH. by carefully choosing a weak acid and its salt
-
this is titration of a __________ base with an acid - polyprotic base
-
This the the titration of
Na2CO3 with HCI in which the polyprotic acid is the ultimate prod-
uct. - H2C03
-
In region I, little acid has been added and the predominant species is - C032-
-
In region Il, more acid has been added and the predominant
species are ______________, in relatively equal concentrations. The flat
part of the curve is the ______________, c -
C032- and HC03,
first buffer region -
Region Ill contains the _______________ , at which all of the C032 is
titrated to HC03. As the curve illustrates, a rapid change in pH occurs at
the ____________ - equivalence point
-
Region Ill contains the equivalence point, at which all of the ? -
C032 is
titrated to HC03. -
In region IV, the acid has neutralized approximately half of the HC03,
and now _____________are in roughly equal concentrations. - H2C03 and HC03
-
In region IV the flat
region is the? - second buffer region of the titration curve.
-
In region V. the
equivalence point for the entire titration is reached, as all of the HCO3-,
is converted to?. - H2C03
-
In region V. the
___________________for the entire titration is reached, as all of the HCO3-,
is converted to H2C03. Again, a rapid change in pH is observed near the
equ - equivalence point
-
Blood pH is maintained in a relatively small range (slightly
above 7) by a -
bicarbonate buffer
system. This homeostasis can be upset, leading to a condition known as acidosis. -
in a body you have the bicarbonate buffer
system. This homeostasis can be upset, leading to a condition known as ?. - acidosis
-
What volume of a 3 M solution of NaOH is required to titrate 0.05 L
of a 4 M solution of HCl to the equivalence point?
what equat should you use? - NaVa=NbVb
-
At equilibrium, a certain acid, HA, in solution yields 0.94 M [HA] and
0.060 M [A-].
Calculate Ka. -
HA ---> H+ + A-
The molar ratio of A- to H+ is 1:1, so [H+] must also be
0.060 M at equilibrium. It follows, then, that:
Ka [A-][H+]/[HA] = (0.060)(0.060)/(0.94) = 3.8 X 10-3 -
13. Which of the following combinations would produce a buffer solution
of pH = 4?
(Ka HN02=4.5 X 10-4)
A. 0.30 M HNO2, 0.22 M NaNO2
8. 0.22 M HNO2, 0.30 M NaNO2
C. 0.11 M HNO2, 0.50 M NaNO2
D. 0.50 M HNO2, 0.11 M NaN -
pH = PKa + log [A-/[HA]
4 = 3.35 + log [A-]/[HA]
0.65 = log [A-]/[HA]
[A-]/[HA] = 4.5
Only Choice C fulfills this criterion as 0.50/0.11 = 4.5.