408984
Description
Mind Map by sadieburgess0, updated more than 1 year ago
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Created by sadieburgess0
almost 11 years ago
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Acids and Bases
- Calculating the pH of
strong acids
- A mono-protic acid donates one mole of
protons per mole of acid eg: HCl--> H+ + Cl-
- a diprotic acid donates 2 moles of proton per
mole of acid eg: H2SO4 --> 2H+ + SO4(2-)
- a diprotic acid donates 2 moles of proton per
mole of acid eg: H2SO4 --> 2H+ + SO4(2-)
- Strong acids fully
dissociate and have
more H+ ions
- to calculate the pH we use
: pH-log[H+] and
[H+]=10(-pH)
- to calculate the pH we use
: pH-log[H+] and
[H+]=10(-pH)
- A mono-protic acid donates one mole of
protons per mole of acid eg: HCl--> H+ + Cl-
- Bronsted-Lowry
theory
- an acid is a proton donor
where as a base is a proton
acceptor
- some substances are Amphoteric
eg H2O, they can act as an acid or a
base depending on the reaction
- some substances are Amphoteric
eg H2O, they can act as an acid or a
base depending on the reaction
- an acid-base reaction is the direct
transfer of a proton from an acid to a
base
- if calculating pH to
concentration for diprotic
DIVIDE by 2, if calculating
concentration to pH TIMES
by 2
- an acid is a proton donor
where as a base is a proton
acceptor
- Kw (=10 to the -14)
- the ionic
product of
water: Kw-[H+][OH-]
- has the units mol-2dm6
- Kw increases as temperature
increases because its an
endothermic process
- this causes the pH to decrease
because an increase in temperature
causes equilibrium to shift to the
right, increasing the yield of H+ and
OH- therefore decreasing the pH
(when subbed into equation)
- this causes the pH to decrease
because an increase in temperature
causes equilibrium to shift to the
right, increasing the yield of H+ and
OH- therefore decreasing the pH
(when subbed into equation)
- has the units mol-2dm6
- the ionic
product of
water: Kw-[H+][OH-]
- Calculating the pH of strong bases
- mono basic- accepts one mole
of H+ per mole of base eg:
NaOH
- dibasic- accepts 2
moles of H+ per mole
of base eg: Ba(OH)2
- dibasic- accepts 2
moles of H+ per mole
of base eg: Ba(OH)2
- we need to find the OH- ion
then use Kw/OH- to find [H+]
- For dilutions. find how much the volume
has changed by and apply this to the
concentration of OH- first.
- for a reaction between a strong acid
and a strong base: calculate moles of
each, find which is in excess and by how
much, calculate conc of excess and find
pH accordingly
- For dilutions. find how much the volume
has changed by and apply this to the
concentration of OH- first.
- mono basic- accepts one mole
of H+ per mole of base eg:
NaOH
- Weak Acids
- partially dissociate
- in order to determine the pH we need to
find H+ we use Ka= [H+][X-]/[HX]
- therefore to calculate the [H+] =square
root of {Ka[HX]}
- therefore to calculate the [H+] =square
root of {Ka[HX]}
- weak acid and strong base
- for every mole of OH- added, one mole of
HA is used up and one mole of A- is
formed
- calculate the moles of HX and OH, see which is in excess.
- if HA is in excess: use table to work
out how much HA and A- is in excess,
divide each of the mol by the total vol
to find concentration. sub into [H+]= Ka
x [HA] / [A-]. then sub this answer into
pH=-log 10[H+}
- if OH- is in excess,calculate [OH],
then Kw/ [OH] (Kw is 10-14). sub
into pH=-log[H+]
- if HA is in excess: use table to work
out how much HA and A- is in excess,
divide each of the mol by the total vol
to find concentration. sub into [H+]= Ka
x [HA] / [A-]. then sub this answer into
pH=-log 10[H+}
- calculate the moles of HX and OH, see which is in excess.
- for every mole of OH- added, one mole of
HA is used up and one mole of A- is
formed
- partially dissociate
- pH curves and indicators
- indicators are weak acids so this
occurs in solution: HIn <> H+ + In-
- in order to be useful, HIn and In must have different colours
- the equivalence point shows exactly the same
amount of acid and base, the indicator which
goes through the equivalence point is the
suitable one.
- what makes a good indicator? the colour
change must be easily observed and colour
change must be easy
- the equivalence point shows exactly the same
amount of acid and base, the indicator which
goes through the equivalence point is the
suitable one.
- indicators are weak acids so this
occurs in solution: HIn <> H+ + In-
- Buffer Solutions
- solutions which
can resist
change in acidity
or alkalinity
- by keeping [OH-] and
[H+} almost unchanged
- by keeping [OH-] and
[H+} almost unchanged
- Acidic Buffers- made in
one of 2 ways
- mixing a weak acid and the salt of
that acid A- + H+ --> HA (so HA
increases when A- decreases
- by partly neutralising weak acid with a strong
alkali HA + OH- --> H2O + A- (HA decreases as
A- increases)
- mixing a weak acid and the salt of
that acid A- + H+ --> HA (so HA
increases when A- decreases
- Basic buffers
- Acid to the buffer removes H+ so
NH3+ H+ --> NH4+
- add alkali removes OH- so
NH4+ + OIH---> H2O +
NH3
- Acid to the buffer removes H+ so
NH3+ H+ --> NH4+
- see notes in book for buffer reactions.
- solutions which
can resist
change in acidity
or alkalinity
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