For bases normally given conc of hydroxide ion , use Kw as strong bases completely dissociate into their ions
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How do you find Ph of pure water
Because pure water, neutral solutions have [H+]=[OH-] Kw=[H+]^2
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At diff temps to 25 what happens to ph of pure water
changes, le chateliers can be used to predict change as dissociation is endothermic so increasing temperature pushes E to right giving bigger conc of H+ ions and lower Ph
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So what does Conc H+ =
root Kw
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What is pOH
equal and opposite scale to Ph
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pOH=
-log[OH-]
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[OH-]=
10^-pOH
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Ph+pOH=
14
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2 methods of calculating Ph of strong bases
Kw and pOH
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What does the -OH ions in bases enable them to act as
Proton acceptors
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What do strong bases do, POE
fully dissociate to release OH- ions, Right
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Kw method
use con base to work out conc OH-, sub OH- into Kw calc, rearrange Kw for H+ then sub into ph=-log[H+]
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Using pOH
Use conc of base into conc OH, sub conc OH into pOH to calc pOH then sub into ph+pOH=14 to get ph
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What are all carboxylic acids
weak acids
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so what do weak acids do
only slightly partially dissociate when dissolved in water giving an equilibrium mixture
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What is any organic acid=
weak eg meth,eth oic acid
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What do weak acids and weak bases do
dissociate only slightly in aqueous solution
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eg with CH3COOH
CH3COO- =not acidic proton is part of methyl group and doesn't dissociate whereas H+ is an acidic proton
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As it only partially dissociates where is POE
LHS making calculating PH challenging as conc H+ unknown
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What is Ka
Acid dissociation constant for a weak acid
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HA(weak acid)=
A- + H+ -paratially dissociates
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What does Ka tell you
Acid dissociating not Kc set out like Kc but put ka
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What do all weak acids have
VALUE of Ka which indicates how much the acid dissociates
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If Ka greater than 1
POE right
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If Ka less than 1
POE LEFT
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Larger the Ka
stronger the acid dissociated more
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What are Ka values sometimes quoted as
Pka values
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pKa=
-log Ka (poe)
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Ka=
10^-Pka
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WHY IS ph used
H+ too big so log makes scale easier to work with
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What does pKa mirror
pH scale
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Stronger acid
lower Ph, lower pKa
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pKa=
ph
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Ka=
[H+]
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2 assumptions to simplify Ka expression
concH+= conc A- BECAUSE dissociated according to 1;1 ratio
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and
Amount of dissociation is small we assume initial conc of undissolved acid remained constant
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so
[HA]equil;ibrum=[HA] initial
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Ka=
[H+]^2/[HA]initial
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Strong acid strong base neutralisation
Work out moles of original acid hence mol H+ work out moles are added and hence moles of OH- and work out which one is in excess
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If excess acid
Work out new conc of excess H+ ions [H+]=moles excess H+/total volume (dm3)
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What it total volume
vol acid+vol of base
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If excess alkali
Work out new conc of excess OH- ion [oh-]= moles of excess OH-/ total volume the work out H+= Kw/ conc OH- then Ph
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Other cards in this set
Card 2
Front
How do you find Ph of pure water
Back
Because pure water, neutral solutions have [H+]=[OH-] Kw=[H+]^2
Card 3
Front
At diff temps to 25 what happens to ph of pure water
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