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Unit 4

Terms in this set (25)

Missed days
10//// 23, 25, 27, 30?

11//1?,!!!!!!!!!!!!!!!!3, last friday,
Weak acid/ vs weak base
WEAK= rice table
no rice for strong acid/bases

Using Ka equation
-if PH is known, canc alculate Ka
-if Ka is known, solve equilibrium problem to calculte [H+] and PH

X use to calculate PH or Poh
x= hydrionium ion concentrateion= PH
Kb solve for x
x= hyrdoxide ions
use Kw

weak base
to calculate Ph: obtin[OH-} at equilibrium, use Kw


pH=-log[H3o+]
strong acid/ base
dissociate completely
so teh initial molarity is used for molarity of ions formed= based on stoichiometry


POlyprotic Acid
H3A

3 pdonatable protons
1st= H3A +H2O<-->H3O+ + H2A-
*HA2-
*A3-

forward reactions Kaq,Ka2,Ka3

Congigate Bases for each reacted acid is the previos product

reverse reactions going bottom to top ar
Kb1,Kb2,Kb3
order of gaining proton



H3O+=Kw/Oh
PH of salts
salt is inorganic compouhnd
msot time cal A and B reactions neutralization reactions bevause they form watwr and something else,,, water at 25C will have neutral PH, but the solution left after reaction may not be neutral, the PH at teh end is related to the salt-- some salts are neutral

Basic salt if splt into ions, one ion reactes with water, and increases the OH- in a solution

strongest wins

compare the ions sto if they ar strong base or acid
both strong no one wins

SrCL2
both form strong

AlBR3
Br is strong
acidic salt

Both are weak it depends on Ka values

Ka>Kb= acidic


REMMEBR IF IT gives you Ph during a rice problem you can find H+ or OH ions wich may be the X and dont have to use quadratic formula


The conjugate base of a weak acid acts as a weak base.

The weaker a weak acid, the stronger its conjugate base.

Low Ka for bases = stronger base
That means, for example, that since the acid dissociation constant
Ka
of
HCH3CO2
is lower than the
Ka, of HF,CH3CO2−
is a stronger base than
F−. That is, a solution containing
CH3CO2−
will have a higher pH than a solution containing the same molarity of
F−
.
The conjugate acid of a weak base acts as a weak acid.
stong acids
the stronger the acid
Ka larger than 1
as Ka increases Pka decreases, the Ph gets lower, the POH increases, [H+] ncreases,{OH-] decrease, ?%diss increasess?,,,, oposite for bases

Ka follows trends of Ph
relationship between Ka and Kb
Kw=Ka x Kb
Kw= 1.0 x 10^-14
smartwork notes
pKa=-log(ka)

pKb=-log(kb)

Kb=Kw/Ka

Pka increases then acitity decreases

degree of ionization decreses-= smaller ka

For % question= % of molarityand solve for Ka

larger the
pKa
the smaller the
Ka
, and the weaker the acid


PH= consitancy (lower great thn 7)
%dissacitiation = X/[Ci]
protics
monoprotic= 1 mole of H+ leased
diprotic= 2 moles
polyprotic acids =

HCL
H2SO4
H3AsO4

water gains proton
thwn 2nd reaction happens
but second reaction is creating less hydronium ions
what is affecting ionization percent= ionization energy increases because 1 ionization is easier(weak bond) now there is more electron density (electronegative environement) its harder for water to gain that protonm,,,, as energy increases dissasociation decrreases

Ka2 is teh same as concentration formed for the second ionization reaction, so the actual concentration of hydronium ions taht affect the Ph is the concentration from 1st ionization reaction (for these acids)

Ph affected by forst ionization reaction--- forms the most hydronium ions
common ion effect
if you add a salt to acid or base, then the soluion has 2 different equilibrium,,, 1 for A &B and 1 for salt

by adding ions it shifts equilibrium towrads products or reactants and is consumed,,,, mjst be added to rice table

usually will give the concentration of one of the other moleculess
acid and base titrations
Titrant= know concentration
Analyte

add titrant to the analyte

steps
-acuuret;y transfer known volume oof sample to flask or beaker
- add few drops of indicator or insert probe of PH metere
-fill buret with solution of known concentration (titrant) to react with solute sample (analyte)
-Slowely add titrant until reaction is complete (equivilence point)



equivilence point= forming a salt

stong witha strong equvilence point is 7, neutral


weak acid strong base

initial ph a little high- higher than 2
-doesnt dissociate completely
- also have molecule because no dissociation compeltely
-ph @ equiveinece= higher than seven- forming a basic salt
-then hydrolizes with water to make basic solution
-Ph gets stable as acids and bases keep reacting together but at ph constant


opposite for strong acid, weak base

MIDPOINT
1/2 volume of the equivilence point
- the Ph is equivilent to the pKa of the acid
-buffer region

The smaller Ka= weaker= higher the inititial ph, the higher the equivilence point

weaker the base the lower the ph, lower ph at equivilence point



never titratate weak with a weak because you are putting 2 buffers together

strong acid and strong base no buffer region
That means, for example, that since the acid dissociation constant

.
Buffers
at mid point
solution has neutralizing acid or base
-can use a buffer solution to keep ph constant

buffer- solution tat resists changes in PH when acids or bases are added to it
in a titration we keep adding the titrant to react with the ]buffer until it is not viable as a buffer

Need a weak A or B and its conjigate
or we add salt of congitage


Henderson- hasselbalch equation

Ka= [H3O+][A-]/ [HA]

-logKa=-log[H3O]+log[A-]/[HA]
pKa=PH-log [A-]/[HA]

Ph= pKa+log[A-]/[HA]
if the concentration are equal= log 1=o
ph=pKa
pKa+- 1= buffer range

REMEMBER
POH=pKb+[A-]/[HA]

so if a solution has a PH of 4 becayse the Pka is 4 then the biuffer range will be between 3 and 5


remeber that addidng the salt will react to make the strong acid or base


when doing buffers remember a product is some times the oher buffer and thus added to tthe base/ acid side
buffer
-resist changes in PH
- necessityis
-weka acid and conjigate base i order to be a buffer
- reduces affect of adding stong acid or base, because already have acid and base
-henderson hosle bok equation to vcalculate ph of buffer

Ph=pKa +log([A-]/[HA])

pKA= constant
more acid tahn base Ph= lower
if acid base are equal then = to pKa
pH=pKa


Pka=-log(Ka)

solve pKa seperate for refernce of answer maing sense



wk acid+ conj bs= buffer
wk acid+strong base= no buffer (just titrating the acid) (@ equal M)!!!!!

wk acid + .5 M strong base= yes buffer (.5 M of wk acid remqin and .5 formed conjigate base

BASE OVER ACID

also remember buffer can fluxiate pH +-1 above or below that it destroyes buffer



Ph=pKa+log([A-]/[HA])
[A-]/[HA]=10^pH-pKa
ten flip 1/X if finding other ratio ---- or do it my way and get right answer!!!


acif- base buffer
weak a or b and conjigate
rsist Ph change
enviromental buffer
carbonate in river, lakes, protects natural waters from acid rain (carbonic acid and its conjigate, bi carbinate)
Physiological buffer
PH blood= 7.4

too much CO2- reaction shifts right, PH decreases, acidosis

-decrease in blood Co2- reaction shidts left, PH increases, alkalosis
Buffer Capacity
-measure of teh ability of the buffer to resist pH changes
- dpeends on he concentration of the buffer components
- teh more concentrated teh components of a buffer, teh greater the bufefr capacity. A buffer has the highest capacity when the components concentratons are equal, when [A]/[HA]=1

A buffer whose Ph is equal or near the pKa of its acid component has the highest capacity

when choosing buffer choos one closest to pKa value
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