Common Ion effect in the ionization of Acid & base - Ionic Equilibrium Video Lecture | Physical Chemistry

hello friends this video an equilibrium
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you have watched all the previous parts
so we have discussed the effects of
changing concentration temperature
pressure or equilibrium change discuss
the effect of ionization effects of eye
addition I am you have you add some ions
in the already ionized acid base so you
see this is a equilibrium actually this
is equilibrium and this is achieving the
concentration by adding ions so if you
add some let's suppose acetate ions so
adding and what will happen so you're
adding more ions here according to each
other principle the stress of addition
of acetate anion he'll be released only
when the reaction movements this
direction right the reaction is in this
direction what happened the H+ ions also
is used so the concentration of H+ ions
also goes down right so you want to
decrease the pH the amount will decrease
the acidity of this solution if you add
this s89
what does guy window this will make all
this for the equilibrium to move in this
direction and thus it will consume h
plus and the H+ ions will also similarly
now if you add this h+ ions from
external source then also if you see it
will move in this direction because the
stress of adding H+ ions can be relieved
only when they give Vidia moves in the
left direction and then there also if
you see this behind
concentration is less than their sitting
s it is so the effect of mine
concentration you don't have to think a
lot on this just assume this as the
effect of concentration change in
concentration changing it will be we
have discussed this use the same logic
ions is nothing but a species in the
reaction assume that as a specie and
just add this it's apple h+ what in band
it will have it will obviously move
equilibrium to the left hand side in
this concentration of acetate will
decrease and st casting
right so common effect fine if we
defined it as nothing but set of the
equilibrium on a nigga's of sin that
provides more of our I need species
already present in the equilibrium so if
this is equilibrium there's a mining
species which is then equilibrium and it
is part of the equilibrium you provide
some more ions in that case the
equilibrium sips in the left side and if
you remove it's a political science here
they clearing the city the orbit
direction testify for examples of the
effect of ion defendants so we have to
calculate the pH of the ammonium point
the one in dunya and can you calculate
the pH after 50 ml of this solution is
treated with 25 ml of s CL and the
dissociation constant of ammonia is
given so I have a Maria this is the base
it reacts with water it forms a medium
and it forms NH 4 plus and OH minus like
so at T is equal to zero it was point 1
M so it was putting M this was not
required this is zero and this and so at
T is equal to equilibrium this becomes
point 1 minus X not required it becomes
X and seducers right
first thing is we defined up gather at
the pH first first beach initial PHA
let's talk about this lightly you find
that the each of this so you define the
co h minus concentration let's do that
so we know that K D's is equal to one
ammonia concentration is X OS -
penetration is X by n it's mo neighbors
will hear the point 1 minus X so NH 4
plus concentration OHP response
attrition and it's 3 concentration so
this is nothing but one point seven
seven tickets bar - give it a key so if
you solve this is nothing but X square
by 0.1 minus X is this so if you solve
this you get X as 1 point 3 3 K to the
power minus 3
I think what os- fascination but I do
find pH I have to my express
concentration so H+ concentration is 1
10 to the power minus 14 Yoh means
concentration right because h plus ion
consolation to which mass concentration
only one shade up so with this it
becomes cinema - 14 by 1.33 in the power
minus 3 and this is nothing but 7.5 1
into 10 to the power minus so pH should
be got pH will be minus log of 1 into P
1 minus 2 L that is 11 point this is my
initial piece this is my first part
correctly see now the trick begins now
what happens is in this we are adding 50
ml of this solution with 25 ml of 0.1 m
HC L so let us take the fresh page now
let's this part is done now come the
trick part so you define the pH of the
50 ml of this solidness to pivot 25 ml
of 0.1 MSC so we have 25 ml of 0.1 m SL
this is nothing but my 25 ml by 1,000
into point 1 this is 2.5 by 1,000 moles
that is 2 point 5 millivolts so this of
my I have s plus activism right through
much here and this reacts with I appoint
one in 250 right so 50 ml of 0.1 m n is
3
this is going what it's P by thousand in
two point one that is fine by Hallows n
Mo's that is five melee words meaning
moves off and it's so nh3 plus STN if
you see scintillation reaction is in
comes in it's for CL right so that means
I can say that by 5 millivolts of NSC
and I had two point five million moles
of HCl Ashley right Sen any have so this
neutralization reaction will happen and
will get only two point five moles of
it's from this sickness react we get net
2.5 million holes off in this thing why
because we have we had five millimoles
of nsv although that 2 point 5
millivolts than Dubai SEL correct so
let's say two reaction once again I have
an X 3 plus same reaction is 3 plus h2o
on a plane theory and it's 3 plus h2o
gives a nice or plus o H minus miles
right so this is now my new point 5
millivolts now I have to find it in
molarity in moles per volume so total
volume is what two per volume was 25 ml
from HCl 50 mm from ammonia that is 75
ml correct so let me find the
concentration of NH 4 plus no and it's
identified in history and that's the
initial molarity
this will be what moles per liter what
is the most 2 point 5 millivolts and
volume is 75 ml right that is nothing
word or
three three you know that has millivolts
Hey all the good this reaction with the
anus on NSC when this and it's
foreseeable this becomes NH four plus
hey this I'm thirsty and it's for
Sheehan so let's find the concentration
of NH four plus also this will also be
same right moles per liter doesn't
become two point five he knows where 75
as point zero he cheated because you see
in this case only the same because two
point five million balls of energy left
and 2.5 millimoles of the is used right
so this is 2 point 5 millivolts used so
we got this concentration of NH 4 plus
not so now we have TV let let's have
this reaction on in s 3 gives NH 4 plus
plus so T is equal to 0 we have n s3 was
point zero three three and just ignore
and this also we had problems around to
be 3 this is 0 at t is equal to
equilibrium this will become point zero
three three - why this becomes point
zero three three plus finding this Y so
write KB KB is nothing but point zero 3
3 plus y into y by point zero 3 3 minus
1 right so let us assume these are
almost same point 0 3 3 plus y minus y
almost same value will be very less
actually and this KB is given as 1 point
7 7 minus so with this I got Y as 1.75
inch very small very correct so you got
the value of y and Y is what my Oh H
minus concentration correct I have to
find what H this concentration and that
is what
10 to the power minus 14 by oh s minus
concentration and that is what 10 to the
power minus 14 by 1.77 integral over
minus Phi this value you solve this you
get 0.56 10 to power minus 9 I have to
find pH pH is nothing but minus log of
this value X plus 9 is fine 561 you
solve this you get 9.2 for advanced
little tricky here the trick here was
the moment you add HCl to 8 it has
consumed the Oh H minus from here hey
and treated NH 4 plus also so we now
have when you are starting this reaction
we have zooming there M is 3 has this
value now and NH 4 plus also has this
that initial concentration okay thank
you
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