Solubility Equilibria Video Lecture - NEET

all right how does equilibrium play a
role if the solubility here I'm giving
that calcium phosphate has a certain
equilibrium constant when it's
dissolving in water but what does that
need and how to use an ice table to
solve it well when a solid dissolves in
water we write it like this in terms of
chemistry or chemical reactions the
solid is in equilibrium with its aqueous
ions what ions did this originate from
well if you remember the criss-cross
rule of where these subscripts came from
or if you just look at a periodic table
and notice that calcium is in the second
column you'll know that it breaks up
into a calcium 2 plus ion let's see
really the only charge I've ever seen
calcium 1/2 and its aqueous because it's
now dissolved in water phosphate ions
Pio force have a charge of minus 3
they're also aqueous but the deal is
that we have three calcium's and two
phosphates so this is my balanced
chemical equation for the dissolving of
calcium phosphate to figure out the
solubility at equilibrium we're going to
need to do an ice table as soon as ice
well this is an equilibrium arrow that
I've drawn here as soon as you know that
it's at equilibrium you should probably
do an ice table well we're going to
start here with my initial
concentrations the initial concentration
of this doesn't even really matter
because it's a solid and solids don't
really play a role in equilibrium the
initial concentration of each of these
ions is clearly zero because we're
starting only with solid and how much of
like we start with none of these in the
solution the change well we lose some of
this solid but again it doesn't matter
we end up gaining some calcium I'm going
to say that the number or a number of
moles per liter of this that dissolved
is X
and I get three times as many calcium's
as I lost of this so I'm going to say
this is plus 3x the three comes from the
coefficient here and I gain two
phosphates for each of those steps so my
equilibrium concentrations of calcium is
3x moles per liter now my equilibrium
concentration two phosphate is 2x why
does this matter well because we were
given K the equilibrium constant of two
times ten to the minus twenty nine we
can use these concentrations in a KSP
expression and solve for x KSP as you
create an equilibrium expression as
products over reactants calcium to the
power of its coefficient phosphate to
the power of its coefficient and we
don't include the solid in the
denominator because it's a solid and we
don't include solids of the equilibrium
expression now we fill in what we know
for the final row of our ice table the
KSP was given in the question two times
ten to the minus twenty nine our calcium
concentration is 3x that's cubed at our
phosphate concentration is 2x that's
squared let me just take a moment to
simplify that for you three cubed times
two squared is twenty seven times four
which is a hundred and eight X cubed
times x squared is X to the five look at
this algebra is useful how do we solve
for X well the basic math says we divide
both sides by 108 that causes to get
cancelled out here that gives me two
times ten to the minus twenty nine
/ 108 gives me one point 8 5 times 10 to
the minus 31 and how do I do it to the
power of 5 it's called the fifth root
that fifth root cancels it to the to the
power of five on your calculator you may
prefer to just use your to the power of
button it's to the power of 1/5 or again
if you don't like fractions it's one
point eight five times ten to the minus
thirty-one to the power of zero point
two let's just do that on my calculator
here to the power of zero point two or
1/5 or fifth root and what I end up with
is seven point one three times ten to
the power of three six seven cool I've
solved for my X and it's seven point one
three times ten to the minus seven that
is the solubility in moles per liter of
calcium phosphate if they asked you for
the equilibrium concentrations of
calcium it's three times that number if
they asked you for the equilibrium
concentration of phosphate it's two
times that number but the point I'm
trying to get across to you is a
solubility question if you're giving KSP
is the same as all other equilibrium
problems use a nice table use your
expression for K plug in your values and
solve for X it's a beautiful thing just
like you are my friend take care and
best of luck