The acids
which are capable to furnish more than one protons to water are called polyprotic
acids or polyprotic acids are those acids which have more than one acidic hydrogen.
For examples
Polyprotic acids are dissociate in steps wise with
different value of dissociation constant for each step for example carbonic
acid is dissociate in two steps as-
Step-1
Step-2
Over all
reaction
Experimentally
known that for typical Polyprotic acids Ka1>> Ka2 as result all
the H+ is due to the first acid ionization.
These
dissociation constant also illustrate simultaneous equilibria in the acid
dissociation of poly acids both are happening at same time , so both anions
HCO3-1 and CO3-2 are
present in equilibrium mixture in solution.
PH CALCULATION OF POLYPROTIC ACIDS:
Consider a
general triprotic acid (H3A) and dissociate in three steps
Experimentally
we know that Ka1>>Ka2>>Ka3,
hence x >>y>>Z so y and z can be neglected with respect
to x
This is the
quadratic equation solve by following formulae
Step-2
Experimentally
we know that Ka1>>Ka2>>Ka3, hence
x>>y>>Z So y and z can be neglected with respect to
x and the x present in denominator and numerator both are cancelled .
Step-3
Experimentally
we know that Ka1>>Ka2>>Ka3 hence x >>y>>Z, So y and z present in numerator can be
neglected with respect to x and the z present in denominator is also neglected
with respect to y.
Finally -
And concentration of different species:
In such cases following assumption are applicable. In generally
Ka1 >> Ka2 >> Ka3
x >> y >> z [H+] = x, [H2A¯ ] = x, [HA2–] = y, [A3–] = z, [H3A] = c – x
ILUUSTRATIVE
EXAMPLE (1): 0.1M H3X,
find the
concentration of (X-3 ) ?.
SOLUTION: We
know that:
Calculate the
value of (z) by solving these three equations (z) =10-19
ILUUSTRATIVE EXAMPLE (2): Calculate
the concentration of all species of significant contribution present in 0.1M H3PO4
Solution.(Given ka1 7.5×10-3,Ka2 6.2×10-8
and Ka3 3.6×10-13)
ILUUSTRATIVE EXAMPLE (3):
Calculate
concentration of (a) H+ (B) HCO3-1 (C) CO3-2
(D) DOD of HCO3-1 in 0.1M aqueous solution.(Given for Ka1=1.6x10-7
Ka2=4x10-11).
ILUUSTRATIVE EXAMPLE (4): Find concentration of (a) H+
(B) HCO3-1 (C) CO3-2 in 0.01M
aqueous solution of carbonic acid if pH
of the solution is 4.18, .(Given for Ka1=4.45x10-7 Ka2=4.69x10-11).
ILUUSTRATIVE EXAMPLE (5):
Consider the acid
dissociation of 0.25 M H2CO3. What are the concentration
of all species in equilibrium in each stage and pH?
Answer key
(2) (H+=HPO3-1)
= 0.0273 M, H3PO4= 0.1- x= 0.0727M, PO4-3=8.26x10-19
(3) H+= 4.0x10-5
M, HCO3-1= x-y=4.0x10-5 M DOD= 10-6
(4) H+=6.6x10-5 x=2.1095x10-5,
y =CO3 -2=ka2=4.69x10-11 M
(5) H+=x=3.24x10-4 M , HCO3-1=x=
3.24 x10-4 M
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