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.

find the
concentration of (X

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 Ka

_{1}>> Ka_{2}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 (H

_{3}A) and dissociate in three steps
Experimentally
we know that Ka

_{1}>>Ka_{2}>>Ka_{3,}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 Ka

_{1}>>Ka_{2}>>Ka_{3, }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 Ka

_{1}>>Ka_{2}>>Ka_{3 }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, [H

_{2}A¯ ] = x, [HA^{2–]}= y, [A^{3–]}= z, [H_{3}A] = c – x**ILUUSTRATIVE EXAMPLE (1):**0.1M H

_{3}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 H

_{3}PO

_{4}Solution.(Given ka

_{1}7.5×10

^{-3},Ka

_{2}6.2×10

^{-8}and Ka

_{3}3.6×10

^{-13})

**ILUUSTRATIVE EXAMPLE (3):**Calculate concentration of (a) H

^{+}(B) HCO

_{3}

^{-1 }(C) CO

_{3}

^{-2}(D) DOD of HCO3-1 in 0.1M aqueous solution.(Given for Ka

_{1}=1.6x10

^{-7 }Ka

_{2}=4x10

^{-11}

**).**

**ILUUSTRATIVE EXAMPLE (4):**Find concentration of (a) H

^{+}(B) HCO

_{3}

^{-1 }(C) CO

_{3}

^{-2}in 0.01M aqueous solution of carbonic acid if pH of the solution is 4.18, .(Given for Ka

_{1}=4.45x10

^{-7 }Ka

_{2}=4.69x10

^{-11}).

**ILUUSTRATIVE EXAMPLE (5):**Consider the acid dissociation of 0.25 M H

_{2}CO

_{3}. What are the concentration of all species in equilibrium in each stage and pH?

**Answer key**

(2) (H+=HPO

_{3}^{-1}) = 0.0273 M, H_{3}PO_{4}= 0.1- x= 0.0727M, PO_{4}^{-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 =CO_{3 }^{-2}=ka_{2}=4.69x10^{-11}M
(5) H

^{+}=x=3.24x10^{-4 }M , HCO_{3}^{-1}=x= 3.24 x10^{-4 }M
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