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PH OF POLYPROTIC ACIDS:

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
Step-1 
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>>Ka 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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