DETERMINATION Acids have some characteristics which are as follows-

DETERMINATION OF pH OF A POLYPROTIC ACIDTo determine the pH of a monoprotic acidwe just take the concentration of H+ ions (at equilibrium in case ofa weak acid) formed in the solution and put the H+ value in thebelow formula and easily get the pH value.pH=-log10H+but in case of di, tri or more generallypolyprotic acid the pH determination method is something lengthy. For this weneed the step wise dissociation constant (Ka) values. PolyproticAcids have some characteristics which are as follows-a.

can give more than one proton.b. dissociation of 2nd andconsecutive protons are more difficult as after dissociation the acid turns toa negatively charged ion and it is increasingly difficult to lose proton from anegatively charged species compared to a neutral molecule.

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c. having n (n>1) no. of ionizableproton, the 1st dissociation step should be faster than all othersteps i.e. Ka1 should be greater than that of Ka2, Ka3and so on.We can discuss an example-Q.

Calculate the pH of 0.05 (M) H2SO4 solution.(Given Ka1= Very Large, Ka2=1.2 X 10-2)Approach: As here it is given that Ka1is very large so we can easily assume that H2SO4 is fully ionized to H+ andHSO4- ion. So, we can write it as-                                                   H2SO4   ?      H+      +      HSO4-                                                  0.05 (M)      0.

05 (M)    0.05 (M)That means the concentration of H+ion, H+ after first dissociation is 0.05 (M).Now coming to 2nd dissociationstep. But it is involving comparatively low Ka value than the 1stone, (Ka2 = 1.

2 X 10-2). So, it can be treated asequilibrium as follows-                                                    HSO4-   ?      H+      +      SO42-Suppose x (M) HSO4-has dissociated. So, at equilibrium the picture will be like below-                                                     HSO4-   ?      H+      +      SO42-At equilibrium                          0.05 – x      0.05 + x           x              (As already 0.

05 M H+                                                                                                                 ion has been produced                                                                                                                  in the 1st step) Thus,                                      So, the total H+ ionconcentration, H+ at equilibrium is (0.05+0.0085) (M) = 0.0585 (M)Therefore, the pH value will be-pH = -log10 H+ = -log10 (0.

0585) =1.232Below, there is a similar problem which isopen to all of you. Try to solve it own.Q.

Calculate the pH of 0.02 (M) H3PO4 solution.(Given Ka1 = 7.5 X 10-3, Ka2 = 6.

2 X 10-8and Ka3 = 4.8 X 10-13).