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Q.

For the titration of weak dibasic acid H2A (conical flask) against NaOH, graph is given  below. To the one litre of 0.1MH2A, when 2g, 4g and 6 g of NaOH is added, pH of the mixtures are respectively X, Y and Z. (X+Y+Z) value is ( [H2A]0 is the initial concentration of H2A. [H2A] is concentration of H2A at that instant,  [HA] and [A2] are the concentrations of HA and A2 at that instant.)

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answer is 27.

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Detailed Solution

0.1 mole H2A+0.05 mole NaOH 0.05 mole H2A+0.05 mole HA

Now it is buffer PH=PKa1+log[HA][H2A]=PKa1=7=X

0.1 mole H2A+0.1 mole NaOH 0.1 mole NaHA

PHofNaHA=PKa1+PKa22=7+112=9=Y

0.1 mole H2A+0.15 mole NaOH 0.05 mole HA+0.05 mole A2

Now it is buffer PH=PKa2+log[A2][HA]=11=Z

X+Y+Z=7+9+11=27

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For the titration of weak dibasic acid H2A (conical flask) against NaOH, graph is given  below. To the one litre of 0.1 MH2A, when 2g, 4g and 6 g of NaOH is added, pH of the mixtures are respectively X, Y and Z. (X+Y+Z) value is ( [H2A]0 is the initial concentration of H2A. [H2A] is concentration of H2A at that instant,  [HA−] and [A2−] are the concentrations of HA− and A2− at that instant.)