For the titration of weak dibasic acid $H_{2} A$ (conical flask) against $N a O H,$ graph is given below. To the one litre of $0.1 M H_{2} A,$ 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 ( ${[H_{2} A]}_{0}$ is the initial concentration of $H_{2} A . [H_{2} A]$ is concentration of $H_{2} A$ at that instant, $[H A^{-}]$ and $[A^{2 -}]$ are the concentrations of $H A^{-}$ and $A^{2 -}$ at that instant.)

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answer is 27.

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Detailed Solution

0.1 mole $H_{2} A + 0.05 mole N a O H \overset{}{\to} 0.05 mole H_{2} A + 0.05 mole H A^{-}$

Now it is buffer $P^{H} = P^{K a_{1}} + \log \frac{[H A^{-}]}{[H_{2} A]} = P^{K a_{1}} = 7 = X$

0.1 mole $H_{2} A + 0.1 mole N a O H \overset{}{\to} 0.1 mole N a H A$

$P^{H} o f N a H A = \frac{P^{K a_{1}} + P^{K a_{2}}}{2} = \frac{7 + 11}{2} = 9 = Y$

0.1 mole $H_{2} A + 0.15 mole N a O H \overset{}{\to} 0.05 {mole HA}^{-} + 0.05 mole A^{2 -}$

Now it is buffer $P^{H} = P^{K a_{2}} + \log \frac{[A^{2 -}]}{[H A]} = 11 = Z$

$∴ X + Y + Z = 7 + 9 + 11 = 27$

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