In 50 mL of 0-2 M - H 2 A solution- 30 mL of 0-5 M - NaOH solution is added at 25- C The pH of resulting solution is for H 2 A - K a 1- -10-6 - K a 2-10-12

M mole of H_2A = 50 × 0.2 = 10 m mole of NaOH = 15 H_2A + NaOH → NaHA 10 15 0 5 10 NaHA + NaOH → Na_2A 10 ...

In 50 mL of 0...

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In $50 m L$ of $0.2 M - H_{2} A$ solution, $30 m L$ of $0.5 M - N a O H$ solution is added at $25^{\circ} C$ The pH of resulting solution is (for $H_{2} A, K_{a_{1}} = 10^{- 6}; K_{a_{2}} = 10^{- 12})$

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