In 50 mL of 0-2 M H 2 A solution- 25 mL of 0-4 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

[ amp; H_2A amp; + amp; NaOH → amp;NaHA; t = 0 amp; 50 mL × 0.2 am ...

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In 50 mL of 0...

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In $50 m L$ of $0.2 M H_{2} A$ solution, $25 m L$ of $0.4 M N a O H$ solution is added at $25^{o} C$ . The $p H$ of resulting solution is (for $H_{2} A, K_{a_{1}} = 10^{- 6}; K_{a_{2}} = 10^{- 12}$ )

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