10 mL of 0.2M HA is titrated with 0.2M NaOH solution. Calculate the change in pH of the solution between 50% of the equivalence point to the equivalence point ( K a of HA = 10 -5 )

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\displaylines p H a t 50% o f e q u i v a l e n tp o in t p H = p K a + lo g [ Acid ] [ Salt ] ​ p H = 5 A t e q u i v a l e n tp o in t N a A − + H 2 ​ O ( l ) ⇌ H A + O H − Difference = 9 − 5 = 4 ​ K H ​ = x 2 = 1 0 − 10 x = 1 0 − 5 p H = 9 ​ K a ​ K W ​ ​ = 1 0 − 9 = 0.1 x 2 ​ ​

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