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JEE Advanced levelNumerical answerDifficulty 3 of 3ie156

150 mL of 0.20 M HA ( $K_{a} = 2.0 \times 1 0^{- 5}$ ) is mixed with 100 mL of 0.10 M NaOH. Using $lo g 2 = 0.301$ , find the pH to two decimals.

Numerical answer type, tolerance 0.01.

Answer: 4.3979

HA = 30 - 10 = 20 mmol; A

^{-}

= 10 mmol.

p K_{a} = 5 - 0.301 = 4.699

. pH =

4.699 + lo g (10/20) = 4.699 - 0.301 = 4.398

, i.e. 4.40.

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