100 mL of 0.10 M Na2CO3 is mixed with 60 mL of 0.10 M HCl. For HCO3-, pK_a2 = 10.33. With log 2 = 0.301 and log 3 = 0.477, the pH of the resulting solution is

JEE Main levelSingle correct optionDifficulty 2 of 3ia098

$100 mL$ of $0.10 M$ $Na X_{2} CO X_{3}$ is mixed with $60 mL$ of $0.10 M$ $HCl$ . For $HCO X_{3} X^{-}$ , $p K_{a 2} = 10.33$ . With $lo g 2 = 0.301$ and $lo g 3 = 0.477$ , the pH of the resulting solution is

CO X_{3} X^{2 -} + H X^{+} \to HCO X_{3} X^{-}

10 mmol

carbonate and

6 mmol

acid give

6 mmol

HCO X_{3} X^{-}

and

4 mmol

CO X_{3} X^{2 -}

left.

pH = 10.33 + lo g (4/6) = 10.33 + 0.301 - 0.477 = 10.15

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