20.0 mL of 0.10 M Na2CO3 is titrated with 0.10 M HCl (pK_a1 = 6.35, pK_a2 = 10.33 for H2CO3). Select the correct statements. (Use the amphiprotic formula where appropriate.)

After 10.0 mL of HCl the pH is 10.33

The first equivalence point is at 20.0 mL and its pH is about 8.34

The second equivalence point is at 40.0 mL

After 10.0 mL of HCl the pH is 10.33

The first equivalence point is at 20.0 mL and its pH is about 8.34

The second equivalence point is at 40.0 mL

After 30.0 mL of HCl the pH is 8.34

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ia053

20.0 mL of 0.10 M $Na X_{2} CO X_{3}$ is titrated with 0.10 M HCl ( $p K_{a 1} = 6.35$ , $p K_{a 2} = 10.33$ for $H X_{2} CO X_{3}$ ). Select the correct statements. (Use the amphiprotic formula where appropriate.)

Carbonate

= 2.0

mmol. After 10 mL (1.0 mmol HCl): equal

CO X_{3} X^{2 -}

and

HCO X_{3} X^{-}

, pH

= p K_{a 2} = 10.33

. After 20 mL, only

HCO X_{3} X^{-}

remains, pH

= \frac{1}{2} (6.35 + 10.33) = 8.34

. After 40 mL, all is

H X_{2} CO X_{3}

. After 30 mL (3.0 mmol HCl) there are equal

HCO X_{3} X^{-}

and

H X_{2} CO X_{3}

(1.0 mmol each), so pH

= p K_{a 1} = 6.35

, not 8.34. Answer A, B, C.