100 mL of 0.200 M NaOH is left open to air and exactly 25.0% of the NaOH is converted into Na2CO3 by CO2. The solution is titrated with 0.100 M HCl using phenolphthalein; the volume of acid needed to reach the end point is

JEE Advanced levelSingle correct optionDifficulty 3 of 3sb201

100 mL of 0.200 M $NaOH$ is left open to air and exactly 25.0% of the $NaOH$ is converted into $Na X_{2} CO X_{3}$ by $CO X_{2}$ . The solution is titrated with 0.100 M HCl using phenolphthalein; the volume of acid needed to reach the end point is

Total NaOH

= 0.100 \times 0.200 = 0.0200

mol; 25.0% is

0.00500

mol, giving

0.00250

mol

Na X_{2} CO X_{3}

(

2 NaOH + CO X_{2} Na X_{2} CO X_{3} + H X_{2} O

). Remaining NaOH is

0.0150

mol. At the phenolphthalein end point NaOH is fully neutralised and carbonate goes only to hydrogencarbonate:

0.0150 + 0.00250 = 0.0175

mol HCl, i.e. 175 mL. A methyl orange end point would need 200 mL.