Verified Solution • 2025–26 Session moderate

QJEEAdv20Ch16b9: JEE Advanced 2020 Chemistry - Class 11 JEEAdv20Ch16b9 Solution: 5.00 mL of 0.10 M oxalic acid solution taken in a conical flask …

5.00 mL of 0.10 M oxalic acid solution taken in a conical flask is titrated against NaOH from a burette using phenolphthalein indicator. The volume of NaOH required for the appearance of permanent faint pink color is tabulated below for five experiments. What is the concentration, in molarity, of the NaOH solution?

Exp. No. Vol. of NaOH (mL)
1 12.5
2 10.5
3 9.0
4 9.0
5 9.0

Step-by-Step Verified Solution

Oxalic acid solution titrated with NaOH solution using phenolphthalein as an indicator.

H2C2O4+2NaOH⟶Na2C2O4+2H2O

Equivalent of H2C2O4 reacted = Equivalent of NaOH reacted

=5×2×0.11000=9×M(NaOH)×11000= {{5 \times 2 \times 0.1} \over {1000}} = {{9 \times {M_{(NaOH)}} \times 1} \over {1000}}

M(NaOH)=19=0.11{M_{(NaOH)}} = {1 \over 9} = 0.11

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