JEE MAIN - Chemistry (2021 - 16th March Evening Shift - No. 4)

The exact volumes of 1 M NaOH solution required to neutralise 50 mL of 1 M H₃PO₃ solution and 100 mL of 2 M H₃PO₂ solution, respectively, are :

100 mL and 50 mL

100 mL and 200 mL

100 mL and 100 mL

50 mL and 50 mL

JEE Main 2021 (Online) 16th March Evening Shift Chemistry - Some Basic Concepts of Chemistry Question 136 English Explanation 1

Millimoles of H₃PO₃ = M $$\times$$ V = 1 $$\times$$ 50 = 50

For 1 millimole of H₃PO₃, we require 2 millimoles of NaOH.

For 50 millimole of H₃PO₃, we require (2 $$\times$$ 50) = 100 millimoles of NaOH.

Millimoles of NaOH = M $$\times$$ V = 100

1 $$\times$$ V = 100

V = 100 mL

JEE Main 2021 (Online) 16th March Evening Shift Chemistry - Some Basic Concepts of Chemistry Question 136 English Explanation 2

Millimoles of H₃PO₂ = M $$\times$$ V = 2 $$\times$$ 100 = 200

For 1 millimole of H₃PO₂, we require 1 millimoles of NaOH.

For 200 millimoles of H₃PO₂, we require 200 millimoles of NaOH.

So, volume of NaOH = 200 mL