Question

How many milliliters of a 1.0 M sodium hydroxide is added to reach the second half equivalence point in the titration of 25 mL of a 1.0 M acid titration of H2A?

Answer 1

Molarity = moles of solute/Volume of solution in L

Molarity of H₂A = 1 M

Volume of H₂A= 25ml = 25 x 10^-3 L

Moles of H₂A = 1M x 25 x 10^-3 L = 25 x 10^-3 moles

1 mole of H₂A produces 2 moles of H⁺.

Moles of H⁺ = 2 x 25 x 10^-3 moles = 50 x 10^-3 moles

Given, we need to find the volume of 1 M NaOH such that second half equivalence point is reached.

To reach the second half equivalence point, we need to add NaOH such that we reach first equivalence and then half of the next equivalent point.

The first equivalence point will occur at 25 x 10^-3 moles of H⁺ and the second equivalence at 50 x 10^-3 moles of H⁺.

So, we need NaOH to react 25 x 10^-3 + (25 x 10^-3 )/2 moles = 37.5 x 10^-3 moles of H⁺.

1 mole of H⁺ needs 1 mole of NaOH.

So, 37.5 x 10^-3 moles of H⁺ needs 37.5 x 10^-3 moles of NaOH.

Volume of NaOH = Moles of NaOH/Molarity

= 37.5 x 10^-3 L/1M

=37.5 ml