   Chapter 23.5, Problem 3.5ACP

Chapter
Section
Textbook Problem

What quantity of 0.050 M NaOH would be required to react with 300 mg of BPA in 1.00 L of water? (Assume both —OH groups react with NaOH.)

Interpretation Introduction

Interpretation: The volume of NaOH has to be calculated for the reaction between BPA and 0.050 M NaOH.

Concept introduction:

The reaction between NaOH and BPA is a weak acid vs strong base titration. BPA has two hydroxyl groups which will be neutralized by NaOH. The volume of NaOH is calculated by using following expression,

M1V1 = M2V2 (1)

Here, M1 and M2 are concentration values of acid and base respectively. And V1 and V2 are volume of acid and base respectively.

Molarity=mass(molar mass)(1 L volume of solvent)M

Explanation

The calculation of NaOH volume is given below.

Given:

The concentartion of NaOH is 0.050 M.

Let V2 is the volume used of NaOH.

The amount of BPA dissolved in 1 L water is 300 mg.

Therefore, concentration of BPA is calculated as,

Molarity=mass(molar mass)(1 L volume of solvent)M=0.300 g(228 gmol1)(1 L)M=0.001315 M

As BPA has two hydroxyl groups so its concentartion will be twice of 0.001315 M.

Therefore, concentartion of BPA to be used is 0.002631 M

Volume of BPA solution is 1 L.

Volume of NaOH is calculated by using equation (1).

M1V1 = M2V2

Substitute, 0

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