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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
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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Chapter 23 Solutions

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Sect-23.5 P-1.3ACPSect-23.5 P-2.1ACPSect-23.5 P-2.2ACPSect-23.5 P-2.3ACPSect-23.5 P-3.1ACPSect-23.5 P-3.2ACPSect-23.5 P-3.3ACPSect-23.5 P-3.4ACPSect-23.5 P-3.5ACPCh-23 P-1PSCh-23 P-2PSCh-23 P-3PSCh-23 P-4PSCh-23 P-5PSCh-23 P-6PSCh-23 P-7PSCh-23 P-8PSCh-23 P-9PSCh-23 P-10PSCh-23 P-11PSCh-23 P-12PSCh-23 P-13PSCh-23 P-14PSCh-23 P-15PSCh-23 P-16PSCh-23 P-17PSCh-23 P-18PSCh-23 P-19PSCh-23 P-20PSCh-23 P-21PSCh-23 P-22PSCh-23 P-23PSCh-23 P-24PSCh-23 P-25PSCh-23 P-26PSCh-23 P-27PSCh-23 P-28PSCh-23 P-29PSCh-23 P-30PSCh-23 P-31PSCh-23 P-32PSCh-23 P-33PSCh-23 P-34PSCh-23 P-35PSCh-23 P-36PSCh-23 P-37PSCh-23 P-38PSCh-23 P-39PSCh-23 P-40PSCh-23 P-41PSCh-23 P-42PSCh-23 P-43PSCh-23 P-44PSCh-23 P-45PSCh-23 P-46PSCh-23 P-47PSCh-23 P-48PSCh-23 P-49PSCh-23 P-50PSCh-23 P-51PSCh-23 P-52PSCh-23 P-53PSCh-23 P-54PSCh-23 P-55PSCh-23 P-56PSCh-23 P-57PSCh-23 P-58PSCh-23 P-59PSCh-23 P-60PSCh-23 P-61PSCh-23 P-62PSCh-23 P-63PSCh-23 P-64PSCh-23 P-65PSCh-23 P-66PSCh-23 P-67PSCh-23 P-68PSCh-23 P-69PSCh-23 P-70PSCh-23 P-71PSCh-23 P-72PSCh-23 P-73PSCh-23 P-74PSCh-23 P-75GQCh-23 P-76GQCh-23 P-77GQCh-23 P-78GQCh-23 P-79GQCh-23 P-80GQCh-23 P-81GQCh-23 P-82GQCh-23 P-83GQCh-23 P-84GQCh-23 P-85GQCh-23 P-86GQCh-23 P-87GQCh-23 P-88GQCh-23 P-89GQCh-23 P-90GQCh-23 P-91GQCh-23 P-92GQCh-23 P-93GQCh-23 P-94GQCh-23 P-95GQCh-23 P-96GQCh-23 P-97GQCh-23 P-98GQCh-23 P-99GQCh-23 P-100GQCh-23 P-101GQCh-23 P-102GQCh-23 P-103ILCh-23 P-104ILCh-23 P-105ILCh-23 P-106ILCh-23 P-107ILCh-23 P-108ILCh-23 P-109ILCh-23 P-110ILCh-23 P-111ILCh-23 P-112ILCh-23 P-113ILCh-23 P-114ILCh-23 P-115SCQCh-23 P-116SCQCh-23 P-117SCQCh-23 P-118SCQCh-23 P-119SCQCh-23 P-120SCQCh-23 P-121SCQ

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