EX 2.Given->Volume of Na+ = 500 mlMolarity of Na+= 0.0100MMolar mass of Na2CO3 = 105.99 gm/mole Millimole of Na+ = molarity × volumeNumber of millimole = 0.0100 × 500= 5 millimoleNa2CO3 ---> 2Na+ + CO32-Millimole of Na2CO3 = millimole of Na+/2Millimole of Na2CO3 =5/2 = 2.5 millimoleMole of Na2CO3 = 2.5 × 10-3mole (1 mole = 10^3 millimole)Weight of Na2CO3 required = mole × molar mass= 2.5 × 10-3 × 105.99=0.26 gm Hence, 0.26 gm Na2CO3 must dissolve in 500 ml of water. 13-3. How would you prepare 50.0-mL portions of SS that are 0.00500 M,0.00200 M and 0.00100 M in Na* from the soln in Ex. 2?

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Answered: 13-3. How would you prepare 50.0-mL…

13-3. How would you prepare 50.0-mL portions of SS that are 0.00500 M, 0.00200 M and 0.00100 M in Na* from the soln in Ex. 2?

Chemistry by OpenStax (2015-05-04)

1st Edition

ISBN:9781938168390

Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Chapter15: Equilibria Of Other Reaction Classes

Section: Chapter Questions

Problem 11E: The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities...

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Precipitation Titration

Precipitation analysis is used to analyze drugs, and to measure the salt content in food and beverages. It is an important concept from the examination point of view many practical questions are asked on the topic. This topic is significant in the professional exams for both undergraduate and graduate courses, especially for

Question

EX 2.

Given->

Volume of Na⁺= 500 ml

Molarity of Na⁺= 0.0100M

Molar mass of Na₂CO₃= 105.99 gm/mole

Millimole of Na⁺ = molarity × volume

Number of millimole = 0.0100 × 500

= 5 millimole

Na₂CO₃---> 2Na⁺ + CO₃^2-

Millimole of Na₂CO₃= millimole of Na⁺/2

Millimole of Na₂CO_{3 =}5/2 = 2.5 millimole

Mole of Na₂CO₃ = 2.5 × 10^-3mole (1 mole = 10^3 millimole)

Weight of Na₂CO₃required = mole × molar mass

= 2.5 × 10^-3 × 105.99

=0.26 gm

Hence, 0.26 gm Na₂CO₃ must dissolve in 500 ml of water.

13-3. How would you prepare 50.0-mL portions of SS that are 0.00500 M,
0.00200 M and 0.00100 M in Na* from the soln in Ex. 2?

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