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To electrodeposit all the Cu and Cd from a solution of

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 132E Chapter 18

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 132E

To electrodeposit all the Cu and Cd from a solution of CuSO4 and CdSO4 required 1.20 F of electricity (1 F = 1 mol e-). The mixture of Cu and Cd that was deposited had a mass of 50.36 g. What mass of CuSO4 was present in the original mixture?

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Note Set #3 – week 5 Analytical Chemistry SOLUTIONS OF INTERMEDIATE SPECIES of DIPROTIC SYSTEMS + 2- - o Charge balance eqn.  [H ] + [H A] = [A 2 + [OH ] - 2- o Mass Balance  F = [H A] + [HA2] + [A ] - + 2- o HA  H + A (deprotonation of intermediate) o HA + H O 2H A + OH 2 - (protonation of intermediate) TITRATION o Titration of a Strong Acid with a Weak Base o e.g.) HI + KOH  H O + 2I *goes to completion* 1. Before equivalence point: pH is dependent on excess, unreacted/un-neutralized H+ 2. At equivalence point: pH = 7.00; [H+] is equivalent to [OH-] 3. After equivalence point: pH is dependent on excess, unreacted/un-neutralized OH- Useful eqn.: C V1= 1 V 2 2 o Titration of a Weak Acid with a Strong Base o e.g.) HF + KOH  H O + K2 2 + - 1. Before eq.pt: No base added, treat as a weak acid problem. K = a − , x = [H ] = [A ] 2. Before eq.pt: Some base added, pH = pK + log ([A a/[HA]). Moles HA = Moles HA – i Moles OH added. Moles A = Moles A + Moles OH added. Make sure to convert units back to molar concentrations (if you have to). 3. At eq.pt: All weak acid conv

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Chapter 18, Problem 132E is Solved
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Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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