Exactly 1 litre of solution contains 0.018 mole of Ba²+ ions and 0.0018 mole of Pb²* ions. You wish to separate these ions by selective precipitation using the drop-wise addition of a solution of K2SO4. Explain, by using a full calculation, which of the salts BaSO, or PbSO, will precipitate first? Given: Kp (BaSO4) = 1.1 × 10-10 and Ksp (PbSO,) = 1.6 × 10-8

Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Write down the fully balanced cell-reaction equation.
What must the value of the ratio :
[Mn2+]6
be if [Br03] = 0.90 M, [Br¯] = 1.11 M, ailud the
[MnO,]6
pH = 0.60?
Transcribed Image Text:Write down the fully balanced cell-reaction equation. What must the value of the ratio : [Mn2+]6 be if [Br03] = 0.90 M, [Br¯] = 1.11 M, ailud the [MnO,]6 pH = 0.60?
Exactly 1 litre of solution contains 0.018 mole of Ba?+ ions and 0.0018 mole of Pb²+ ions.
You wish to separate these ions by selective precipitation using the drop-wise addition of a
solution of K2SO4.
Explain, by using a full calculation, which of the salts BaSo, or PbSO4 will precipitate first?
Given: Kep (BaSO4) = 1.1 x 10-10 and Kẹp (PbSO4) = 1.6 x 10-8
A voltaic cell is set up at 25 °C and delivers 0.060 V. One half cell contains an acidified solution
of permanganate and manganese(II) ions, and the other half-cell is based on the following
reaction:
Br03 (aq) + 6H30*(aq) + 6e
Br¯(aq) + 9H20 (f), Eed
= 1.44 V
Transcribed Image Text:Exactly 1 litre of solution contains 0.018 mole of Ba?+ ions and 0.0018 mole of Pb²+ ions. You wish to separate these ions by selective precipitation using the drop-wise addition of a solution of K2SO4. Explain, by using a full calculation, which of the salts BaSo, or PbSO4 will precipitate first? Given: Kep (BaSO4) = 1.1 x 10-10 and Kẹp (PbSO4) = 1.6 x 10-8 A voltaic cell is set up at 25 °C and delivers 0.060 V. One half cell contains an acidified solution of permanganate and manganese(II) ions, and the other half-cell is based on the following reaction: Br03 (aq) + 6H30*(aq) + 6e Br¯(aq) + 9H20 (f), Eed = 1.44 V
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