Let us assume that Cu(OH)2(s) is completely insoluble, which signifies that the precipitation reaction with NaOH(aq) (presented in the transition) would go to completion.Cu2+(aq)+2NaOH(aq) → Cu(OH)2(s)+2Na+(aq)If you had a 0.150 L solution containing 0.0250 M of Cu2+(aq), and you wished to add enough 1.31 M  NaOH(aq) to precipitate all of the metal, what is the minimum amount of the NaOH(aq) solution you would need to add? Assume that the NaOH(aq) solution is the only source of OH−(aq) for the precipitation.Express the volume to three significant figures and include the appropriate units.

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Asked Nov 18, 2019
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Let us assume that Cu(OH)2(s) is completely insoluble, which signifies that the precipitation reaction with NaOH(aq) (presented in the transition) would go to completion.

Cu2+(aq)+2NaOH(aq) → Cu(OH)2(s)+2Na+(aq)

If you had a 0.150 L solution containing 0.0250 M of Cu2+(aq), and you wished to add enough 1.31 M  NaOH(aq) to precipitate all of the metal, what is the minimum amount of the NaOH(aq) solution you would need to add? Assume that the NaOH(aq) solution is the only source of OH−(aq) for the precipitation.

Express the volume to three significant figures and include the appropriate units.
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Expert Answer

Step 1

Given that,

Volume of Cu2+ is 0.150 L

Molarity of Cu2+ is 0.0250 M

Molarity of NaOH is 1.31 M

The given reaction is

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Cu2 (aq)+2NaOH->Cu(OH), (s) +2Na" (aq)

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Step 2

It means 1 mole of Cu2+ react with 2 mole of NaOH

Now, you calculate the no.of mole of Cu2+ and NaOH

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no.of mole molarityxvolume mole x.150L L no.of mole of Cu2-0.0250 -0.00375 mole 1 mole of Cu2 react with 2 mole of N2OH Hence,0.00375 mole Cu2 react with no.of mole of N2OH-0.00375x2 -0.00750 mole

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Step 3

Now, you calculate the...

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no.of mole of NaOH Volume of NaOH= Molarity of NaOH 0.00750 mole mole 1.31 L -0.00286 L

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