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Sunny_sXe [5.5K]
3 years ago
7

Of the metals pb, mg, au, and na, which will not spontaneously donate electrons to copper in solution?

Chemistry
2 answers:
Schach [20]3 years ago
7 0
Au :)

Pb + Cu(2+)---> Pb(2+) + Cu
Mg + Cu(2+)---> Mg(2+) + Cu
2Na + Cu(2+) ---> 2Na(+) + Cu

Au + Cu(2+) ---> (this reaction is not possible)
QveST [7]3 years ago
3 0

Answer:

Au will not spontaneously donate electrons to copper in solution.  

Explanation:

Based on the activity series of metals:

K > Na > Ca> Mg> Al> Zn>Fe> Sn > Pb> H> Cu> Hg> Ag> Au  

Among the given metals Pb, Mg, Au and Na, the most reactive metals is Na, followed by Mg, Pb and Au.  All three metals Pb, Mg, and Na are above Cu in the activity series. So these metals will spontaneously donate electrons to Cu in solution. Au being lower than Cu in the activity will not spontaneously donate electrons to copper in solution.  

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Consider a solution prepared by mixing the following:50.0 mL of 0.100 M Na3PO4100.0 mL of 0.0500 M KOH200.0 mL of 0.0750 M HCl50
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Answer:

you must add 50 mL

Explanation:

Hi

KOH is a strong base and by adding 100mL 0.05M you will have an amount of 5 millimol.

NaCN is a base and by adding 50 mL 0,150 M you will have an amount of 7,5 mmol.

HCl is a acid and by adding 200 mL 0,075 M you will have an amount of 15 mmol.

The acid reacts with the bases leaving 2.5 mmol unreacted.

Na3PO4 is a base and by adding 50 mL 0,1 M you will have an amount of 5 mmol.

The 2.5 mmol of acid react with the base PO4 ^ -3 forming a regulatory solution of PO4 ^ -3 and HPO4 ^ -2 of pKa 2.12

5 mmol of acid (HNO3) must be added to obtain a regulatory solution formed by the same amount of HPO4 ^ -2 (2.5 mmol) and H2PO4 ^ -1 (2.5 mmol) with pKa 7.21

Considering a quantity of 5 mmol of HNO3 of concentration 0.1 M, 50 mL must be added.

To calculate the pH of the regulatory solution you should consider pH = pKa × Ca / Cb pH = 7.21 × 2.5 / 2.5 = 7.21 Being in the same solution the volume is the same and can be simplified to achieve a faster calculation.

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