Copper Solubility in Water with pH 5

In summary, the conversation is about a task involving the calculation of the amount of copper dissolved in water due to corrosion of a copper pipe. The person asking for help has tried various methods and equations, including the Nernst equation, but has been unsuccessful. They provide a figure and their attempted steps for solving the problem. Another person joins the conversation, but is also unsure of how to help and questions the relevance of pH in the task.
  • #1
Kristina_elt
3
0
Hello,

I have a problem with this task:

"Water with pH 5 and reduction potential +0.30 V flows through copper pipe at 25 °C. Calculate how much copper dissolves (in ppm) in water due to corrosion of this pipe. Standard reduction potential of copper Cu0/Cu2+ is 0.34 V."

I spent lot of hours with lot of calculations without any glory :( Please, is there anybody who could try to help me?

This figure could be helpful:

http://www.advancedaquarist.com/2003/5/chemistry_album/figure3.gif

I tried to use Nernst equation:

Eh = E0 - (RT / nF)*(lnQ).

The main equation I used was:

Cu + 2H+ -> Cu2+ + H2.
 
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  • #2
Write Nernst equations for both half cells. At the equilibrium both reduction potentials will be identical. Potential of the hydrogen cell is easy to calculate, that leaves you with just one unknown - concentration of the copper.
 
  • #3
I do not think so :( ...
 
  • #4
Have you tried to follow what I wrote?

If so, can you show the steps?
 
  • #5
Borek said:
Have you tried to follow what I wrote?

If so, can you show the steps?

Ok. Firstly I wrote this:

Cu -> Cu2+ + 2e- (E0 = -0.34 V)
H2 -> 2H+ + 2e- (E0 = 0.00 V)

The following reaction was:

Cu + 2H+ -> Cu2+ + H2 (E0 cell = 0.34 V)

Next step:

0.3 V = 0.34 V - (RT / 2F) * ln ([Cu2+] / [10^-10])

Then I got:

3.11 = ln ([Cu2+] / [10^-10])

And for [Cu2+]:

22.42 * 10^-10 = approx. 2.24E-09

In ppm:

2.24E-09 * 1000 * 63.5 = 1.42E-04 ppm (I think this result is incorrect)
 
  • #6
Sorry. On the second read question no longer makes any sense to me. Apparently I was way too fast in my answer now (and I realized why I didn't answer the first time I have seen the question elsewhere).

I have no idea what they mean by "Water with pH 5 and reduction potential +0.30 V". Unless pH is completely irrelevant and the only thing that is important is the potential - regardless of its source (most likely some not mentioned, but present, redox system). If so, question can be solved by ignoring pH (just by applying Nernst equation to copper). If not - I have no idea what to do.
 

1. What is the solubility of copper in water with pH 5?

The solubility of copper in water with pH 5 is approximately 1.0 mg/L.

2. Why is the solubility of copper lower in acidic water?

Copper is less soluble in acidic water because the lower pH causes the formation of insoluble copper compounds, such as copper oxide or copper hydroxide, which precipitate out of the solution.

3. Can the solubility of copper in water with pH 5 be increased?

Yes, the solubility of copper in water with pH 5 can be increased by adding chelating agents, such as EDTA, which bind to the copper ions and prevent them from forming insoluble compounds.

4. What are the potential health risks of copper solubility in water with pH 5?

The solubility of copper in water with pH 5 is within the safe limit for drinking water set by the World Health Organization. However, high levels of copper in drinking water can cause gastrointestinal and liver problems in humans.

5. Does the temperature of water affect the solubility of copper at pH 5?

Yes, the solubility of copper in water with pH 5 increases with higher temperatures, as warmer water can hold more dissolved solids. However, this increase is not significant and does not greatly impact the overall solubility of copper in water with pH 5.

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