"Grumpy" wrote in message
. au...

"Daniel Abranko" wrote in message
...
In article ,
"LLBrown" wrote:

Here is what goes on.... I bend a thin wall brass tube to shape then I
have
to melt the bending medium out of the pipe with a torch. This causes a
black scaling to show up on the outside of the pipe. To remove this I
dip
the pipe in a 50/50 mix of muratic acid and water. This in turn plates
the
pipe with very thin copper that I have to buff off.

I know the acid is leaching copper and plating later pipes but what is
the
science behind this? Why does it plate?

LLB in Laredo
\
I suspect that it really doesn't plate. What I think happens is that the
acid dissolves the zinc out of the surface of the brass, and leaves the
copper behind. Buffing it up removes the zinc depleted layer so it gets
back to original brass colour.

Hi Grumpy,
Here is what one of the earlier letters said:


I guess I admit to being a chemist but I'm really rusty on this kind of
stuff. First, pure hydrochloric acid will not dissolve copper metal, but
hcl with oxygen dissolved in it will, so that is one difference between
fresh acid and "used" stuff. The used stuff most likely does have copper
ions dissolved in it. Then, since zinc is more electrochemically active
than copper, when you put the copper ions in contact with the zinc metal the
zinc is oxidized and the copper will plate out. Now, at the same time the
hcl is chewing up the zinc as fast as it can so it comes down to how "used
up" the acid is. If the hcl is mostly consumed so it doesn't react with the
zinc very fast, and if the solution has picked up lots of dissolved air
(oxygen) in use and then used that to dissolve lots of copper along the way,
that copper will have time to plate out on the zinc before the remaining hcl
eats away the zinc. If you took fresh hcl and just added pure zinc until
the acid was mostly used up, it should not plate copper out if you then put
it on brass. This is all my take on this without doing any research to back
it up, ymmv, it's worth what you paid for it, etc ;-).

-----
Regards,
Carl Ijames