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April 19, 2014

April 19, 2014

Posted by **sp** on Wednesday, September 8, 2010 at 8:52pm.

- chemistry -
**DrBob222**, Wednesday, September 8, 2010 at 11:06pmTechnically, I don't think the problem can be worked exactly because the volumes may/may not be additive but here is what I would do.

mL x N = mL x N

11.0 x 0.1 = 5N

Solve for N = 0.22N which is

0.22 me/mL or 0.22 me/mL x 1000 mL = 220 me in the 1 L.

We want 0.25N which is

0.25 me/mL x 1000 or 250 me in the 1L.

The difference is 250-220 = 30 milliequivalents.

Now back to the mL x N = mL x N

?mL x 15.8 = 30

solve for ?mL.

The glitch is that if this is ADDED to 1L of the 0.22N stuff, the total volume will not be 1L, rather it will be 1 L + the volume of 15.8N HNO3 added. If I work it out the final concn ends up being 0.2495, which to two significant figures (the 0.10N has two s.f.) that rounds to 0.25 N.

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