posted by sp on .
One liter of approximetely 0.25N HNO3 has been prepared. Upon titration it was found that 5.0 ml of the acid required 11.0 ml of 0.10N NaOH for neutrality. How much concetrated HNO3 at 15.8N must be added to one liter to make it exactly 0.25N?
Technically, 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.