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.

To solve this problem, we need to use the concept of titration and the equation for the neutralization reaction between HNO3 and NaOH.

The equation for the neutralization reaction is:

HNO3 + NaOH → NaNO3 + H2O

From the given information, we know that 5.0 ml of the 0.25N HNO3 solution requires 11.0 ml of 0.10N NaOH for neutrality.

Step 1: Calculate the amount of HNO3 in the 5.0 ml of 0.25N HNO3 solution.

The concentration (N) of a solution is defined as the number of equivalents (moles) of solute present in one liter of solution.

Given:
Volume of HNO3 solution = 5.0 ml = 0.005 L
Concentration of HNO3 solution = 0.25N

Number of equivalents (moles) of HNO3 = Concentration (N) × Volume (L)
= 0.25N × 0.005 L

Step 2: Calculate the amount of NaOH required to neutralize the HNO3 in the 5.0 ml of 0.25N HNO3 solution.

Given:
Volume of NaOH required = 11.0 ml = 0.011 L
Concentration of NaOH = 0.10N

Number of equivalents (moles) of NaOH = Concentration (N) × Volume (L)
= 0.10N × 0.011 L

Step 3: Calculate the ratio of HNO3 to NaOH in the neutralization reaction.

From the balanced equation, the stoichiometric ratio of HNO3 to NaOH is 1:1. This means that 1 mole of HNO3 reacts with 1 mole of NaOH.

Step 4: Calculate the amount of concentrated HNO3 (15.8N) needed to make the solution exactly 0.25N.

Let x be the volume of concentrated HNO3 (15.8N) required in liters.

Number of equivalents (moles) of HNO3 from the concentrated solution = Concentration (N) × Volume (L)
= 15.8N × x L

From the stoichiometric ratio, the number of equivalents (moles) of HNO3 is equal to the number of equivalents (moles) of NaOH.

Therefore:
15.8N × x L = 0.10N × 0.011 L

Solve for x:
x = (0.10N × 0.011 L) / (15.8N)

Now, substitute the given values:
x = (0.10 × 0.011) / 15.8
x ≈ 0.0000699 L

To convert the volume to milliliters:
x ≈ 0.0000699 L × 1000 ml/L
x ≈ 0.0699 ml (rounded to 4 decimal places)

Therefore, approximately 0.0699 ml of concentrated HNO3 (15.8N) must be added to one liter of the 0.25N HNO3 solution to make it exactly 0.25N.