Edited: I was confused as whell, but I think I figured it out...
My understanding is that you're adding some specific number moles of NaOH (s) to a solution of 0.0001M Mg(NO3)2 (aq) ... and you only add enough moles so the NaOH(aq) concentration will be equal to 0.001M.
I'm assuming it's NaOH solid even though they didn't specifically put that -- that took me a few minutes to realize. If it's not solid, then, I dunno. I guess it's all just black magic lab techniques. Or maybe you just assume 0.0001M Mg(NO3)2 is the final concentration after adding NaOH. From the wording of the question, I would think it's a 0.0001M solution Mg(NO3)2 before you add NaOH. Or maybe I'm dumb. Anyways, assuming NaOH is added as a solid...
In this instance, you wouldn't have to worry about diluting Mg(NO3)2. The volume of the solution would barely change since it's such a tiny amount of solute being added. The solution volume could technically change if the solute(s) weren't so dilute. This was confusing me -- I mean, how would I know they're dilute enough to not affect overall solution volume? So, I worked some numbers out for this problem, just to help give myself a picture of what's happening...
For the NaOH, you're adding 0.04g NaOH per 1 liter solution. The solution you're adding it to has 0.0156g Mg(NO3)2 dissolved per 1 liter H2O.
(0.0156g Mg(NO3)2 * (1 mole Mg(NO3)2 / 148 g Mg(NO3)2 ) = 0.0001 moles Mg(NO3)2.
0.0001 moles Mg(NO3)2 / 1 Liter H2O to make a 0.0001M solution Mg(NO3)2.
1 Liter H2O is 1000g/L (~56moles of H2O). 0.0156grams solute just isn't going to affect volume to the extent that you'd have to take it into account. In terms of moles, 0.0001 moles Mg(NO3)2 per ~56 moles H2O.
And adding the 0.04g NaOH/1L solution just isn't going to affect the solution volume to any extent we'd have to worry about. Knowing that these are such tiny amounts is what made it, at least intuitively, click for me.
We know each solute has their own concentration, independent of other solutes, with respect to the overall solution volume. And, we know adding a pinch of the solutes won't affect the volume of the solution. So, adding this small amount of solute won't dilute the other one's concentration.
As you add more and more solute (like, a lot), then it'll start to make a difference in the volume of solution. If you had like 0.0001M Mg(NO3)2 and you added NaOH to a final concentration of 9bajillionM then I think we would need to need into account volume of solution and the dilution of Mg(NO3)2.
I do know in the cases where you're taking, like, 0.5M 500mL Mg(NO3)2 (aq) and 0.5M 500mL NaOH (aq) ... and combining them -- then that's when you need to take into account the dilution of the solution concentration. (Using M1V1 = M2V2). The reason you've gotta do that is because, ultimately, the volume of the solution is significantly changing (in this example, it's doubling 0.5L -> 1L) and the moles of the solutes is staying constant.
