Neutralization and Percipitation, URGENT HELP PLEASE!?

Question

Ok I've got less than 1 hour to answer this question, I've got 6.42x10^-3 mol of Cu^+2, it takes 1.61 mL of 16.0 M HNO3 to completely dissolve this.

If I add 5 mL 16.0 M HNO3 to the already dissolved Cu in HNO3, how much 4.00 M NaOH will be required to neutralize the acid AND turn all the Cu+2 into Cu(OH)2???
Answers I know are incorrect:
2.32x10^-2 L
1.32x10^-2 L
2.96x10^-2 L

Help within the next hour would be SOOO amazing, I would be so grateful!

Ok I've got less than 1 hour to answer this question, I've got 6.42x10^-3 mol of Cu^+2, it takes 1.61 mL of 16.0 M HNO3 to completely dissolve this.

If I add 5 mL 16.0 M HNO3 to the already dissolved Cu in HNO3, how much 4.00 M NaOH will be required to neutralize the acid AND turn all the Cu+2 into Cu(OH)2???
Answers I know are incorrect:
2.32x10^-2 L
1.32x10^-2 L
2.96x10^-2 L

Help within the next hour would be SOOO amazing, I would be so grateful!

edit: 2.625x10^-2 L is incorrect, thanks anyways cattbarf, but all my chances have run out, now I would just like to know the answer to know the answer... and please don't use milimoles, it gets confusing with too many unit conversions.

Answer 1

Inventory time: (all results are appx)
We added 6.61 mL of 16 M HNO3 to the Cu+2.
This is appx 105 millimoles of HNO3.
To provide nitrate for 6.42 millimoles of Cu+2 requires 12.8 millimoles HNO3
So 92.2 millimoles is in solution as acid

To achieve goals,
92.2 millimoles of NaOH neutralizes nitric acid.
12.8 millimoles to "turn Cu+2 cation from the nitrate to the hydoxide"
You need 105 millimoles of NaOH. Given 4 molar solution, you would need about 2.6x10-2 L of solution.