The flask shown here contains 10.0 ml of HCl and a few drops of phenolphthalein indicator. The buret contains 30ml of 0.130M NaOH.

What Volume of NaOH is needed to reach the end point of the titration?

I've used
m1v1=m2v2,

(x)(10.0ml HCl) = (0.130 M/L NaOH) (0.03L)

and have gotten 3.9ml. But I know that I am wrong.

m1v1 = m2v2 technically isn't the one to use but it will work in this case. The REAL problem is that you don't have the concentration of the HCl. Without that you're dead in the water. You work the problem this way IF you had the molarity of the HCl.

mols HCl = M x L = ?M x 0.0100
mols NaOH = mols HCl from line 1.
M NaOH = mols NaOH/L NaOH. YOu know M and mols, solve for L and convert to mL if desired.

To find the volume of NaOH needed to reach the end point of the titration, you can use the equation molarity1 x volume1 = molarity2 x volume2.

In this case, you need to determine the volume of NaOH (molarity2 x volume2) required to neutralize the given volume of HCl (molarity1 x volume1).

Let's break it down step by step:

1. Determine the number of moles of HCl using the given volume and concentration:
Moles of HCl = volume1 x molarity1

In your case, volume1 = 10.0 mL and since HCl is assumed to be a strong acid, the molarity can be assumed to be 1.00 M.

Moles of HCl = 10.0 mL x (1.00 mol/L) = 0.0100 mol

2. Since HCl and NaOH react in a 1:1 ratio according to the balanced equation, the number of moles of NaOH required to react completely with the HCl is also 0.0100 mol.

3. Now, let's calculate the volume of NaOH needed using its molarity and the number of moles required:
volume2 = moles NaOH / molarity2

In your case, moles NaOH = 0.0100 mol and molarity2 = 0.130 M.

volume2 = 0.0100 mol / (0.130 mol/L) = 0.0769 L = 76.9 mL

Therefore, the correct volume of NaOH needed to reach the end point of the titration is 76.9 mL.

It seems like you made a calculation error using your given values, which is why you obtained an incorrect answer of 3.9 mL.