Can anyone help? Calculate the [s2-] needed to precipitate CU2+ from a solution in which the [CU2+] is 0.0020 M.?

Refer to: http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf
Ksp for CuS = 8 × 10⁻³⁷
(Different sources may give different values of Ksp, and this may result in slightly different answer.)

CuS(s) ⇌ Cu²⁺(aq) + S²⁻(aq) …… Ksp = 8 × 10⁻³⁷

Precipitation occurs when [Cu²⁺][S²⁻] = Ksp
(0.0020 M) [S²⁻] = 8 × 10⁻³⁷
[S²⁻] = (8 × 10⁻³⁷) / 0.0020 M = 4 × 10⁻³⁴ M

First write down the BALANCED reaction equation
Cu^2+ + S^2- = CuS (s)
The molar ratios are 1:1::1 Hence it follows that the molar amounts are
0.002 : 0.002 :: 0.002
So you need '0.002 moles of sulphide ions to precipitate the copper ions.
Mass of (S) is = 0.002 x 32 = 0.064 g

Cu+2 at 0.002 M means you have 0.002 moles/L. You need an equivalent amount of S-2 to precipitate out this much Cu+2.

If you had, for example 0.04 M S-2, you would need 50 mL to get all of the S-2 in 1 L