If the original sodium tartrate solution had a concentration of 1.00 M instead of 0.100 M, what would the concentration of copper be in a solution created by combining 10 mL of it with 10 mL of the 0.100 M copper (II) sulfate solution.
The Ksp is 1.37 x 10 ^-3

Here is the work i tried to do:
CuSO4 is highly soluble so we know that it dissociates to Cu2+ and SO4 2- in equal amounts, 0.001moles Cu2+ and 0.001moles SO4 2-

for CuC4H4O6, Ksp = 0.0013 = [Cu2+][C4H4O6 2-] / [CuC4H4O6]
0.0013 = x^2 / 1M - x
x^2 + 0.0013x - 0.0013 = 0
x = 0.036M Cu 2+