posted by OLGA .
Some solid copper(II) sulfate is dissolved in sufficient water to make 100.0 mL of solution. Enough of a sodium hydroxide solution of concentration 0.1500 mol/L is added just to react with all of the first solution, with none left over. This turns out to be a volume of 47.08 mL. (a)calculate the mass of the solid copper(II) sulfate. (b)calculate the end concentrations of all dissolved ions.
Part a is a stoichiometry problem. Just follow the steps at this link.
The equation is
CuSO4 + 2NaOH ==> Cu(OH)2 + Na2SO4
Make an ICE chart. The equilibrium conditions will be moles/L. Don't forget that the final volume is 147.08 mL. Also, the (CU^+2) will be determined from the solubility product of Cu(OH)2.
Post your work if you get stuck.