Question

Many metal ions are precipitated from a solution by the sulfide ion. As an example, consider treating a solution of copper(II) sulfate with sodium sulfide solution: CuSO4(aq) + Na2S(aq) → CuS(s) + Na2SO4(aq) What volume of 0.104 M Na2S solution would be required to precipitate all of the copper(II) ion from 13.4 mL of 0.120 M CuSO4 solution? Volume = mL Submit Answer Try Another Version 7 item attempts remaining

Answer 1

From the given molar concentration of Cusou solution we can calculate number of moles of cuson that will react with As Molarity - Number of moles of solute . - Volume of solution in So No. of moles of cuso, can be calculated as: No. of moles of cusou (olarity) + (volume in) Given: Molarity of cusou solution 0.120m oro.120 mol/L (M= mol/L is unit of molarity) . Volume of Cuson solution : 13 um - - 134 L30.0134 L 1000 sor Because IL=1000ml) No. of moles of cusou (0.120 mol/L) x (0.01344) 20:0016 moles i According to the given equation: - Cusont was cast na sou Imole of Cusou reacts with mole of was to produce I mole of cus. If mole of Cusly reacts with Imole of was then 0.0016 moles of casou will react completely with o.2016 moles of was.

Thus all of Cufiion of cusou will precipitate if all 0.0016 moles of Casca react with 0.0016 moles of Nags قلعه هعنهعقطکیوند Molarity of Thus, using Was solution - 0.104m we can write

Volume of ways Solution - No. of moles of ways - Molarity = 0.0016 mot 0.104 mol/L = 0.0154 L = 0.0154 x 1000 ml =lis.umL