Answer to Question #247264 in General Chemistry for banne

Question

Calculate the number of milliliters of 0.618 M KOH required to precipitate all of the Cr³⁺ ions in 194 mL of 0.621 M CrCl₃ solution as Cr(OH)₃. The equation for the reaction is:

CrCl₃(aq) + 3KOH(aq) Cr(OH)₃(s) + 3KCl(aq)

Answer 1

CrCl₃(aq) + 3KOH(aq) $\to$  Cr(OH)₃(s) + 3KCl(aq)

Mole ratio of CrCl₃ and KOH is 1:3

Moles of Cr³⁺ = Molarity × Volume

= 0.621× 194 = 120.474 moles

Moles of KOH = 3× 120.474 = 361.422 moles

Hence Volume of KOH = moles of KOH/Molarity of KOH

= 361.422 moles/0.618M = 584.83 milliliters