Question

A sample is prepared by adding 750 mL of 4.00 × 10–3 M Ce(NO3)3 solution to 300 mL of 2.00 × 10–2 mol L–1 KIO3 solution. Will any Ce(IO3)3 (Ksp = 1.9 × 10–19) precipitate be observed for the final solution? Show all working.

Answer 1

Ce(NO3)3 +3 KIO3 =Ce(IO3)3 +3KNO3 .........................................................................1 that means one molecule of Ce(NO3)3 nutralised ( consumed) 3 molecules of KIO3.. and produce one molecule of Ce(IO3)3

so from the given data how much Ce(IO3)3 can be prepare we can measure .

750 ml of 4.00x 10^-3 M/L = (4.00x 10^-3 X750)/1000 = 3X10^-3 Mol . of Ce(NO3).............................2

300 ml of 2x10^-2 M/L = (300x2x10^-2)/1000 = 6x10^-3 mol of KIO3 ...........................................3 (as in one liter contain 2x10^-2 moles)

Ce(NO3)3 and KIO3 both are strong electrolyte . so they are fully ionised in sol .

and produced Ce(NO3)3 Ce³⁺ + 3 NO3 ^1-

KIO3 K¹⁺ + IO3^1-

  concentration of Ce³⁺ is (3x10^-3)/(750+300) = 2.8x10^-6 (eq2 ) M and concentration of IO3^- is (6x10^-3)/(750+300)= 5.7x10^-6 M ion multiplication [Ce][3IO3]³ = (2.8x10^-6)(3x5.7x10^-6) ³ =0.14x10^-19

as ion multiplication< K_sp .. so it will not precipitate .. as 0.41x10^-19,< 1.9x10^-19 (ksp)