It is difficult to reliably filter, dry and weighthe solid CuI produced. Therefo

Question

It is difficult to reliably filter, dry and weighthe solid CuI produced. Therefore an alternative (volumetric)method is used in this analysis for copper.

The potassium triiodide (KI3) produced,can be quantitatively converted back to potassium iodide (KI) bythe addition of exactly 19.28 mL of0.8634-M potassium thiosulfate(K2S2O3) solution.

This volume is determined by the controlledaddition of the standardized potassium thiosulfate solution. Theend-point of this titration is detected by thedisappearance of the colour associated withI3- ion. This is facilitated by the additionof a small quantity of dilute starch solution. Starch reversiblyforms a complex with I3- which is dark bluein colour.

What was the molarity of the initialCuSO4 solution?

What is the mass of the cuprous iodide (CuI)precipitate?

2CuSO4(aq) + 5KI(aq) 2CuI(s) + KI3(aq) +2K2SO4(aq)

Explanation / Answer

The reaction between potassium triiodide and potassiumthiosulfate can be represented as          KI3(aq) + 2K2S2O3(aq) 3KI (aq) +K2S4O6(aq) The number of moles of potassium thiosulfate consumed inthe titration =( 19.28 mL * 1 L / 1000 mL) * 0.8634 M                                                                                                           = 0.0166 mol K2S2O3 From the equation 1 mol KI3 reacts with 2 molK2S2O3 The number of moles of KI3 that would reactwith 0.0166 mol K2S2O3                                                    = (0.0166 mol K2S2O3 * 1mol  KI3) / 2 molK2S2O3                                                    =0.0083 mol KI3   2CuSO4(aq) + 5KI(aq) 2CuI(s) + KI3(aq) +2K2SO4(aq) 2 mol CuSO4 produced 1 molKI3 The number of moles of CuSO4 that would berequired to produce 0.0083 mol KI3                                                     =(0.0083 mol KI3 * 2 mol CuSO4) / 1 molKI3                                                    = 0.0166 mol The volume of CuSO4 solution taken = 20.0mL The molarity of CuSO4 solution = 0.0166 mol /20.0 * 10-3 L                                                  =0.83 M 2 mol CuSO4 produced 2 mol CuI 0.0166 mol CuSO4 will produce 0.0166 molCuI The molar mass of CuI = 190.45 g/mol The mass of CuI = 0.0166 mol * 190.45 g/mol                          = 3.16 g CuI                                                    = 0.0166 mol The volume of CuSO4 solution taken = 20.0mL The molarity of CuSO4 solution = 0.0166 mol /20.0 * 10-3 L                                                  =0.83 M 2 mol CuSO4 produced 2 mol CuI 0.0166 mol CuSO4 will produce 0.0166 molCuI The molar mass of CuI = 190.45 g/mol The mass of CuI = 0.0166 mol * 190.45 g/mol                          = 3.16 g CuI

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