A 3.650-g sample containing bromate and bromide was dissolved in sufficient wate

Question

A 3.650-g sample containing bromate and bromide was dissolved in sufficient water to give 250.0 mL. After acidification, silver nitrate was introduced to a 25.00-mL aliquot to precipitate AgBr , which was filtered, washed, and then redissolved in an ammoniacal solution of potassium tetracyanonickelate(II):

            Ni(CN)42- + 2AgBr(s) à 2Ag(CN)2- + Ni2+ + 2Br-

The liberated nickel ion required 21.23 mL of 0.02089 M EDTA. The bromate in a 10.00-mL aliquot was reduced to bromide with arsenic(III) prior to the addition of silver nitrate. The same procedure was followed, and the released nickel ion was titrated with 25.61 mL of the EDTA solution. Calculate the percentages of NaBr and NaBrO3 in the sample.

Explanation / Answer

The amount of NaBr + NaBrO3 = 3.65 grams

The amount of water = 250mL

the aliquot taken = 25mL

The liberated Ni+2 required = 21.23mL of 0.02089 M EDTA

So moles of EDTA = molarity X volume = 21.23 X 0.02089 = 0.443 millimoles of EDTA

Moles of EDTA = Moles of Ni+2

As per stoichiometry of given reaction

Ni(CN)42- + 2AgBr(s) --> 2Ag(CN)2- + Ni2+ + 2Br-

For each mole of N+2 released 2 moles of Br- are produced

So moles of Bromie ion = 2 X 0.443 millimoles

These moles are present in 25mL of aliquot so moles present in 250mL of aliquot = 8.86 millimoles

Mass of bromide ion = millimoles X atomic weight = 8.86 X 79.9 = 707.9 milligrams = 0.708 grams

Mass of NABr = 8.86 X 102.9 = 911.69 millimgrams = 0.912 grams

The reaction of arisenic (III) with bromate ion will be

BrO3- + 3As(III) --> Br- + 3As(V)

Again the bromide ion is reacted with nickel complex

Moles of EDTA used = 25.61 X 0.02089 = 0.535 millimoles

Ni+2 = 0.535 millimoles

Corresponding Br- = 0.535 X 2

Bromide ion already present in 10mL = 0.354 millimoles

corresponding BrO3- = 0.535 X 2 - 0.354 = 0.716 millimoles in 10mL

So bromate ion in 250 mL =17.9 millimoles

Mass of bromate ion = moles X molecular weight = 17.9 X 127.9 = 2289.41 milligrams = 2.289 grams

Mass of sodium bormate = 17.9 X 150.9 = 2701 milligram = 2.701 grams (approx)

The % of sodium bromide = Mass of sodium bromide X 100 / total mass = 0.912 X 100 / 3.65 = 24.98 %

The % of sodium borate = Mass X 100 / total mass = 2.701 X 100 / 3.65 = 73.97 % (approx)

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