Single-reagent analyte solutions: 0.20 M NaBr (aq) 0.20 M Nal (aq) 0.20 M NaCl (

Question

Single-reagent analyte solutions: 0.20 M NaBr (aq) 0.20 M Nal (aq) 0.20 M NaCl (aq) 0.20 M Na2S04 (aq) 2.0 M Na2COs (aq) Multiple-reagent analyte solutions Known: aqueous solution containing a mixture of equal volumes of all five analytes Unknown A: aqueous solution of only two, three or four of the five analytes Unknown B (numbered): aqueous solution of only two, three or four of the five analytes Testing reagent solutions 6 M HNO3 (aq) 0.2 M Ba(NOs)2 (aq) 0.2 M AgNO3 (aq) 6 M NH3 (aq) 15 M NH3 (aq

Explanation / Answer

Question 4 is incomplete

Questio 3:

As the solution is prepared by mixing equal volume of each of the given solution to make a total volume of 1mL

it means we have taken 0.2mL of each solution

So moles of each salt taken = Molarity X volume

a) concentration of bromide ion :

Moles of NaBr added = Moles of bromide ion = Molarity X volume = 0.2 X 0.2 mL = 0.04 millimoles

[Br-] = Moles / total volume = 0.04 mmoles / 1 mL = 0.04 M

b) concentration of chloride ion :

Moles of NaCl added = Moles of chloride ion = Molarity X volume = 0.2 X 0.2 mL = 0.04 millimoles

[Cl-] = Moles / total volume = 0.04 mmoles / 1 mL = 0.04 M

c) concentration of iodide ion :

Moles of NaI added = Moles of Iodide ion = Molarity X volume = 0.2 X 0.2 mL = 0.04 millimoles

[I-] = Moles / total volume = 0.04 mmoles / 1 mL = 0.04 M

d) concentration of sulphate ion :

Moles of Na2SO4 added = Moles of sulphatee ion = Molarity X volume = 0.2 X 0.2 mL = 0.04 millimoles

[SO4-2] = Moles / total volume = 0.04 mmoles / 1 mL = 0.04 M

e) concentration of carbonate ion :

Moles of Na2CO3 added = Moles of carbonate ion = Molarity X volume = 2 X 0.2 mL = 0.4 millimoles

[CO3-2] = Moles / total volume = 0.4 mmoles / 1 mL = 0.4 M

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