Nitrate (NO_2^-) can be determined by oxidation with excess Ce^4+, followed by b
ID: 944561 • Letter: N
Question
Nitrate (NO_2^-) can be determined by oxidation with excess Ce^4+, followed by back titration of the unreacted Ce_4^+. A 30.47 g of solid sample containing NaN02 (FW 68.995 g/mol) and NaNO_3 (FW 84.994 g/mol) was dissolved in 500.00 mL. A 25.00 mL of the solution was treated with 50.00 mL of 0.1106 M Ce^4+ in strong acid for 5 min, and the excess Ce^4+ was back titrated with 31.13 mL of 0.04289 M FeSO_4. Please calculate the weight percent of NaNO_2 in the solid. Here are the two reactions involved in the titration: 2 Ce^4+ + NO_2^- + H_2O -> 2 Ce^3+ + NO_3^- + 2H^+ Ce^4+ + Fe^2+ -> Ce^3+ + Fe^3+ 85.42% 30.23% 11.70% 9 498%Explanation / Answer
Millimoles of Ce4+ taken for titration = MxV = 0.1106M x 50.00 mL = 5.53 millimol
The reaction of Ce4+ with Fe2+ is
Ce4+ + Fe2+ ----- > Ce3+ + Fe3+
1 mol, 1 mol, ------- 1 mol, 1 mol
millimoles of Fe2+ required for back titration = MxV = 0.04289 M x 31.13 mL = 1.33517 millimol
Also from the above reaction it is clear that,
millimoles of excess Ce4+ = millimoles of Fe2+ required for back titration = 1.33517 millimol
Hence millimoles of Ce4+ actually reacted with NO2- ion
= initial millimol of Ce4+ - millimoles of excess Ce4+
= 5.53 millimol - 1.33517 millimol = 4.195 millimol Ce4+
The reaction of Ce4+ with NO2- ion is
2Ce4+ + NO2-+ H2O ---- > 2Ce3+ + NO3- + 2H+
2 mmol, 1 mmol--------------- 2 mmol, 1 mmol
2 millimoles of Ce4+ reacts with 1 millimol of NO2- ion.
Hence 4.195 millimol Ce4+ that will react with the moles of NO2- ion
= 4.195 millimol Ce4+ x (1 millimol of NO2- ion / 2 millimoles of Ce4+)
= 2.0975 millimol of NO2- ion
Hence moles of NaNO2 in 25.00 mL of the sample = 2.0975 millimol of NaNO2 = 2.0975x10-3 mol
Hence mass of NaNO2 in 25.00 mL of the sample = 2.0975x10-3 mol x 68.995 g/mol = 0.1447 g
Hence mass of NaNO2 in 500.00 mL of the sample
= (0.1447 g / 25.00 mL) x 500.00 mL
= 2.8943 g
Total weight of solid = 30.47 g
Hence weight percent of NaNO2 in the solid = (2.8943 g / 30.47 g) x 100 = 9.498 % (answer)
Hence last option is correct.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.