Conductometric titration is a type of titration in which the electrical conductivity of the reaction — Physical Chemistry — Kinetics Chemistry Question
Conductometric determination of ammonium nitrate and nitric acid
Conductometric titration is a type of titration in which the electrical conductivity of the reaction mixture is continuously monitored as one reactant is added. The equivalence point in such titration is determined by the change in electrical conductivity of the solution. Marked jumps of conductance are primarily associated with changes of concentrations of the two most highly conducting species, hydrogen and hydroxyl ions. The method can be used for titrating colored solutions or suspensions, the latter being impossible with color indicators. Electrical conductivity measurement is used as a tool to locate the endpoint.
Industrial production of ammonium nitrate involves the acid-base reaction of ammonia with nitric acid. Conductometric titration can be used to control the residual concentration of nitric acid in the solution after the reaction with ammonia.
In this work you will perform a conductometric titration of a mixture of nitric acid and ammonium nitrate.
Chemicals
• HNO3, solution in water, ~1 mol dm-3
• NH3(aq), solution in water, ~1 mol dm-3
• NaOH(aq), solution in water, ~1 mol dm-3
• NaCl, solid, 0.6 g
Equipment and Glassware
• Conductivity meter
• Analytical balance (± 0.0001 g)
• Burette
• Volumetric pipettes, 10, 15 and 25 cm3
• Pipette bulb or pump
• Magnetic stirrer
• Stirring bar
• Volumetric flasks, 100 cm3 (5 each)
• Glass beaker, 100 cm3
Procedure
a) Place ammonia and nitric acid solutions into three 100- cm3 volumetric flasks marked A, B, and C in quantities indicated in the hereunder table. Fill the flasks with deionized water up to the mark and mix thoroughly.
[VISUAL]
b) Transfer 25.0 cm3 of solution A into a glass beaker using a 25 cm3 transfer pipette.
c) Titrate the sample solution with a standardized solution of NaOH (~1 mol dm–3, known exactly) by adding 0.2 cm3 portions of the titrant. After adding each titrant portion, stir the solution. Record the value of the electric conductivity when it becomes constant.
d) Titrate the sample solution until the conductivity starts to rise (add a few more titrant portions to be able to draw a straight line).
e) Repeat steps (b – d) for solutions B and C.
f) Transfer 20 cm3 of HNO3 and 10 cm3 of NH3 solutions into each of volumetric flasks D and E. Fill the flasks up to the mark and mix thoroughly. For flasks filling, use distilled (instead of deionized) water for D and deionized water containing 0.6 g of NaCl for E.
g) Repeat steps (b – d) for solutions D and E.
Give balanced chemical equations for the reactions taking place when the titrant is added.
Model Answer
Equilibria in the system can be described by the following equations:
H+ + OH– ⇆ H2O (1)
NH4+ + OH– ⇆ NH3 + H2O (2)
Draw the titration curve in the coordinates “electrical conductivity – volume of titrant” for all the solutions studied (A – E). How many breaks of titration curves should be observed? Explain the resulting dependences. Which curves are practically the same and why?
Model Answer
Conductivity of a solution is primarily dependent on the concentration of H+ and OH– ions (species with the highest mobility) as well as on that of salts. Solutions A and B contain the same amount of NH4NO3 (solution A with an excess of ammonia reveals a bit higher conductivity). On the titration curves, there are monotonously descending portions reflecting the displacement of the weak base (NH3) from its salt (reaction 2). Minimum conductivity is reached when the concentration of protons appearing from NH4+ hydrolysis is minimal (reaction 2 completed). This is further changed by a sharp rise corresponding to the increasing excess of alkali.
In the case of solution C, the first descending portion is steeper (than those for A and B) and is associated with diminishing concentration of free protons coming from HNO3. The first equivalence point causes a sharp break of the curve (reaction 1 completed). The second descending portion characterized by a lower slope reflects the displacement of the weak base from its salt (reaction 2). Minimum conductivity is also reached when reaction 2 is completed, which is followed by a sharp rise of conductivity due to the alkali excess.
The difference between cases C, D, and E is due to various levels of conductivity caused by the salts that are not titrated with NaOH.
[VISUAL]
Draw straight lines through the linear portions of the titration curves. Find the inflection points as the abscissa values corresponding to the intersections of the lines.
Model Answer
Refer to the plots provided in the solution of part 29.2. Straight lines are drawn through the linear portions of the titration curves to determine the inflection points as the intersections of these lines.
Calculate the concentrations of nitric acid and ammonium salt using these inflection points for each case. Compare the results with those calculated from the known amounts of HNO3 and NH3.
Model Answer
Calculations can be done in the same way as for a regular acid-base titration, using titrant volumes in inflection points VNaOH(1), VNaOH(2):
cH+Vsample = cNaOHVNaOH(1),
cNH4+ Vsample = cNaOH · (VNaOH(2) – VNaOH(1))
Examples:
For A and B: If 2.45 cm3 of NaOH solution (c = 0.9987 mol dm–3) were spent until the inflection point was reached, then cNH4+ = 0.9987 × 2.45 / 25 = 0.0979 mol dm–3.
For C – E: If 2.40 cm3 of NaOH solution (c = 0.9987 mol dm–3) were spent until the first inflection point was reached (the neutralization of HNO3) in a 25.0 cm3 sample aliquot, then cHNO3 = 0.9987 × 2.40 / 25 = 0.0895 mol dm–3. If the second inflection point was reached at 4.85 cm3, then cNH4+ is: 0.9987 × (4.85 – 2.40) / 25 = 0.0979 mol dm–3.
Using the obtained results, predict the curve shape for the titration of a mixture of sodium hydroxide and free ammonia with HCl.
Model Answer
HCl first neutralizes the strong base, which is followed by neutralization of the weak one. So, the titration curve of a mixture of two bases reveals two breaks. NaOH neutralization is accompanied by a linear decrease of conductivity due to lowering concentration of highly mobile hydroxyl ions. After the first equivalence point, conductivity starts increasing due to the formation of a well dissociating salt (a strong electrolyte) as a result of ammonia (a weak electrolyte) neutralization. After the second equivalence point, conductivity of the solution sharply increases due to the excess of hydrogen ions.
[VISUAL]