'A' Level Chemistry Problem Analysis: Titrating Hydroxylamine Chloride Against Sodium Hydroxide

Hydroxylamine NH2OH is a toxic substance to many aquatic organisms, and it has chemical properties similar to ammonia. For example, it reacts with hydrochloric acid to form the salt hydroxylammonium chloride NH3OHCl.

In an experiment, 25.0 cm3 sample of the hydroxylammonium chloride solution was titrated against a solution of 0.100M sodium hydroxide, and 14.40 cm3 of sodium hydroxide was needed for complete neutralisation.

(i) With the aid of an equation, determine if the nature of the end point of titration is neutral, basic or acidic.

(ii) Calculate the PH value at the equivalence point of the titration. The base dissociate value, Kb of hydroxylamine is 1.10×10^-8 mol/dm3.

Thought process:

(Don’t worry about the counterbalancing Cl- ion as it's just a spectator ion; it doesn’t undergo hydrolysis much because of its low charge density and stability from its strong ion-dipole interactions with water).

The HONH3+ ion is obviously acidic, as it has an available proton to donate, and no lone pairs available to accept protons. It’s a relatively weak acid though, not a strong one, just like NH4+ losing protons to become NH3 (since HONH2, like NH3, will again have the tendency to re-accept protons, thus becoming HOHN3+ or NH4+ again; to properly understand why this happens, we have to look at a variety of factors including both conjugate species, charge densities, stabilities, solute-solvent interactions and thermodynamics - obviously not the focus of this particular problem. Suffice to say that students familiar with NH3 and NH4+ system should be aware this is a weak base and its weak* conjugate acid respectively, and consequently be able to extrapolate its similiarities to the HONH2 / HONH3+ system).

(*Many students unfortunately simply memorize without having a deeper understanding of the implication “weak base = strong conjugate acid”, which is symptomatic of Singapore's rote learning culture ingrained deep within learners ). I always make it a point to ensure my tuition students fully appreciate underlying key concepts, by extension the limitations of applying them. For instance, understanding why Markonikov’s Rule usually works (ie. stability of carbocation intermediate species), will enable you to recognize and understand when it actually fails (ie. when the major product is actually the anti-Markonikov product). Back to the “weak base = strong conjugate acid” idea, it’s definitely true but only to a relative degree, this being based on relative stabilities. The irrefutable fact however is, while NH3 is a weak base, NH4+ certainly isn’t a strong acid. If you truly understand Chemistry, rather than blindly memorizing stuff, you shouldn't feel confounded at all by what I just mentioned.)

Titrating HONH3+ (again, you can ignore the counterbalancing spectator Cl- ion) against NaOH, a strong alkali, we end up with HONH2 (ie. deprotonated conjugate base form of HONH3+) at equivalence point (regarding them spectator ions Na+Cl-, ignore these harmless buggers) which is obviously basic given its 2 remaining protons are not at all acidic at this stage/form (note the conjugate base of HONH2, which is HONH-, is exceedingly unstable), with an available lone pair on the N atom being able to accept a proton from water.

Hence at the equivalence point, HONH2 will undergo hydrolysis (ie. reaction with water), in a Bronsted-Lowry acid-base proton transfer reaction, where a proton is transferred from water to HONH2 to generate HONH3+ and OH-.

Therefore, your Initial Change Equilibrium (ICE) table will focus on to what extent (as specified by the Kb value) hydrolysis will occur to generate OH- ions; you can thus calculate the molarity of these OH- ions based on their quantity determined herein, subsequently their pOH and pH as well.

As a general guideline for acid-base equilibria questions, start off by drafting an ICF (Initial Change Final) table in terms of number of moles; thereafter do an ICE (Initial Change Equilibrium) table, in terms of molarities. Bear in mind when applying the Ka or Kb formula, all species are always expressed in molarities, and exercise care to factor in changing volumes at different stages (eg. the 1st equivalence point, the 2nd equivalence point, etc) when calculating molarities.

Also note that the hydroxy group of HONH2 / HONH3+ is neither strongly acidic nor basic. Although there can be reaction pathways to forcefully protonate or deprotonate the hydroxy group (after being done with the amine group), no significant hydrolysis of the hydroxy group will occur in water. Treat this hydroxy group as you would an alcohol – neither strongly acidic or basic; inert as far as this titration problem is concerned.

Also, in case an exam problem requires you to calculate the initial pH, you would therefore need to be able to compute the initial molarity of HONH3+.

Quoting from the problem: "In an experiment, 25.0cm3 sample of the hydroxylammonium chloride solution was titrated against a solution of 0.100M sodium hydroxide, and 14.40 cm3 of sodium hydroxide was needed for complete neutralisation."

Find the number of moles of OH- ions required for complete neutralization.

This is thus equivalent to the number of moles of the weak monoprotic HONH3+ acid present.

Using the formula Molarity = Number of Moles / Volume (dm3), you can discover the initial molarity of the species HONH3+.

At equivalence point, the number of moles of the species HONH2 is exactly the same as the initial number of moles of HONH3+, note though owing to the difference in volume (since NaOH solution was added), the molarity of HONH2 is certainly going to be different.

Providing an analogy (for a diprotic acid) to understand why number of moles of the species always remain the same, just not its molarity : You began with 10 teddy bears with 2 arms (representing protons) each. During titration neutralization, you’re chopping the arms off the teddy bears. At the 1st equivalence point, there are still 10 teddy bears, only now they’re all one-armed. At the 2nd equivalence point, there are still 10 teddy bears, only now they all have no arms left. Bear in mind while the number (of moles) of teddy bears remain the same throughout, the molarity of the teddy bears keep changing owing to ever increasing volume of solution (as you continually add an alkali).

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The above content is contributed by Mr Heng, owner and 'A' Level Chemistry tutor at Bedok Funland JC. He also goes by the handle UltimaOnline on various online popular homework forums.