A Level Chemistry. Ionic Equilibria Qn

lein SG

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.

(a) In an experiment, 25.0cm^3 sample of the hydroxylammonium chloride solution was titrated against a solution of 0.100M sodium hydroxide, and 14.40 cm^3 of sodium hydroxide was needed for complete neutralisation.

(i) With the aid of an eqn, 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 x 10^(-8) M

Thanks in advance! please help me :(

UltimaOnline SG

~ 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.
(a) In an experiment, 25.0cm^{3 sample of the hydroxylammonium
chloride solution was titrated against a solution of 0.100M sodium
hydroxide, and 14.40 cm}3 of sodium hydroxide was needed for complete neutralisation.
(i) With the aid of an eqn, 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^(- Cool M ~

Points to note :

(Don’t worry about the counterbalancing ion Cl-, its 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 accept protons back, and become HOHN3+ or NH4+ again; to properly understand why, we have to look at both conjugate species, charge densities, stabilities, solute-solvent interactions and thermodynamics; not the focus of your question here. Suffice that students familiar with NH3 and NH4+ system are aware this is a weak base and its weak* conjugate acid, and extrapolate its similiarities to the HONH2 and HONH3+ system).

(*Many students just memorize without understanding “weak base = strong conjugate acid” without understanding, which is unfortunate (that most Sg students just memorize without understanding and appreciating what they study). I make sure my tuition students understand why, and therefore the limitations of applying such ideas. 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 to a relative degree, based on relative stabilities. But the fact is, while NH3 is a weak base, NH4+ certainly isn’t a strong acid. If you truly understand Chemistry, rather than blindly memorizing, this fact shouldn’t confuse you at all.)

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

Hence at equivalence point, the species present HONH2, will undergo hydrolysis (ie. reaction with water), in a Bronsted-Lowry acid-base proton transfer reaction, proton 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 how much OH- ions; so you can calcuate molarity of OH- ions, and thus pOH, and therefore pH.

As a general guideline for acid-base equilibria questions, do an ICF (Initial Change Final) table first, in number of moles; then do an ICE (Initial Change Equilibrium) table next, in molarities. Bear in mind when applying Ka or Kb expression, it is always in molarities of all species, and take care to factor in changing volumes at different stages (eg. 1st equivalence point, 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), but in water, no significant hydrolysis of the hydroxy group will occur. Treat this hydroxy group as you would an alcohol – neither strongly acidic or basic; inert as far as this titration question is concerned.

UltimaOnline SG

In case you’re still confused and need more tips,

~ (a) In an experiment, 25.0cm^{3 sample of the
hydroxylammonium chloride solution was titrated against a solution
of 0.100M sodium hydroxide, and 14.40 cm}3 of sodium hydroxide was needed for complete neutralisation. ~

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

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

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

For equivalence point, the no. of moles of the species HONH2 is exactly the same as the no. of moles of HONH3+ at initial, but the volume will be different (since you added NaOH solution), hence the molarity of HONH2 at equivalence point is certainly different.

Analogy (for a diprotic acid) to understand why no. of moles of the species is always the same, just not molarity :
You begun with 10 teddy bears with 2 arms (representing protons) each. During titration neutralization, you’re chopping the arms off the teddy bears. At 1st equivalence point, there are still 10 teddy bears, only now they’re all one-armed. At 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 the teddy bears remain the same throughout, the molarity of the teddy bears keep changing, as the volume of the solution keeps increasing (as you keep adding the alkali).