Question

This is a hydrolysis practice, and you have to use the decision table to write the net ionic equation.

For 0.10 NH3, the pH is around 10 making it a weak base. I dont understnad why we determine that Nh3 cannot donate a H+ to H20, but can only accept H+ from water? Aside from the reason that Nh3 being an acid will create a H30 (which is what we don't want because the pH of Nh3 is 10), the conceptual reason is that NH3 has a lone pair and thus doesn’t want to donate a H cause that would make it more unstable. It also makes sense cause the pH is 10 so NH3 wouldn’t want to donate H+ and create a H30+. But why is HSO4- able to donate H+, when it has many lone pairs? Obviously, it has a H here and thus it can donate, but I don't want to reason this out using the H availability, cause Nh3 obviously has H to donate, but it cannot here.

I have also seen NH2 too, so jsut basically, why cannot NH3 donate H+ into NH2 + H30 (aside from the reason-its pH 10)?

H H acceptor from H2O donor to H20 Molecule YesNo 0.10 M NH3 eme NIE

Answer 1

your question is at pH = 10 , Why acts as a weak base. in Hb. But it cannot donate Ht to Ho to make it H₂ot & become NH , but can accept 4 from water Clearly, the reason is, in H₂O, NH₃ is a stronger base than 40 [pky of Witz is 4.75]. Also if we see the stability of the conjugate acide Conjugate pha value Base Conjugate bases, ph Acid Conjugate base H2O -1.74 9-24 H₂O + NH3 & a WHAT ve value of pka of hot implies it is a strong acid its conjugate bare H₂O is stable e value of NH, oka of will implies it's weak acid to boos its conjugate base Nl is less stable compared to the value of NH, pka of In other words, no in reverse way we can say availakility of love pair of H₂O for donation is less compared to that of love pair of NH3 Since it is more prone to donnte lone pair (or accept H+) it is a stronger a base compared to H2o. So casi simply it won't just lose an He & become NH3 & make thot waste as both NH & Hot are unstable compared to NH₂ & H₂O In seeper level, electronegativity of N (3:0) is less than that of o(3:5). So positive charge on o is less stable them on N. (Although there are two in H₂O love pairs on Oxygend, it is still more electronegative)

In case of ASO , the donation of Ht can be explained by the stability of its Coningate base. The lone pair Oxygen can delocalise into vacant a orbitals of sulphur. HSO HP + 509 of 2- ge koos soos etc. The Conjugate base is resonance stabilized. Also, altor first broton donation the of H₂SO4 , the acidity to becomes quite less. ie. HSO4 is a fairly a weak base compared to H₂SO4.