Question

why second ionization constant of sulphuric acid in water much less than it's first ionization constant?​

Answer 1

Answer:

For the second de-protonation you are trying to remove a proton from something that is already negatively charged, a process far more difficult than removing a proton from something that is neutral.

Alternatively you can think that the −2 charged SO42− anion attracts really strongly the H+ resulting in the equilibrium being towards HSO4−, making it a weak acid compared to H2SO4

It is similar thinking to why the second ionization energies are always much higher than the first.

Note that even before HSO4− dissociates, there is an equal concentration of H+ ions already present in the solution. We know, by Le Chatelier's principle, that the increased concentration of the products favors the reactants. Hence for the second dissociation, since H+ is one of the products, and is in large concentration, the second dissociation is actually favored in the reverse direction. Thus, deprotonation of HSO4−is not favorable.Hence, the Ka2 of H2SO4 is significantly lower than its Ka1.

H2SO4(aq)⟶H+(aq)+HSO4−(aq);Ka1=1×103

HSO4−(aq)⟶H+(aq)+SO42−(aq);Ka2=1.2×10−2

Answer 2

Second ionization constant of sulphuric acid in water is much less than it's first ionization constant:

• The second ionization constant of sulphuric acid in water is much less than its first ionization constant.
• It is due to the fact that in the latter it already donated a proton in the water while in the former it is quite difficult to take another proton from the negatively charged anion $HSO4^-$.
• According to Lewis concept of acids and bases, acids are good proton donors and accepts electron pair.

To know more:

While preparing dilute sulphuric acid from concentrated sulphuric acid in laboutry the concentrated sulphuric acid is added slowly to water with constant stirring because?

https://brainly.in/question/3613262