# Is common ion effect applicable on two strong electrolytes

Consider the nitration of benzene using mixed conc. $$\ce{H2SO4}$$ and $$\ce{HNO3}$$.If a large amount of $$\ce{KHSO4}$$ is added to the mixture the rate of nitration will be :

1. Faster
2. Slower
3. Unchanged
4. doubled

The answer given in my textbook is option 2 , i.e. the reaction will become slow. my teacher told me that it is due to common ion effect.

But in common ion effect, a strong electrolyte decrease the dissociation of weak electrolyte. So,how can common ion effect possibly be the reason here($$\ce{KHSO4}$$ and $$\ce{H2SO4}$$ both are strong electrolytes). In my view the answer should be no effect and I'd love to have my concepts about Ionic Equilibrium be corrected if I'm wrong.

• I just want to commend you for self-removing the homework tag.
– A.K.
Mar 10 '19 at 19:40
• I removed it because then people won't even read my question and just move forward even if I would have mentioned my thoughts in the question. I can add it again though but before I do that I'll ask you for a bit of help. It's been quite a while using stack exchange and I'm still not able to figure out that how do you guys subscript and superscript and how do you make sentence appear in yellow box to highlight it. Please help me with it and I'll add the Homework tag again. Mar 11 '19 at 3:21
• Don't add the Homework tag. WE are working to perminately remove it from the site. Here are some Meta posts that may help you get started with equations: chemistry.meta.stackexchange.com/q/86/23561 and a more advanced markup post chemistry.meta.stackexchange.com/q/3044/23561
– A.K.
Mar 11 '19 at 3:37

Certainly not.

Suppression of dissociation of a strong electrolyte is not a very rational thing to discuss, owing to the unmeasurably large value of $$K$$.

To see what I mean, try looking up the $$K_{a1}$$ of $$\ce{H2SO4}$$. (At least I couldn't find it!)

The equilibrium suppressed by the introduction of $$\ce{KHSO4}$$ is not the dissociation of $$\ce{H2SO4}$$, but rather this one:
$$\ce{HNO3} \ + \ \ce{H2SO4} \rightleftharpoons \ce{NO2+} \ + \ \ce{HSO4-} \ + \ \ce{H2O}$$
PS. Conclusion: $$\ce{KHSO4}$$ sucessfully decreases the concentration of $$\ce{NO2+}$$, which happens to be the electrophile and a key participant in the rds(rate determining step) of Nitration of benzene.
$$Rate=k[\ce{NO2+}][\ce{C6H6}]$$