A) faster
B) slower
C) unchanged
D) doubled
Correct Answer: B
Solution :
Slower, as large amount of \[\text{HS}{{\text{O}}_{4}}^{-}\]will decrease ionisation of \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\] that result in lesser ionization of nitric acid and lesser formation of nitronium ion \[[N{{O}_{2}}^{+}]\]You need to login to perform this action.
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