In a solution of 0.10 M H2SO4, the ions present in order of decreasing order.

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In a 0.10 M H2SO4 solution, the predominant ions present are H3O+, HSO4-, SO4^2-, and OH-, in that order. The strong acid dissociates completely, leading to a high concentration of H3O+ and HSO4-. The large Ka value for H2SO4 indicates that it fully ionizes, making an ICE table unnecessary for this calculation. The logic behind the order of ions is based on their concentrations rather than exact calculations. Understanding the dissociation of strong acids clarifies the ion distribution in the solution.
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Homework Statement


The answer is [ H3O+ ] > [HSO4-]>[SO4-2]>[OH-], but I do not understand how they got it?


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The Attempt at a Solution


First I tried to put H2SO4 into water and got HSO4-. Then I tried to make an ice table, but it didn't work out since Ka for H2SO4 = very large?

H2SO4 + H2O <====> HSO4- + H3O+
 
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It is a problem in logic. You don't need to do an exact calculation.
 
Thread 'Confusion regarding a chemical kinetics problem'

Thread 'Confusion regarding a chemical kinetics problem'

TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
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