Question

Question #d1234

Answer 1

Sulfuric acid is a strong diprotic acid that dissociates in aqueous solution in two steps, each with a different acid dissociation constant

`H_2SO_(4(aq)) + H_2O_((l)) rightleftharpoons HSO_(4(aq))^(-) + H_3^(+)O_((aq))`, `K_(a1) = 2.4 * 10^(6)`

and

`HSO_(4(aq))^(-) + H_2O_((l)) rightleftharpoons SO_(4(aq))^(2-) + H_3^(+)O_((aq))`, `K_(a2) = 1.0 * 10^(-2)`

Since no mention of these constants is made, I assume that the second step is ignored and sulfuric acid's dissociation is expressed as

`H_2SO_(4(aq)) rightleftharpoons 2H_((aq))^(+) + SO_(4(aq))^(2-)`

One mole of sulfuric acid will produce twice as much moles of protons in aqueous solution, which means that

`[H^(+)] = 2 * [H_2SO_4]`

SInce pH is calculated as

`pH_("sol") = -log([H^(+)])`, you can easily determine that

`[H^(+)] = 10^(-pH_("sol")) = 10^(-3.5) = "0.000316 M"`

This means that molarity of the sulfuric acid is

`[H_2SO_4] = ([H^(+)])/2 = "0.000136 M"/2 = 1.58 * 10^(-4)"M"`

Answer 2

The concentration = `1.58xx10^(-4)mol//l`

`H_2SO_4rarr2H^++SO_4^(2-)`

`[pH]=3.5`

So

`-log[H^+]=3.5`

`log[H^+]=-3.5`

`10^(-3.5)=[H^+]`

`[H^+]=3.16xx10^(-4)mol//l`

Since sulfuric acid is 2 molar with respect to `H^+` the concentration of `H_2SO_4=(3.16xx10^(-4))/2=1.58xx10^(-4)mol//l`