Question #d1234

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"#

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#