Question #3478a

Sulfuric acid, #H_2SO_4#, is a diprotic acid, which means that it has two hydrogens that can get ionized.

Sulfuric acid's first ionization is considered strong, the compound being fully dissociate into the hydrogen sulfate ion, #HSO_4^(-)# and hydronium ion, #H_3O^(+)#.

#H_2SO_(4(aq)) + H_2O_((l)) -> HSO_(4(aq))^(-) + H_3O_((aq))^(+)#

As you can see, 1 mole of sulfuric acid will produce 1 mole of hydronium ions in solution. As a result, the concentration of hydronium and hydrogen sulfate ions after the first hydrogen is ionized will be

#[H_3O^(+)] = [HSO_4^(-)] = [H_2SO_4]#

The acid dissociation constant for the second ionization, #K_(a2)# is listed as being equal to #1.2 * 10^(-2)#. This implies that #HSO_4^(-)# will act as a weak acid, the following equilibrium being established

#HSO_(4(aq))^(-) + H_2O_((l)) rightleftharpoons SO_(4(aq))^(2-) + H_3O_((aq))^(+)#

To solve for the concentration of hydronium ions, you have to use an ICE table

#" "HSO_(4(aq))^(-) + H_2O_((l)) rightleftharpoons SO_(4(aq))^(2-) + H_3O_((aq))^(+)#
I........0.056.................................0...................0.056
C.........(-x)..................................(+x)...................(+x)
E......0.056 -x ..............................x..................0.056 + x

Notice that the starting concentration for the hydronium ions is not zero, since this equilibrium follows the strong first ionization, which set the concentration of #H_3O^(+)# to the initial concentration of the sulfuric acid.

By definition, the acid dissociation constant will be equal to

#K_(a2) = ([H_3O^(+)] * [SO_4^(2-)])/([HSO_4^(-)]) = 1.2 * 10^(-2)#

#1.2 * 10^(-2) = ((0.056 + x) * x)/(0.056 - x)#

#x^2 + 0.068*x - 6.72 * 10^(-4) = 0#

Solving this equation for #x# will produce two values, one positive and one negative. Since #x# is meant to express concentration, the negative solution cannot be used. As a result,

#x = 0.008755#

Therefore, the total concentration of the hydronium ions will be

#[H_3O^(+)]_"total" = [H_3O^(+)]_"1st ionization" + [H_3O^(+)]_"2nd ionization"#

#[H_3O^(+)]_"total" = = 0.056 + 0.008755 = "0.064755 M"#

The pH of the solution will be

#pH_"sol" = -log([H_3O^(+)]) = -log(0.064755) = color(green)(1.19)#