# Question #3478a

Apr 20, 2015

Sulfuric acid, ${H}_{2} S {O}_{4}$, 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, $H S {O}_{4}^{-}$ and hydronium ion, ${H}_{3} {O}^{+}$.

${H}_{2} S {O}_{4 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} \to H S {O}_{4 \left(a q\right)}^{-} + {H}_{3} {O}_{\left(a q\right)}^{+}$

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

$\left[{H}_{3} {O}^{+}\right] = \left[H S {O}_{4}^{-}\right] = \left[{H}_{2} S {O}_{4}\right]$

The acid dissociation constant for the second ionization, ${K}_{a 2}$ is listed as being equal to $1.2 \cdot {10}^{- 2}$. This implies that $H S {O}_{4}^{-}$ will act as a weak acid, the following equilibrium being established

$H S {O}_{4 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s S {O}_{4 \left(a q\right)}^{2 -} + {H}_{3} {O}_{\left(a q\right)}^{+}$

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

$\text{ } H S {O}_{4 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s S {O}_{4 \left(a q\right)}^{2 -} + {H}_{3} {O}_{\left(a q\right)}^{+}$
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 ${H}_{3} {O}^{+}$ to the initial concentration of the sulfuric acid.

By definition, the acid dissociation constant will be equal to

${K}_{a 2} = \frac{\left[{H}_{3} {O}^{+}\right] \cdot \left[S {O}_{4}^{2 -}\right]}{\left[H S {O}_{4}^{-}\right]} = 1.2 \cdot {10}^{- 2}$

$1.2 \cdot {10}^{- 2} = \frac{\left(0.056 + x\right) \cdot x}{0.056 - x}$

${x}^{2} + 0.068 \cdot x - 6.72 \cdot {10}^{- 4} = 0$

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

${\left[{H}_{3} {O}^{+}\right]}_{\text{total" = [H_3O^(+)]_"1st ionization" + [H_3O^(+)]_"2nd ionization}}$

${\left[{H}_{3} {O}^{+}\right]}_{\text{total" = = 0.056 + 0.008755 = "0.064755 M}}$

The pH of the solution will be

$p {H}_{\text{sol}} = - \log \left(\left[{H}_{3} {O}^{+}\right]\right) = - \log \left(0.064755\right) = \textcolor{g r e e n}{1.19}$