# Question d1234

Feb 4, 2015

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

${H}_{2} S {O}_{4 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s H S {O}_{4 \left(a q\right)}^{-} + {H}_{3}^{+} {O}_{\left(a q\right)}$, ${K}_{a 1} = 2.4 \cdot {10}^{6}$

and

$H S {O}_{4 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s S {O}_{4 \left(a q\right)}^{2 -} + {H}_{3}^{+} {O}_{\left(a q\right)}$, ${K}_{a 2} = 1.0 \cdot {10}^{- 2}$

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

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

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

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

SInce pH is calculated as

$p {H}_{\text{sol}} = - \log \left(\left[{H}^{+}\right]\right)$, you can easily determine that

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

This means that molarity of the sulfuric acid is

$\left[{H}_{2} S {O}_{4}\right] = \frac{\left[{H}^{+}\right]}{2} = \text{0.000136 M"/2 = 1.58 * 10^(-4)"M}$

Feb 4, 2015

The concentration = $1.58 \times {10}^{- 4} m o l / l$

${H}_{2} S {O}_{4} \rightarrow 2 {H}^{+} + S {O}_{4}^{2 -}$

$\left[p H\right] = 3.5$

So

$- \log \left[{H}^{+}\right] = 3.5$

$\log \left[{H}^{+}\right] = - 3.5$

${10}^{- 3.5} = \left[{H}^{+}\right]$

$\left[{H}^{+}\right] = 3.16 \times {10}^{- 4} m o l / l$

Since sulfuric acid is 2 molar with respect to ${H}^{+}$ the concentration of ${H}_{2} S {O}_{4} = \frac{3.16 \times {10}^{- 4}}{2} = 1.58 \times {10}^{- 4} m o l / l$