# Question 8bf92

Jun 18, 2017

$\text{pH} = 2.4$

#### Explanation:

The thing to remember about sulfuric acid is that you can assume that it acts as a strong acid in both ionizations, not just in the first one.

${\text{H"_ 2"SO"_ (4(aq)) + 2"H"_ 2"O"_ ((l)) -> color(red)(2)"H"_ 3"O"_ ((aq))^(+) + "SO}}_{4 \left(a q\right)}^{2 -}$

This implies that, for all intended purposes, you can say that a solution of sulfuric acid will always contain

$\left[{\text{H"_3"O"^(+)] = color(red)(2) * ["H"_2"SO}}_{4}\right]$

In your case, the concentration of hydronium cations will be equal to

["H"_3"O"^(+)] = color(red)(2) * "0.002 M"

["H"_3"O"^(+)] = "0.004 M"

Now, the $\text{pH}$ of the solution is given by the following equation

color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#

This means that your solution will have

$\text{pH} = - \log \left(0.004\right)$

$= - \log \left(4 \cdot {10}^{- 3}\right)$

$= - \left[\log \left(4\right) + \log \left({10}^{- 3}\right)\right]$

$= - \log \left(4\right) - \left(- 3\right) \cdot {\overbrace{\log \left(10\right)}}^{\textcolor{b l u e}{= 1}}$

$= 3 - \log \left(4\right)$

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{2.4}}}$

The answer is rounded to one decimal place, the number of sig figs you have for the concentration of the acid.