# Question 418b5

May 29, 2017

$\text{pH} = 1.618$

#### Explanation:

You know that

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

so your ultimate goal here is to figure out the concentration of hydronium cations in this solution.

Start by calculating the number of moles of sulfuric acid present in your solution by using the compound's molar mass

3.25 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"SO"_4)/(98.08color(red)(cancel(color(black)("g")))) = "0.03314 moles H"_2"SO"_4

Now, the molarity of the sulfuric acid is calculated by looking at how many moles of acid are present for every $\text{1 L}$ of solution.

1 color(red)(cancel(color(black)("L solution"))) * ("0.03314 moles H"_2"SO"_4)/(2.75color(red)(cancel(color(black)("L solution")))) = "0.01205 moles H"_2"SO"_4

This means that the sulfuric acid has a molarity of

["H"_2"SO"_4] = "0.01205 mol L"^(-1)

Now, it's very important to realize that sulfuric acid is a strong acid that is capable of donating two protons.

For all intended purposes, you can assume that every mole of sulfuric acid will ionize completely to produce hydronium cations and sulfate anions

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

Notice that every $1$ mole of sulfuric acid that ionizes produces $\textcolor{red}{2}$ moles of hydronium cations. This means that your solution will have

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

In your case, this will be equal to

["H"_3"O"^(+)] = color(red)(2) * "0.01205 mol L"^(-1)

["H"_3"O"^(+)] = "0.02410 mol L"^(-1)#

Therefore, you can say that the $\text{pH}$ of the solution will be equal to

$\text{pH} = - \log \left(0.02410\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.618}}}$

The answer is rounded to three decimal places, the number of sig figs you have for your values.