# Question #418b5

##### 1 Answer

#### 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**

#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

#"H"_ 2"SO"_ (4(aq)) + 2"H"_ 2"O"_ ((l)) -> color(red)(2)"H"_ 3"O"_ ((aq))^(+) + "SO"_ (4(aq))^(2-)#

Notice that every **mole** of sulfuric acid that ionizes produces **moles** of hydronium cations. This means that your solution will have

#["H"_3"O"^(+)] = color(red)(2) * ["H"_2"SO"_4]#

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

#"pH" = - log(0.02410) = color(darkgreen)(ul(color(black)(1.618)))#

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