# Question #fee36

##### 1 Answer

Here's what you could do.

#### Explanation:

The first thing to do in order to solve this problem is figure out what concentration of *hydronium ions*,

To do that, use the equation

#["H"_3"O"^(+)] = 10^(-"pH") = 10^(-1) = "0.1 M"#

Since *nitric acid*,

#"HNO"_text(3(aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "NO"_text(3(aq])^(-)#

This means that *every mole* of nitric acid will produce *1 mole* of hydronium ions in solution. This means that you need the nitric acid solution to have a molarity of

To make the calculations easier, let's assume that your target solution will have a volume of **0.1 moles** of nitric acid.

Use nitric acid's molar mass to determine how many grams of nitric acid would contain that many moles

#0.1color(red)(cancel(color(black)("moles HNO"_3))) * "63.01 g"/(1color(red)(cancel(color(black)("mole HNO"_3)))) = "6.301 g HNO"_3#

Now, a **35 g** of nitric acid *for every* **100 g** of solution. Use the solution's percent concetration by mass to determine what mass of stock solution would contain **6.301 g** of nitric acid

#6.301color(red)(cancel(color(black)("g HNO"_3))) * "100 g solution"/(35color(red)(cancel(color(black)("g HNO"_3)))) = "18.003 g solution"#

You can now weigh out about **one liter**.

Alternatively, you can figure out what volume of the stock solution would contain that many grams of nitric acid. Use the solution's density, which at room temperature is given as

http://www.handymath.com/cgi-bin/nitrictble2.cgi?submit=Entry

#18.003color(red)(cancel(color(black)("g solution"))) * "1 mL"/(1.214color(red)(cancel(color(black)("g solution")))) = "14.83 mL"#

So, you would take out a volume of **enough water** to make the volume of the target solution equal to **one liter**.

And now you have a

**SIDE NOTE** *As practice, try getting different volumes for the target solution and see how that affects how much stock solution you need to use.*