# Question fee36

Oct 12, 2015

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, ${\text{H"_3"O}}^{+}$, is needed in order to hav a solution with a pH of $1$.

To do that, use the equation

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

Since nitric acid, ${\text{HNO}}_{3}$, is a strong acid, it will dissociate completely in aqueous solution to give

${\text{HNO"_text(3(aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "NO}}_{\textrm{3 \left(a q\right]}}^{-}$

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 $\text{0.1 M}$, since that would produce the desired concentration of hydronium ions.

To make the calculations easier, let's assume that your target solution will have a volume of $\text{1 L}$. That means that it must contain 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% ${\text{HNO}}_{3}$ solution will contain 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 $\text{18 g}$ of solution and add water until the volume of the target solution is equal to 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 $\text{1.214 g/mL}$

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 $\text{14.83 mL}$ out of your stock solution, and add enough water to make the volume of the target solution equal to one liter.

And now you have a $\text{1-L}$ ${\text{HNO}}_{3}$ solution that has a pH of $1$.

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.