# How many milliliters of sulfuric acid is required to make an aqueous solution with a pH of 6 if the solution is 0.065 L from a bottle of 12M sulfuric acid?

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To solve for millileters of sulfuric acid, use these measurements:

Molar mass of sulfuric acid: #98.09# #g# /#mol#

Density of sulfuric acid: #1.84# #g# /#cm^3#

#1# #cm^3 = # #1# #mL#

To solve for millileters of sulfuric acid, use these measurements:

Molar mass of sulfuric acid:

Density of sulfuric acid:

##### 1 Answer

Here's what I got.

#### Explanation:

The way I understand the problem, you have to determine what volume of a

If this is the case, then you don't really need the density of the stock solution. You can treat this problem as a classic dilution calculation.

The first thing to do here is determine exactly how many moles of sulfuric acid are needed to make the target solution. To do that, use the solution's pH to determine the concentration of protons in solution

#"pH" = -log(["H"^(+)]) implies ["H"^(+)] = 10^(-"pH")#

You will thus have

#["H"^(+)] = 10^(-6)"M"#

Assuming that you're supposed to take *both* of sulfuric acid's ionizations are being complete, you can say that

#"H"_2"SO"_text(4(aq]) -> color(red)(2)"H"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#

Notice that **every mole** of sulfuric acid produces **moles** of hydrogen ions in solution. This means that the concentration of sulfuric acid, so to speak, in the target solution will be **half** of that of the hydrogen ions.

#color(red)(2)["H"^(+)] = ["H"_2"SO"_4] implies ["H"_2"SO"_4] = 1/2 * ["H"^(+)]#

So, you need the concentration of sulfuric acid to be equal to

#["H"_2"SO"_4] = 0.5 * 10^(-6)"M"#

Use the molarity and volume of the target solution to find how many moles of sulfuric acid you have in the target solution

#color(blue)(c = n/V implies n =c * V)#

#n = 0.5 * 10^(-6)"M" * "0.065 L" = 0.0325 * 10^(-6)"moles H"_2"SO"_4#

Now all you have to do is figure out what volume of the stock solution would contain this many moles of sulfuric acid

#color(blue)(c = n/V implies V = n/C)#

#V_"stock" = (0.0325 * 10^(-6)color(red)(cancel(color(black)("moles"))))/(12color(red)(cancel(color(black)("moles")))/"L") = 0.00271 * 10^(-6)"L"#

Expressed in *milliliters* and rounded to two sig figs, the answer will be

#V_"stock" = color(green)(2.7 * 10^(-3)"mL")#