# Given the following information and assuming the final solution will be diluted to 1.00L, how much more HCl should you add to achieve the desired pH?

## Imagine that you are in chemistry lab and need to make 1.00L of a solution with a pH of 2.50. You have in front of you 100 mL of 6.00×10−2M HCl, 100 mL of 5.00×10−2M NaOH, and plenty of distilled water. You start to add HCl to a beaker of water when someone asks you a question. When you return to your dilution, you accidentally grab the wrong cylinder and add some NaOH. Once you realize your error, you assess the situation. You have 81.0mL of HCl and 89.0mL of NaOH left in their original containers.

Aug 15, 2016

$\text{42.9 mL}$

#### Explanation:

The idea here is that adding sodium hydroxide, $\text{NaOH}$, to your hydrochloric acid solution will neutralize some, if not all, depending on how much you've added, of the acid.

$\text{100.0 mL "-> 6.00 * 10^(-2)"M HCl}$

$\text{100.0 mL " -> 5.00 * 10^(-2)"M NaOH}$

Now, you know that after you realize your error, you're left with $\text{81.0 mL}$ of hydrochloric acid and $\text{89.0 mL}$ of sodium hydroxide. This means that you've added to the resulting solution

$\text{100.0 mL " - " 81.0 mL" = "19.0 mL HCl}$

$\text{100.0 mL " - " 89.0 mL" = "11.0 mL NaOH}$

When you mix hydrochloric acid, a strong acid, and sodium hydroxide, a strong base, a neutralization reaction takes place

${\text{HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

Because hydrochloric acid and sodium hydroxide produce hydrogen cations, ${\text{H}}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$, respectively, in a $1 : 1$ mole ratio, you will have

${\text{H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O}}_{\left(l\right)}$

Notice that this reaction consumes hydrogen cations and hydroxide anions in a $1 : 1$ mole ratio, which means that for every mole of hydrochloric acid present it takes one mole of sodium hydroxide to neutralize it.

Use the molarities and volumes of the solutions you've mixed to calculate how many moles of each were added

19.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace((6.00 * 10^(-2)"moles HCl")/(1color(red)(cancel(color(black)("L")))))^(color(blue)(=6.00 * 10^(-2)"M")) = "0.00114 moles H"^(+)

11.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace((5.00 * 10^(-2)"moles NaOH")/(1color(red)(cancel(color(black)("L")))))^(color(blue)(=6.00 * 10^(-2)"M")) = "0.000550 moles OH"^(-)

So, you know that you've accidentally added $0.00114$ moles of hydrogen cations and $0.000550$ moles of hydroxide anions to your solution.

Since you have fewer moles of hydroxide anions present, you can say that the hydroxide anions will be completely consumed by the neutralization reaction, i.e. they will act as a limiting raegent.

Your resulting solution will thus contain

$0.000550 - 0.000550 = {\text{0 moles OH}}^{-} \to$ completely consumed

$0.00114 - 0.000550 = {\text{0.000590 moles H}}^{+}$

Now, the volume of this solution will be equal to the volume of hydrochloric acid and the volume of sodium hydroxide solutions you've mixed

${V}_{\text{sol" = "19.0 mL" + "11.0 mL" = "30.0 mL}}$

This means that the concentration of hydrogen cations in the resulting solution will be

["H"^(+)] = "0.000590 moles"/(30.0 * 10^(-3)"L") = "0.01967 M"

The pH of the solution is given by

color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"^(+)])color(white)(a/a)|)))

$\text{pH} = - \log \left(0.01967\right) = 1.71$

So, you know that your target solution must have a volume of $\text{1.00 L}$ and a pH of $2.50$. Use the above equation to find the concentration of hydrogen cations needed to have this target solution

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

["H"^(+)] = 10^(-2.50) = "0.003162 M"

For a volume equal to $\text{1.00 L}$, this gives you $0.003162$ moles of hydrogen cations -- remember that when dealing with a liter of solution, molarity and number of moles of solute are interchangeable.

Your solution contains $0.000590$ moles of hydrogen cations and needs $0.003162$ moles, which means that you must add

n_("H"^(+)"needed") = 0.003162 - 0.000590 = "0.002572 moles H"^(+)

Use the molarity of the stock hydrochloric acid solution to see what volume would contain this many moles of acid

0.002572 color(red)(cancel(color(black)("moles H"^(+)))) * "1.0 L"/(6.00 * 10^(-2)color(red)(cancel(color(black)("moles H"^(+))))) = "0.0429 L"

This is equivalent to $\text{42.9 mL}$, which means that in order to prepare your target solution, you must add another $\text{42.9 mL}$ of stock hydrochloric acid solution to the existing $\text{30.0 mL}$ solution.

This will give you a volume of $\text{72.9 mL}$ of solution. To get your target solution, you must add enough distilled water to get the total volume to $\text{1000 mL}$.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{volume of HCl needed " = " 42.9 mL}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.