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.
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.
1 Answer
Explanation:
!! LONG ANSWER !!
The idea here is that adding sodium hydroxide,
You know that you start with
#"100.0 mL "-> 6.00 * 10^(-2)"M HCl"#
#"100.0 mL " -> 5.00 * 10^(-2)"M NaOH"#
Now, you know that after you realize your error, you're left with
#"100.0 mL " - " 81.0 mL" = "19.0 mL HCl"#
#"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
#"HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))#
Because hydrochloric acid and sodium hydroxide produce hydrogen cations,
#"H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l))#
Notice that this reaction consumes hydrogen cations and hydroxide anions in a
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
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 = "0 moles OH"^(-) -># completely consumed
#0.00114 - 0.000550 = "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_"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)|)))#
In your case, you have
#"pH" = - log(0.01967) = 1.71#
So, you know that your target solution must have a volume of
#"pH" = - log(["H"^(+)]) implies ["H"^(+)] = 10^(-"pH")#
#["H"^(+)] = 10^(-2.50) = "0.003162 M"#
For a volume equal to
Your solution contains
#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
This will give you a volume of
#color(green)(|bar(ul(color(white)(a/a)color(black)("volume of HCl needed " = " 42.9 mL")color(white)(a/a)|)))#
The answer is rounded to three sig figs.