What volume of 0.1292 M NaOH is needed to neutralize 25.00 mL of HCl of unknown concentration?
This is a titration curve question. I know that the initial pH of the hydrochloric acid is 1.65. I know that you can find the concentration of HCl by plugging it into the pH formula. So,
#-log[H+]=1.65#
#10^-1.65=0.02239=[H+]# .
However, that's as far as I've gotten. I know the volume of the 0.1292 M NaOH is between 22 and 23 mL, but I can't seem to figure out how to find the exact value. Any help is greatly appreciated!
This is a titration curve question. I know that the initial pH of the hydrochloric acid is 1.65. I know that you can find the concentration of HCl by plugging it into the pH formula. So,
However, that's as far as I've gotten. I know the volume of the 0.1292 M NaOH is between 22 and 23 mL, but I can't seem to figure out how to find the exact value. Any help is greatly appreciated!
1 Answer
Here's what I got.
Explanation:
Your first step is correct because the first thing that you need to do here is to use the
As you know, hydrochloric acid is a strong acid, which implies that it ionizes completely in aqueous solution to produce hydronium cations,
This means that the hydrochloric acid solution has
#["HCl"] = ["H"_3"O"^(+)]#
and since
#["H"_3"O"^(+)] = 10^(-"pH")#
you can say that
#["HCl"] = 10^(-1.65) quad "M"#
Now, sodium hydroxide and hydrochloric acid neutralize each other in a
#"HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))# Since sodium hydroxide is a strong base, i.e. it dissociates completely in aqueous solution to produce hydroxide anions, the neutralization reaction can be described using the net ionic equation
#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#
This tells you that in order to neutralize
#25.00 color(red)(cancel(color(black)("mL solution"))) * (10^(-1.65) quad "moles HCl")/(10^3color(red)(cancel(color(black)("mL solution")))) = 2.500 * 10^(-3.65) quad "moles HCl"#
This means that the sodium hydroxide solution must contain
#2.500 * 10^(-3.65) color(red)(cancel(color(black)("moles NaOH"))) * (10^3 quad "mL solution")/(0.1292 color(red)(cancel(color(black)("moles NaOH")))) = 1.935 * 10^(0.35) quad "mL"#
Rounded to four sig figs, the answer would be
#color(darkgreen)(ul(color(black)("volume NaOH solution = 4.332 mL")))#
Based on the values that you have provided here, the volume of the sodium hydroxide cannot be between
#color(white)( (color(black)("the more concentrated solution " ->)/(color(black)("the less concentrated solution " ->)))) (0.1292 color(red)(cancel(color(black)("M"))))/(0.022387color(red)(cancel(color(black)("M")))) = color(blue)(5.771)#
times more concentrated than the hydrochloric acid solution. Since the two reactants neutralize each other in a
In this case, you have
#color(white)( (color(black)("the less concentrated solution " ->)/(color(black)("the more concentrated solution " ->)))) (25.00 color(red)(cancel(color(black)("mL"))))/(4.332color(red)(cancel(color(black)("mL")))) = color(blue)(5.771)#
which confirms that the volume of the sodium hydroxide solution is indeed equal to