# What volume of 0.1292 M NaOH is needed to neutralize 25.00 mL of HCl of unknown concentration?

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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 **mole ratio**.

#"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 thenet ionic equation

#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#

This tells you that in order to neutralize **mole** of hydrochloric acid, you need **mole** of sodium hydroxide. To find the number of moles of hydrochloric acid present in the hydrochloric acid solution, use the molarity and the volume of the sample.

#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 **moles** of sodium hydroxide in order for the neutralization to be complete. Once again, use the molarity of the sodium hydroxide solution to figure out the volume of the sample.

#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 **mole ratio**, it follows that the volume of the **less concentrated** solution must be **times higher** than the volume of the more concentrated solution in order for the two solutions to contain **the same number of moles** of each reactant.

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