Question

Question #0365a

Answer 1

`"50 mL"`

Explanation:

!! RELATIVELY QUICK ANSWER !!

Calculate the `pK_a` of the acid and compare it with the pH of the target solution

`color(blue)(pK_a = - log(K_a))`

`pK_a = - log(1.75 * 10^(-5)) = 4.76`

Since the pH of the target solution is approximately equal to the `pK_a` of the acid, you can assume that the resulting solution will contain equal concentrations of the weak acid and of its conjugate base, the acetate anion `->` think Henderson - Hasselbalch equation.

Since you know that

• acetic acid reacts in a `1:1` mole ratio with sodium hydroxide to produce hydronium cations and acetate anions in `1:1` mole ratios
• the acetic acid and sodium hydroxide solutions have the same molarity
• the volume of the acetic acid solution is equal to `"0.1 L"`

you can say that adding a volume of sodium hydroxide that is half that of acetic acid will neutralize half of the number of moles of acid present in solution and produce equal concentration of weak acid and conjugate base.

Therefore, you must add `"0.050 L " = " 50 mL"` of sodium hydroxide solution to get a pH that is approximately equal to `4.74`.

`color(white)(a)`

!! A LITTLE MORE EXPLANATION !!

The thing to remember about reacting weak acids with strong bases is that the reaction produces the conjugate base of the acid.

In this case, you have

`"CH"_3"COOH"_text((aq]) + "OH"_text((aq])^(-) -> "CH"_3"COO"_text((aq])^(-) + "H"_2"O"_text((l])`

Here one mole of acetic acid reacts with one mole of hydroxide anions provided by the strong base to produce one mole of acetate anions, `"CH"_3"COO"^(-)`, the conjugate base of the acid.

Now, if the neutralization is incomplete, i.e. if you don't add enough strong base to neutralize all the weak acid, you will produce a buffer solution.

The pH of a buffer solution can be calculated using the Henderson - Hasselbalch equation

`color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))`

When you have equal concentrations of weak acid and conjugate base, the log term will be equal to zero, since `log(1) = 0`.

This will give you

`"pH" = pK_a`

The problem tells you that the pH of the target solution is equal to `4.74`. The `pK_a` of the acid is equal to

`pK_a = - log(1.75 * 10^(-5)) = 4.76`

For all intended purposes, and considering the fact that you have one sig fig for the molarities of the two solutions, you can say that

`"pH" ~~ pK_a`

This means that you must have

`["CH"_3"COO"^(-)] = ["CH"_3"COOH"]`

The initial solution contained

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

`n_(CH_3COOH) = "0.1 M" * "0.1 L" = "0.01 moles CH"_3"COOH"`

Now, since adding one mole of strong base will consume one mole of weak acid and produce one mole of conjugate base, you can say that every mole of strong base added to the solution will convert one mole of weak acid to one mole of conjugate base.

In order to have equal concentrations of weak acid and conjugate base, you need to have equal numbers of moles of the two chemical species.

To get equal number of moles, you need to add enough strong base to consume half of the number of moles of weak acid. In this case, you will have

`n_(CH_3COOH) = "0.01 moles" - "0.005 moles" = "0.005 moles"`

`n_(HO^(-)) = "0.005 moles" - "0.005 moles" = "0 moles"`

`n_(CH_3COO^(-)) = 0 + "0.005 moles" = "0.005 moles"`

The volume of sodium hydroxide solution that contains `0.005` moles of hydroxide anions will be

`color(blue)(c = n?V implies V = n/c)`

`V_(OH^(-)) = (0.005 color(red)(cancel(color(black)("moles"))))/(0.1color(red)(cancel(color(black)("moles")))"L"^(-)) = "0.05 L" = color(green)("50 mL")`

SIDE NOTE As practice, you can try using the H-H equation to find the volume of base that would correspond to a pH of 4.74.

You will end up with

`4.74 = 4.76 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))`

`log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = -0.02`

`10^log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^(-0.02)`

`(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 0.955`

This means that you need to have slightly more conjugate acid than weak base, since `"pH" < pK_A`.

If you take `x` the number of moles of strong base added, you can say that your solution will end up with

`n_(CH_3COO^(-)) = 0 + x = x color(white)(a)"moles"`

`n_(OH^(-)) = x -x = "0 moles"`

`n_(CH_3COOH) = 0.01-x color(white)(a)"moles"`

The volume of the solution is the same for both species, so you have

`(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = x/(0.01-x)`

This will give you `x = 0.00488` moles of strong base needed.

The corresponding volume would be

`V = (0.00488color(red)(cancel(color(black)("moles"))))/(0.1color(red)(cancel(color(black)("moles")))"L"^(-1)) = "0.0488 L" = "48.8 mL"`

But since you have one sig fig for your values, the result will once again be

`V_(OH^(-)) = "50 mL"`