Question

Question #0984e

Answer 1

!! LONG ANSWER !!

Here's how you'd go about solving this one.

You're dealing with a reaction between acetic acid, a weak acid, and sodium hydroxide, a strong base.

Start with the balanced chemical equation for this neutralization reaction. Because sodium hydroxide is a strong base, it will dissociate completely in aqueous solution, so I'll write the net ionic equation

`CH_3COOH_((aq)) + OH_((aq))^(-) -> CH_3COO_((aq))^(-) + H_2O_((l))`

Notice the `1:1` mole ratio that exists between acetic acid and sodium hydroxide (written as hydroxide ions). This means that, in order to get a complete neutralization, you need equal numbers of moles of each compound.

If you have equal number of moles that react, both compounds will be completely consumed by the reaction. If you have more moles of acetic acid than of sodium hydroxide, all the base will be consumed and you'll be left with excess acetic acid.

Likewise, if you have more moles of sodium hydroxide, you'll be left with excess base.

Use the molarities and volumes of the two solutions to determine how many moles of each you have

`C = n/V => n = C * V`

`n_(CH_3COOH) = 0.5000"moles"/cancel("L") * 200 * 10^(-3)cancel("L") = "0.1000 moles"`

and

`n_(NaOH) = 0.5000"moles"/cancel("L") * 200 * 10^(-3)cancel("L") = "0.1000 moles"`

So, you have equal numbers of moles of each compound. This means that the acetic acid and the sodium hydroxide will be consumed, and that you'll produce 0.1000 moles of acetate anions - this is because of the `1:1` mole ratio that exists between the rectants and the acetate anion.

Here's where it gets a little tricky. The acetate anion will react with water to reform some of the acid and produce hydroxide anions, `OH^(-)`. This tells you that the pH of the solution will be basic, i.e. greater than 7.

The molarity of the acetate anion will be - don't forget to use the total volume of the solution

`C = n/V = "0.1000 moles"/((200 + 200) * 10^(-3)"L") = "0.25 M"`

Use an ICE table to determine the concentration of hydroxide anions

`" "CH_3COO_((aq))^(-) + H_2O_((l)) rightleftharpoons CH_3COOH_((aq)) + OH_((aq))^(-)`
I..........0.2500..............................................0..........................0
C............(-x)................................................(+x).......................(+x)
E.........0.2500-x............................................x..........................x

Because the acetate anion acts as a base, you'll have to use the base dissociation constant, `K_b`, which is equal to

`K_b = K_W/K_a = 10^(-14)/(1.76 * 10^(-5)) = 5.7 * 10^(-10)`

This will get you

`K_b = ([CH_3CHOOH] * [OH^(-)])/([CH_3COO^(-)])`

`K_b = (x * x)/(0.2500 - x) = x^2/(0.2500-x)`

Because `K_b` is so small, you can approximate (0.2500 - x) with 0.2500 to get

`K_b = x^2/0.2500 = 5.7 * 10^(-10) => x = sqrt(0.2500 * 5.7 * 10^(-10))`

`x = 1.2 * 10^(-5)`

The molarity of the hydroxide ions will be

`x = [OH^(-)] = 1.2 * 10^(-5)"M"`

The pOH of the solution will be

`pOH = -log([OH^(-)]) = -log(1.2 * 10^(-5)) = 4.92`

Therefore, the pH of the solution will be

`pH_"sol" = 14 - pOH = 14 - 4.92 = color(green)(9.08)`