Question

If `15.0mL` of glacial acetic acid (pure `HC_2H_3O_2`) is diluted to `1.50L` with water, what is the pH of the resulting solution? The density of glacial acetic acid is `"1.05 g/mL"`

Answer 1

`pH_"sol" = 2.757`

Explanation:

To be able to solve this problem, you need to know the value of acetic acid's acid dissociation constant, `K_a`, which essentially tells you how how many acid molecules will ionize in solution.

The acid dissociation constant for acetic acid is listed as

`K_a = 1.76 * 10^(-5)`

So, the idea is that you dilute a sample of pure acetic acid by a dilution factor of 100. The density of the pure acetic acid will help you determine how many grams of acid you're diluting.

`15.0color(red)(cancel(color(black)("mL"))) * "1.05 g"/(1color(red)(cancel(color(black)("mL")))) = "15.75 g HAc"`

Use the acid's molar mass to see how many moles you'd get in that much mass

`15.75color(red)(cancel(color(black)("g"))) * "1 mole HAc"/(60.05color(red)(cancel(color(black)("g")))) = "0.2623 moles HAc"`

This means that the acetic acid solution has a molarity of

`C = n/V = "0.2623 moles"/("1.50 L") = "0.175 M"`

Next, use an ICE Table to figure out exactly how many hydronium ions you'd get from the acid's dissociation.

`" "CH_3COOH_((aq)) + H_2O_((l)) rightleftharpoons H_3O_((l))^(+) + CH_3COO_((aq))^(-)`
I............0.175............................................0....................0
C............(-x)..............................................(+x)...............(+x)
E..........0.175-x..........................................x...................x

By definition, `K_a` will be

`K_a = ([H_3O^(+)] * [CH_3COO^(-)])/([CH_3COOH])`

`K_a = (x * x)/(0.175 - x) = 1.76 * 10^(-5)`

Because `K_a` is so small, you can approximate 0.175 - x with 0.175 to get

`K_a = x^2/0.175 = 1.76 * 10^(-5)`

`x = sqrt(3.08 * 10^(-6)) = 1.75 * 10^(-3)`

The concentration of the hydronium ions will thus be

`x = [H_3O^(+)] = 1.75 * 10^(-3)"M"`

The pH of the solution will be equal to

`pH_"sol" = -log([H_3O^(+)])`

`pH_"sol" = -log(1.75 * 10^(-3)) = color(green)(2.757)`