# 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"

Jul 30, 2015

$p {H}_{\text{sol}} = 2.757$

#### 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

${K}_{a} = 1.76 \cdot {10}^{- 5}$

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

$C = \frac{n}{V} = \text{0.2623 moles"/("1.50 L") = "0.175 M}$

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

$\text{ } C {H}_{3} C O O {H}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(l\right)}^{+} + C {H}_{3} C O {O}_{\left(a q\right)}^{-}$
I............0.175............................................0....................0
C............(-x)..............................................(+x)...............(+x)
E..........0.175-x..........................................x...................x

By definition, ${K}_{a}$ will be

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \cdot \left[C {H}_{3} C O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right]}$

${K}_{a} = \frac{x \cdot x}{0.175 - x} = 1.76 \cdot {10}^{- 5}$

Because ${K}_{a}$ is so small, you can approximate 0.175 - x with 0.175 to get

${K}_{a} = {x}^{2} / 0.175 = 1.76 \cdot {10}^{- 5}$

$x = \sqrt{3.08 \cdot {10}^{- 6}} = 1.75 \cdot {10}^{- 3}$

The concentration of the hydronium ions will thus be

$x = \left[{H}_{3} {O}^{+}\right] = 1.75 \cdot {10}^{- 3} \text{M}$

The pH of the solution will be equal to

$p {H}_{\text{sol}} = - \log \left(\left[{H}_{3} {O}^{+}\right]\right)$

$p {H}_{\text{sol}} = - \log \left(1.75 \cdot {10}^{- 3}\right) = \textcolor{g r e e n}{2.757}$