# What is the pH of the resulting solution?

## 20.0 mL of pure HC2H3O2 is diluted to 1.30 L with water. The density of pure HC2H3O2 is 1.05 g/mL.

Apr 7, 2018

See Below

#### Explanation:

First we are gonna get mL of Acetic Acid into grams. Then get it into moles. Then get it into molarity. Then use the ${K}_{A}$ for acetic acid to determine $\left[{H}^{+}\right]$, then pH.

20.0mL of $C {H}_{3} C O O H \times 1.05 \frac{g}{\text{mL}}$ =
21g $C {H}_{3} C O O H$
21g CH_3COOH xx ("1mole"/"60.5g") = 0.347 "moles" CH_3COOH

$\left(0.347 \text{moles" CH_3COOH)/(1.3 "liters}\right)$ = 0.267M $C {H}_{3} C O O H$

$C {H}_{3} C O O H {K}_{A} = 1.8 \times {10}^{-} 5$

$C {H}_{3} C O O H = C {H}_{3} C O {O}^{-} + {H}^{+}$

Using an ICE table (and assuming x << 0.267M)

${K}_{A} = 1.85 \times {10}^{-} 5 = {x}^{2} / 0.267$

x = 0.0022M = $\left[{H}^{+}\right]$
$p H = - \log \left[{H}^{+}\right]$
$p H = - \log \left(0.0022 M\right)$
$p H = 2.66$