# Question 76220

May 28, 2017

$\textsf{p H = 2.36}$

#### Explanation:

You should have been provided with the acid dissociation constant $\textsf{{K}_{a}}$.

Vinegar is essentially a solution of ethanoic acid. It dissociates:

$\textsf{C {H}_{3} C O O H r i g h t \le f t h a r p \infty n s C {H}_{3} C O {O}^{-} + {H}^{+}}$

sf(K_a=([CH_3COO^-][H^+])/([CH_3COOH])

Looking up $\textsf{p {K}_{a}}$ I get 4.76. This =$\textsf{- \log {K}_{a}}$

If the concentration of the acid = c we can set up an ICE table in mol/l:

$\textsf{\text{ "CH_3COOH" "rightleftharpoons" } C {H}_{3} C O {O}^{-} + {H}^{+}}$

$\textsf{I \text{ "c" "0" } 0}$

$\textsf{C \text{ "-x" "+x" } + x}$

$\textsf{E \text{ "(c-x)" "x" } x}$

$\therefore$sf(K_a=x^2/((c-x))#

Because of the small value of $\textsf{{K}_{a}}$ (it is less than $\textsf{{10}^{-} 4}$) we can assume that (c-x)$\Rightarrow$c.

$\therefore$$\textsf{{K}_{a} = {x}^{2} / c}$

$\textsf{{x}^{2} = {K}_{a} c}$

$\textsf{x = \sqrt{{K}_{a} c} = {\left({K}_{a} c\right)}^{\frac{1}{2}} = \left[{H}^{+}\right]}$

$\therefore$$\textsf{- \log \left[{H}^{+}\right] = \frac{1}{2} \left[p {K}_{a} - \log c\right] = p H}$

You have calculated c = 1.1 mol/l so:

$\textsf{p H = \frac{1}{2} \left[4.76 - 0.04139\right]}$

$\textsf{p H = 2.36}$