# Question 7c41a

Feb 1, 2017

$\textsf{p H = 3.64}$

#### Explanation:

I'll derive the expression first.

Let acetic acid be HA:

It is a weak acid and dissociates:

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

For which:

$\textsf{{K}_{a} = \frac{\left[{H}^{+}\right] \left[{X}^{-}\right]}{\left[H X\right]} = 1.76 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

Let the initial concentration be $\textsf{C \textcolor{w h i t e}{x} \text{mol/l}}$.

Set up an ICE table in mol/l:

$\textsf{\text{ "HX" "rightleftharpoons" } {H}^{+} + {X}^{-}}$

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

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

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

$\therefore$$\textsf{{K}_{a} = {x}^{2} / \left(\left(C - x\right)\right)}$

If $\textsf{{K}_{a}}$ lies between about $\textsf{{10}^{-} 4}$ to sf(10^(-13) then we can assume that $\textsf{x}$ is small enough to let $\textsf{\left(C - x\right) \Rightarrow C}$.

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

$\therefore$$\textsf{{x}^{2} = {\left[{H}^{+}\right]}^{2} = {K}_{a} C}$

$\therefore$sf([H^+]=sqrt(K_aC)

sf([H^+]=sqrt(1.76xx10^(-5)xx3xx10^(-3))#

$\textsf{\left[{H}^{+}\right] = \pm 2.297 \times {10}^{- 4} \textcolor{w h i t e}{x} \text{mol/l}}$

Since the -ve root is undefined we get:

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left(2.297 \times {10}^{- 4}\right) = 3.64}$

If you are given the $\textsf{p {K}_{a}}$ value I'll show how to get another useful expression:

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

Taking logs of both sides:

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

Multiply through by -1 $\Rightarrow$

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

This becomes:

$\textsf{\textcolor{red}{p H = \frac{1}{2} \left[p {K}_{a} - \log C\right]}}$

This is quick way of finding the pH of a weak acid given the concentration.