# Question #70bb0

Apr 19, 2015

pH of 0.2M $H {C}_{7} {H}_{5} {O}_{2}$

notice: ${H}^{+}$ is really ${H}_{3} {O}^{+}$ just abbreviated.

Since our reaction would look like this:
$H {C}_{7} {H}_{5} {O}_{2}$(aq)$+ {H}_{2} O$(l)$r i g h t \le f t h a r p \infty n s {H}^{+}$(aq)$+ {C}_{7} {H}_{5} {O}_{2}^{-}$(aq)

ICE TABLES
a)$H {C}_{7} {H}_{5} {O}_{2}$
I) 0.2M
C) -x
E) $0.2 M - x \approx 0.2 M$ This only works with weak acids/bases, because the x you solve for will be extremely small.

b)${H}^{+}$
I) 0
C) +x
E) $0 + x = x$

c)${C}_{7} {H}_{5} {O}_{2}^{-}$
I) 0
C) +x
E) $0 + x = x$

The $K a = \frac{\left[P r o \mathrm{du} c t s\right]}{\left[R e a c \tan t s\right]}$$= \frac{\left[{H}^{+}\right] \cdot \left[{C}_{7} {H}_{5} {O}_{2}^{-}\right]}{\left[H {C}_{7} {H}_{5} {O}_{2}\right]}$$= \frac{x \cdot x}{0.2 M} = {x}^{2} / \left(0.2 M\right)$

Notice: a) $\left[X\right] =$Molar Concentration of X
b) We leave ${H}_{2} O$ out of our Ka because its concentration is constant. This happens with any pure liquids or solids. If gases in equation, use partial pressures.

Since $K a = 6.3 \cdot {10}^{-} 6$$= {x}^{2} / \left(0.2 M\right)$

$x = \sqrt{\left(6.3 \cdot {10}^{-} 6\right) \cdot \left(0.2 M\right)}$$= 1.12 \cdot {10}^{-} 3 M = \left[{H}^{+}\right]$

NOW $p H = - \log \left[{H}^{+}\right] = - \log 1.12 \cdot {10}^{-} 3 M$

$p H = 2.94$

Extra step for weak bases: If you are dealing w/ a weak base and your $x = \left[O {H}^{-}\right]$, this last step will give you $p O H$
($p O H = - \log \left[O {H}^{-}\right]$)

You can use $p O H$ to solve for $p H$ with this simple relationship:
$p O H + p H = p K w$ since $p K w = 14.00$

$p H = 14.00 - p O H$

Apr 19, 2015
1. $\text{pH} = 2.95$

2. $\text{pH} = 9.67$

A quick way is to use:

$p H = \frac{1}{2} \left(p {K}_{a} - \log A\right)$

$A$ = conc. of the acid.

$p {K}_{a} = - \log {K}_{a} = - \log \left(6.3 \times {10}^{- 6}\right) = 5.2$

$p H = \frac{1}{2} \left(5.2 - \log 0.2\right) = \frac{1}{2} \left(5.2 + 0.698\right) = \textcolor{red}{2.95}$

We can use the same idea for weak bases:

$p O H = \frac{1}{2} \left(p {K}_{b} - \log B\right)$

Where $B$ = conc. of the base.

$p {K}_{b} = - \log {K}_{b} = - \log \left(1.1 \times {10}^{- 8}\right) = 7.958$

$p O H = \frac{1}{2} \left(7.958 - \log 0.2\right) = \frac{1}{2} \left(7.958 + 0.698\right) = 4.33$

$p H + p O H = p {K}_{w} = 14$

$p H = 14 - 4.33 = \textcolor{red}{9.67}$