# Use the information below to answer the following question. What is the pH of a 0.100 M solution of benzoic acid?

## Use the information below to answer the following question. What is the pH of a 0.100 M solution of benzoic acid?

Sep 8, 2016

$p H$ $=$ $- {\log}_{10} \left(2.5 \times {10}^{-} 3\right)$ $=$ $2.60$

#### Explanation:

Benzoic acid is a weak acid, whose dissociation we represent as:

${C}_{6} {H}_{5} C {O}_{2} H \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {C}_{6} {H}_{5} C {O}_{2}^{-} + {H}_{3} {O}^{+}$

Using the usual conventions, ${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{C}_{6} {H}_{5} C {O}_{2}^{-}\right]}{\left[{C}_{6} {H}_{5} C {O}_{2} H\right]}$

${K}_{a} = 6.3 \times {10}^{-} 5$ $=$ $\frac{{x}^{2}}{0.100 - x}$

Where $x$ $=$ $\left[{H}_{3} {O}^{+}\right] = \left[{C}_{6} {H}_{5} C {O}_{2}^{-}\right]$

We have a quadratic in $x$, that in principle we could solve directly. I'm not gonna cos I'm lazy. If we say $\left(0.100 - x\right) \cong 0.100$, then $x$ $=$ $\sqrt{6.3 \times {10}^{- 5} \times 0.100}$ $=$ $2.5 \times {10}^{-} 3$.

I am not going to bother to do another approx. but perhaps you should; i.e. plug your first approx. back into the equation to see how the answer evolves.

$x = \left[{H}_{3} {O}^{+}\right]$ $=$ $2.5 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$

And $p H$ $=$ $- {\log}_{10} \left(2.5 \times {10}^{-} 3\right)$ $=$ $2.60$