# Question 7507c

Apr 19, 2015

The weak acid to conjugate base ratio for your buffer will be 8.70.

So, you're dealing with a buffer that consists of acetic acid, $C {H}_{3} C O O H$, a weak acid, and sodium acetate, $C {H}_{3} C O O N a$, its conjugate base.

Once again, the Henderson-Hasselbalch equation will be your tool of choice.

$p {H}_{\text{sol}} = p {K}_{a} + \log \left(\frac{\left[C {H}_{3} C O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right]}\right)$

Start by calculating the $p {K}_{a}$ from the acid dissociation constant, ${K}_{a}$

$p {K}_{a} = - \log \left({K}_{a}\right) = - \log \left(1.8 \cdot {10}^{- 5}\right) = 4.74$

Notice that the pH of the buffer is lower than the $p {K}_{a}$. Even before doing any calculations, you can predict that the concentration of the weak acid will be higher than that of the weak base, since the solution is more acidic than the $p {K}_{a}$ value.

This implies that the ratio you're looking for, weak acid to conjugate base, will be greater than 1.

So, plug your data into the Henderson - Hasselbalch equation

$3.80 = 4.74 + \log \left({\underbrace{\frac{\left[C {H}_{3} C O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right]}}}_{\textcolor{b l u e}{\text{x}}}\right)$

$\log \left(\textcolor{b l u e}{\text{x}}\right) = 3.80 - 4.74 = - 0.94$

This is equivalent to

10^(log(color(blue)("x"))) = 10^(-0.94) => "x" = 0.115#

So, the ratio conjugate base to weak acid ratio is

$\text{x} = \frac{\left[C {H}_{3} C O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right]} = 0.115$

This means that the weak acid to conjugate base ratio will be

$\frac{\left[C {H}_{3} C O O H\right]}{\left[C {H}_{3} C O {O}^{-}\right]} = \frac{1}{x} = \frac{1}{0.115} = 8.6957$

Rounded to three sig figs, the answer will be

$\frac{\left[C {H}_{3} C O O H\right]}{\left[C {H}_{3} C O {O}^{-}\right]} = \textcolor{g r e e n}{8.70}$

The initial prediction turned out to be correct, you do have more weak acid than conjugate base at that pH.