# Question 58f8b

Nov 18, 2015

Here's what I got.

#### Explanation:

Before doing any calculations, try to predict what the volume ratio between the weak acid and its conjugate base will be at that pH.

Notice that the $p {K}_{a}$ of the acid is equal to

$p {K}_{a} = - \log \left({K}_{a}\right) = - \log \left(1.75 \cdot {10}^{- 5}\right) = 4.76$

Now, notice that the difference between the acid's $p {K}_{a}$ and the pH of the buffer solution is approximately one unit. As you know, the pH is calculated using a logarithmic scale.

This means that if you have something like

$\log \left(\frac{x}{y}\right) = 1 \text{ }$ and $\text{ } \log \left(\frac{x}{y}\right) = 2$

then you can say that the increase of the value returned by the log is caused by the fact that $x$ increased ten times compared with $y$.

In the first case, $\frac{x}{y}$ = 10. In the second case, $\frac{x}{y} = 100$. So right from the start, you know that you must have approximately 10 times more conjugate base than weak acid.

Now let's do the calculations to prove this prediction. As you know, the pH of a buffer that contains a weak acid and its conjugate base can be calculated using the Henderson - Hasselbalch equation, which looks like this

color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))

$5.8 = 4.76 + \log \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right)$

Your goal here is to solve this equation to find $\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)$

$\log \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right) = 5.8 - 4.76 = 1.04$

This is equivalent to

$\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right) = {10}^{1.04} = 10.96$

Therefore, you need to have about $11$ times more conjugate base than weak acid.

$\left[\text{CH"_3"COO"^(-)] = 11 * ["CH"_3"COOH}\right]$

This means that you need $1$ unit of volume of weak acid for every $11$ units of volume of conjugate base.

Since the volume of the buffer is set at $\text{1.0-L}$, the volumes of the two species will be

1.00color(red)(cancel(color(black)("L buffer"))) * ("1 L CH"_3"COOH")/((1 + 11)color(red)(cancel(color(black)("L buffer")))) = "0.0833 L"

and

1.00color(red)(cancel(color(black)("L buffer"))) * ("11 L CH"_3"COO"^(-))/((1 + 11)color(red)(cancel(color(black)("L buffer")))) = "0.9167 L"

Expressed in mililiters, the answers will be

V_"weak acid" = color(green)("83 mL")

V_"conj. base" = color(green)("917 mL")#