# Question b2a5b

Oct 13, 2017

None of the above?

#### Explanation:

For starters, you should know that acetic acid has

$\text{p} {K}_{a} = 4.75 \to$ source

Now, the $\text{pH}$ of a weak acid-conjugate base buffer can be calculated using the Henderson - Hasselbalch equation, which looks like this

"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))

Right from the start, the fact that you have

$\text{pH " > " p} {K}_{a}$

tells you that the buffer has a higher concentration of acetate anions, the conjugate base of acetic acid, than of acetic acid.

This implies that

$\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right) > 1$

This means that options (1) and (4) are eliminated.

To find the actual ratio that exists between the conjugate base and the weak acid, rearrange the Henderson - Hasselbalch equation

$6 = 4.75 + \log \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right)$

as

$\log \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right) = 6 - 4.75$

This is equivalent to

${10}^{\log} \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right) = {10}^{1.26}$

which gives you

$\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right) = 17.8$

This means that in order to have an acetic acid-acetate anions buffer of $\text{pH} = 6$, the solution must contain $17.8$ times more conjugate base than weak acid.

If you want, you can take options (2) and (3) and use the given ratios to find the $\text{ph}$ of the buffer,

You will have

(2): " "(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 10

$\text{pH} = 4.75 + \log \left(10\right) = 5.75$

(3): " "(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 100#

$\text{pH} = 4.75 + \log \left(100\right) = 6.75$

As you can see, a $\text{pH}$ of $6$ is much closer to a $10 : 1$ conjugate base/weak acid ratio.