# You need to produce a buffer solution that has pH 5.06. You already have a solution that contains 10 mmols of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution?

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The pKa of acetic acid is 4.74.

The pKa of acetic acid is 4.74.

##### 1 Answer

#### Answer:

#### Explanation:

Your strategy here will be to use the **Henderson - Hasselbalch equation** to find the ratio that must exist between the **concentrations** of the weak acid and of the conjugate in order to have a solution that has a pH of

The *Henderson - Hasselbalch equation* looks like this

#color(blue)(|bar(ul(color(white)(a/a)"pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))color(white)(a/a)|)))#

Your buffer solution contains acetic acid, *weak acid*, and the acetate anion, *conjugate base*. Moreover, you know that the

This means that you have

#5.06 = 4.74 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#

Now, before doing any calculations, try to predict how many mmoles of acetate anions you'd need **relative** to the number of mmoles of acetic acid.

Notice that the pH of the target solution is **higher** than the

#5.06 > pK_a#

This tells you that the target solution must contain **more conjugate base** than weak acid. Implicitly, this solution must contain *more moles* of acetate anions than of acetic acid.

Rearrange the H - H equation to isolate the log term on one side

#0.32 = log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#

This will be equivalent to

#10^0.32 = 10^log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#

which will give you

#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 10^0.32 = 2.09#

As predicted, the concentration of the conjugate base **must be** higher than that of the weak acid

#color(green)(|bar(ul(color(white)(a/a)color(black)(["CH"_3"COO"^(-)] = 2.09 xx ["CH"_3"COOH"])color(white)(a/a)|)))#

Now, the *volume* of the solution is the same for both chemical species, let's say

#n_(CH_3COO^(-))/color(red)(cancel(color(black)(V))) = 2.09 xx n_(CH_3COOH)/color(red)(cancel(color(black)(V)))#

Since the solution is said to contain **mmoles** of acetic acid, it follows that you must add

#n_(CH_3COO^(-)) = 2.09 xx "10 mmoles" = color(green)(|bar(ul(color(white)(a/a)"21 mmoles CH"_3"COO"^(-)color(white)(a/a)|)))#

I'll leave the answer rounded to two **sig figs**.