# Question #58f8b

##### 1 Answer

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

#pK_a = -log(K_a) = -log(1.75 * 10^(-5)) = 4.76#

Now, notice that the difference between the acid's *approximately* one unit. As you know, the pH is calculated using a **logarithmic scale**.

This means that if you have something like

#log(x/y) = 1" "# and#" "log(x/y) = 2#

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

In the first case, **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"])))#

In your case, you have

#5.8 = 4.76 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#

Your goal here is to solve this equation to find

#log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 5.8 - 4.76 = 1.04#

This is equivalent to

#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 10^1.04 = 10.96#

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

#["CH"_3"COO"^(-)] = 11 * ["CH"_3"COOH"]#

This means that you need **for every**

Since the volume of the buffer is set at

#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")#