# Question #ef831

##### 1 Answer

#### Answer:

#### Explanation:

You're dealing with a strong acid-conjugate base **buffer** here, so right from the start, you know that you can use the **Henderson - Hasselbalch equation** to find its

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

Here

#"p"K_a = - log(K_a)#

with **acid dissociation constant** of the weak acid.

In your case, acetic acid, *soluble* sodium acetate in a **mole ratio**, so you know that you have

#["CH"_3"COO"^(-)] = "0.125 M"#

Notice that your solution contains **more weak acid** than conjugate base. This tells you that the **lower** than the

This is the case becase at **equal concentrations** of weak acid and conjugate base, the **equal** to the *more acid than conjugate, the ** the

Plug in your values to find

#"pH"= "p"K_a + log (( 0.125 color(red)(cancel(color(black)("M"))))/(0.25color(red)(cancel(color(black)("M")))))#

#"pH" = "p"K_a + log(1/2)#

This is equivalent to

#"pH" = "p"K_a + overbrace(log(1))^(color(blue)(=0)) - log(2)#

#"pH" = "p"K_a - log(2)#

#"pH" = "p"K_a - 0.30#

Now all you have to do is to use the **here**, to find the **decimal places**, the number of **sig figs** you have for the concentration of acetic acid.

Notice that you have

#"pH" = "p"K_a - 0.30 " " < " " "p"K_a#

which is consistent with the fact that the buffer contains more acetic acid than acetate anions.