# Question #109c2

##### 1 Answer

#### Explanation:

Start by calculating the

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

You will have

#"p"K_a = - log(2.3 * 10^(-9)) = 8.64#

Now, notice that the pH of the buffer is slightly **higher** than the **more conjugate base**, which in this case is the hypobromite anion,

In other words, you should expect to find

#(["BrO"^(-)])/(["HBrO"]) > 1#

The pH of a buffer that contains a weak acid and its conjugate base can be calculated by using the **Henderson - Hasselbalch equation**

#color(blue)(ul(color(black)("pH" = - "p"K_a + log( (["conjugate base"])/(["weak acid"])))))#

For your buffer, you have

#"pH" = "p"K_a + log((["BrO"^(-)])/(["HBrO"]))#

which is

#8.75 = 8.64 + log((["BrO"^(-)])/(["HBrO"]))#

Rearrange the equation as

#log((["BrO"^(-)])/(["HBrO"])) = 8.75 - 8.64#

This will be equivalent to

#10^log((["BrO"^(-)])/(["HBrO"])) = 10^(0.11)#

which gets you

#color(darkgreen)(ul(color(black)((["BrO"^(-)])/(["HBrO"]) = 1.3)))#

The answer is rounded to two **sig figs**, the number of *decimal places* you have for the pH of the solution.

As predicted, the buffer contains more conjugate base than weak acid.