# Question #20274

##### 1 Answer

#### Explanation:

For a buffer that contains **equal concentrations** of a weak base and of its conjugate acid, the

#(["conjugate acid"])/(["weak base"]) = 1 implies color(blue)(ul(color(black)("pH" = 14 - "p"K_b)))#

This is the case because a weak base - conjugate acid buffer that has **equal concentrations** of weak base and of conjugate acid has

#"pOH" = "p"K_b#

As you know, an aqueous solution at

#"pH " + " pOH" = 14#

This means that the buffer has

#14 - "pH" = "p"K_b#

which, of course, gets you

#color(blue)(ul(color(black)("pH" = 14 - "p"K_b)))#

Alternatively, you can use the fact that at

#"pK"_a + "p"K_b = 14#

This means that

#"p"K_b = 14 - "pH"#

which, once again, gets you

#color(blue)(ul(color(black)("pH" = 14 - "p"K_b)))#

When in doubt, remember that the **Henderson - Hasselbalch equation**

#"pOH" = "p"K_b + log( (["conjugate acid"])/(["weak base"]))#

This can be rewritten as

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

to get

#"pH" = 14 - ["p"K_b + log( (["conjugate acid"])/(["weak base"]))]#

Notice that when you have

#(["conjugate acid"])/(["weak base"]) = 1#

you get

#"pH" = 14 - ["p"K_b + log(1)]#

#color(blue)(ul(color(black)("pH"= 14 - "p"K_b)))#

So, to find the

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

to get

#"pH" = 14 - [- log(1.8 * 10^(-5))]#

#"pH" = 14 + log(1.8 * 10^(-5))#