# Question #b75b4

##### 1 Answer

#### Explanation:

You know that when the buffer contains **equal concentrations** of weak acid and of conjugate base, you have

#(["conjugate base"])/(["weak acid"]) = 1 implies "pH" = "p"K_a#

You also know that the **logarithm scale** of base **unit**, then the ratio of the conjugate base to the weak acid must **increase** by

#(["conjugate base"])/(["weak acid"]) = 10 implies "pH" = "p"K_a + 1#

Now, this is the case because the **Henderson - Hasselbalch equation**

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

As you can see, the ratio that exists between the concentration of the conjugate base and the concentration of the weak acid determines the difference between the

Notice that in order o get the **decrease** by **unit**, you need to have

#(["conjugate base"])/(["weak acid"]) = 1/10#

since

#"pH" = "p"K_a + log(1/10)#

#"pH" = "p"K_a + (-1)#

#"pH" = "p"K_a - 1#

So remember, if the concentration of the conjugate base is **times higher** than the concentration of the weak acid, the **unit higher** than the

#{("10 times more conjugate base " implies " pH" = "p"K_a + 1),("10 times more weak acid " implies " pH" = "p"K_a -1) :}#