Question #b75b4

1 Answer
Jan 14, 2018

Answer:

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

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 #"pH"# is calculated using a logarithm scale of base #10#, so if the #"pH"# increases by #1# unit, then the ratio of the conjugate base to the weak acid must increase by #10#.

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

Now, this is the case because the #"pH"# of a buffer that contains a weak acid and its conjugate base can be calculated by using 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 #"pH"# of the buffer and the #"p"K_a# of the weak acid.

Notice that in order o get the #"pH"# of the buffer to decrease by #1# 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 #10# times higher than the concentration of the weak acid, the #"pH"# of the buffer is #1# unit higher than the #"p"K_a# of the weak acid.

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