# If a buffer contains #"0.110 M"# weak base and #"0.440 M"# of the weak conjugate acid, what is the #"pH"#? The #"pK"_b# is #4.96#.

##### 2 Answers

#### Explanation:

This looks like a job for the **Henderson - Hasselbalch equation**, which for a weak base/conjugate acid buffer looks like this

#"pH" = 14 - overbrace(["p"K_b + log( (["conjugate acid"])/(["weak base"]))])^(color(blue)("the pOH of the buffer solution"))#

As you know, you have

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

In your case, you know that the buffer contains **higher** than the

In other words, the **lower** than

Plug in your values into the Henderson - Hasselbalch equation to find

#"pH" = 14 - [-log(1.1 * 10^(-5)) + log((0.440 color(red)(cancel(color(black)("M"))))/(0.110color(red)(cancel(color(black)("M")))))]#

#color(darkgreen)(ul(color(black)("pH" = 8.44)))#

The answer is rounded to two *decimal places*, the number of **sig figs** you have for the base dissociation constant.

Stefan has a good answer, but I thought I'd give another approach to this. For buffers (i.e. weak acid + conjugate base, weak base + conjugate acid), the **Henderson-Hasselbalch equation** applies.

To make it so I only have to know one Henderson-Hasselbalch equation, I use the

#"pH" = "pK"_a + log\frac(["A"^(-)])(["HA"])#

Using the idea that

#-log(K_b) = "pK"_b = -log(1.1 xx 10^(-5)) = 4.96#

#=> "pK"_a = 14 - 4.96 = 9.04#

And thus, noting the difference in notation (treating the base as

#color(blue)("pH") = 9.04 + log (("0.110 M")/("0.440 M"))#

#= 9.04 - log 4 = color(blue)(8.44)#

And this makes physical sense, as we started with a weak base, whose conjugate acid dissociates less (

Furthermore, there is a higher concentration of conjugate acid than the weak base. So, we should expect the

**APPENDIX**

And just so you see, this gives the same equation Stefan has. Recall that:

- At
#25^@ "C"# and#"1 atm"# ,#"pK"_a + "pK"_b = 14 = "pK"_w# #log(a/b) = -log(b/a)#

Therefore:

#"pH" = (14 - "pK"_b) + log\frac(["A"^(-)])(["HA"])#

#= (14 - "pK"_b) - log\frac(["HA"])(["A"^(-)])#

#= 14 - ["pK"_b + log\frac(["HA"])(["A"^(-)])]# which is what Stefan used.

With slightly changed notation, and knowing that

#"pH" = 14 - overbrace(["pK"_b + log\frac(["BH"^(+)])(["B"])])^("pOH")#

#= 14 - "pOH"#

Thus,

#barul(|stackrel(" ")(" " "pOH" = "pK"_b + log\frac(["BH"^(+)])(["B"])" ")|)#