Question

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`.

Answer 1

`"pH" = 8.44`

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 `"0.110 M"` of the weak base and `"0.440 M"` of its conjugate acid, so even without doing any calculations, you should be able to say that the `"pOH"` of the buffer will be higher than the `"p"K_b` of the weak base.

In other words, the `"pH"` of the buffer will be lower than `14 - "p"K_b`, what you would get for a `"pOH"` equal to the `"p"K_b` of the weak base.

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.

Answer 2

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 `"pK"_a` one and recall the relationships to interconvert between `"pK"_a`, `"pK"_b`, `"pH"`, and `"pOH"`.

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

Using the idea that `"pK"_a + "pK"_b = 14`, we get:

`-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 `"A"^(-)` or `"B"` and the conjugate acid as `"HA"` or `"BH"^(+)`), the `"pH"` is alternatively found 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 (`K_a < K_b` if `K_b > 10^(-7)` at `25^@ "C"` and `"1 atm"`).

Furthermore, there is a higher concentration of conjugate acid than the weak base. So, we should expect the `"pH"` to be basic, but also more acidic than the `"pK"_a` of the conjugate acid.

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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" + "pOH" = 14` at `25^@ "C"` and `"1 atm"`, we can get the other form of the Henderson-Hasselbalch equation:

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

`= 14 - "pOH"`

Thus,

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