Question

Consider an ammonia/ammonium chloride buffer. Kb for ammonia is 1.8 χ 10—5. Select all the correct statements from those listed below?

The optimum buffer occurs at a pH of 9.26.

The addition of a small amount of acid to this buffer will shift the equilibrium to the left.

The optimum buffer occurs at a pH of 4.74.

The optimum buffer only occurs when the concentration of ammonia and ammonium chloride equals 1.00 M.

This is a basic buffer.

This is an acidic buffer.

The addition of a small amount of acid to this buffer will shift the equilibrium to the right.

Answer 1

We specify a direction of reaction to the equilibrium first....

Explanation:

`NH_3(aq) + H_3O^+ rightleftharpoons NH_4^+ +H_2O(l)`

And so our geometry is informed as we face the PAGE....

And here...`pH=pK_a+log_10{[[NH_4^+]]/[[NH_3(aq)]]}`

`"The optimum buffer occurs at a"` `"pH"` `"of 9.26."`...why? ...because `pK_a` `NH_4^+=9.26`.

`"This is a basic buffer as it maintains pH > 7 at 25"^@ "C"`

`"The addition of a small amount of acid will indeed shift the"`
`"equilibrium to the right as we face the page."`

Answer 2

Here's what I get.

Explanation:

The buffer calculation

The equation for the equilibrium is

`"NH"_3 + "H"_2"O" ⇌ "NH"_4^"+" + "OH"^"-"`

We can apply the Henderson-Hasselbalch equation:

`color(blue)(bar(ul(|color(white)(a/a)"pH" = "p"K_text(a) + log((["NH"_3])/(["NH"_4^"+"]))color(white)(a/a)|)))" "`

For ammonia,

`K_text(b) = 1.8 × 10^"-5"`

`"p"K_text(b) = "-log"(1.8 × 10^"-5") = 4.74`

`"p"K_text(a) = "14.00 - 4.74 = 9.26"`

`"pH" = 9.26 + log((["NH"_3])/(["NH"_4^"+"]))`

The choices

The buffer has the best capacity when `["NH"_3] = ["NH"_4^"+"]`

Then `log((["NH"_3])/(["NH"_4^"+"])) = log1 = 0` and

`"pH = 9.26 +0 = 9.26"`

`color(blue)("The optimum buffer occurs at pH 9.26")`.

`color(blue)("The optimum buffer does not occur at pH 4.74")`.

`color(blue)("The optimum buffer occurs when the concentrations of ammonia and"`
`color(blue)("ammonium chloride are equal")`.

`color(blue)("This is a basic buffer")`.

`color(blue)("This is not an acidic buffer")`.

Le Châtelier's Principle

Le Châtelier's Principle states that if you apply a stress to a system at equilibrium, the position of equilibrium will shift in the direction that reduces the stress.

The equation for the equilibrium is

`"NH"_3 + "H"_2"O" ⇌ "NH"_4^"+" + "OH"^"-"`

If you add a small amount of acid, it will react with the `"OH"^"-"` to form water.

The system will try to restore equilibrium by producing more `"OH"^"-"`.

`color(blue)("The addition of a small amount of acid to this buffer will not shift the"`
`color(blue)("equilibrium to the left")`.

`color(blue)("The addition of a small amount of acid to this buffer will shift the equilibrium"`
`color(blue)("to the right")`.

Answer 3

Honestly, the shifting left/right choice is not good wording; the equilibrium is just that... double-sided.

I choose to write

`"NH"_4^(+)(aq) + "H"_2"O"(l) rightleftharpoons "NH"_3(aq) + "H"_3"O"^(+)(aq)`

since `overbrace("pK"_a)^(9.26) > overbrace("pK"_b)^(4.74)` and thus the acid component dominates over the base component in the buffer.

And thus the acid is on the LEFT... I could just as easily have written `"NH"_3(aq)` associating in water...

The `K_a` at `25^@ "C"` is found as

`K_w/K_b = K_a = 10^(-14)/(1.80 xx 10^(-5)) = 5.56 xx 10^(-10)`

And hence, `"pK"_a = -log(K_a) = 9.26`. For a buffer with weak acid and weak conjugate base, the Henderson-Hasselbalch equation applies:

`"pH" = "pK"_a + log\frac(["NH"_3])(["NH"_4^(+)])`

`= 9.26 + log\frac(["NH"_3])(["NH"_4^(+)])`

And `9.26` becomes the reference point for the optimal buffer, i.e. one with

• Close to equal `"NH"_3` and `"NH"_4^(+)` concentrations... it must be able to buffer BOTH strong acid AND base. It NEED NOT be both `"1.00 M"`. Why not both `"1.50 M"`? Why not both `"0.500 M"`?
• Large `"NH"_3` and `"NH"_4^(+)` concentrations... it should last long. Which is better, both `"0.200 M"` or both `"0.500 M"`?

At `25^@ "C"`, the optimal `"NH"_3:"NH"_4^(+)` ratio leaves the buffer basic, as neutral `"pH" = 7` at `25^@ "C"`. You should convince yourself that since `log(1) = 0`, `"pH" = "pK"_a` for the optimal buffer.

And if you add strong acid to it, the `"pH"` must drop, so that `"pH" < "pK"_a`... This reacts `"H"^(+)` with the weak base and shifts the equilibrium towards `"NH"_4^(+)`.

`"NH"_3(aq) + "H"^(+)(aq) -> "NH"_4^(+)(aq)`

As we have written it, is that the right or the left in the ORIGINAL equilibrium?