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

#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."#

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")#.

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?