Deep Dive: Bonus Buffer Problem
1. Problem Statement
We need to prepare an acetate buffer with pH = 6.420. On the shelf we have available a flask labeled
0.664 m Acetic Acid
The flask contains 975 mL of solution.
(i) Find the pH of the contents of the flask.
(ii) Create the buffer using two different different methods:
Method A: By combining the entire contents of the flask with pure water.
Method B: By combining the entire contents of the flask with 2.50 m KOH solution.
Find the volumes that would be needed for each method.
Data: For acetic acid is Ka = 1.77 × 10−5 .
2. Initial Flask
The flask contains 975 mL of 0.664 m acetic acid solution. Within that flask the following reactions have reached equilibrium(with their associated equilibrium constant)+–
GGB
HAc(aq) + H2O(`) F
GG H3O (aq) + Ac (aq)+GGB
2 H2O(`) F
GG H3O (aq)+OH–(aq)

Ka = 1.77 × 10−5−14Kw = 10(2.1)(2.2)

What follows is a extremely detailed method of analyzing this problem. No matter how much of the HAc dissociates thefollowing four things will be true.
(1) We will always have conservation of mass
Conservation of Mass:[HAc] + [Ac–] = 0.664 m(2) We will always have electroneutrality (conservation of charge)
Conservation of Charge:
H3O+ = [OH–] + [Ac–](3) We will always have equilibrium expression for the dissociation of HAc
H3O+ · [Ac–]
Equilibrium WA:
Ka =[HAc](4) We will always have equilibrium expression for the autoionization of water
Equilibrium Water:
Kw = H3O+ · [OH–]

(2.3)

(2.4)

(2.5)

(2.6)

The equilibrium constants are known (Ka = 1.77 × 10−5 and Kw = 10−14 ). Thus, the four equations eqs. (2.3) to (2.6)
have these four unknowns

and[OH–]and[HAc]
and[Ac–]
H3O+
Solving numerically for the results we get (an unrealistic number of sig figs are shown and then rounded)≈ 0.661 m[HAc] = 0.660 580 602 88–−3–−12[Ac ] = 3.419 397 119 52 × 10

H3O+ = 3.419 397 122 45 × 10−3
[OH ] = 2.924 492 137 62 × 10

≈ 0.00342 m≈ 0.00342 m≈ 2.92 × 10−12 m

Some observations:(i) The flask initially contains the above (as well as water with [H2O] = 55.5 m). Apart from H2O, the dominate
species present in the flask initially is HAc. Its concentration is about 200 times larger than Ac– and H3O+ andabout 1011 times larger than OH–.
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(ii) The concentration of the Ac– and H3O+ are slightly different because of the autoionization of water. However,they match to eight sig figs which is way, way beyond anything experimental.
(iii) The initial contents of the flask is quite acidic
pH = 2.466
3. Quicker Approximation
Normally these type of weak acid problems are solved more simply. We first note that in this problem
Ka  Kw=⇒Kw ≈ 0

That is, Ka is so much larger than Kw that we could consider Kw to be essentially zero in this problem. In other words,reaction eq. (2.1) for HA dissociation dominates reaction eq. (2.2) for the dissociation of water.
We can use this observation to uncouple the four equations eqs. (2.3) to (2.6). Since Kw is so small we solve eq. (2.6) forthe concentration of OH– and then findKw ≈0[OH–] = H3O+
The concentration of the hydroxide ion is so small that the conservation of charge (eq. (2.4)) gives
H3O+ = [OH–] + [Ac–]=⇒H3O+ ≈ [Ac–](CENTRAL APPROXIMATION)
The concentration of mass (eq. (2.3)) becomes[HAc] = 0.664 − [Ac–]Using these in eq. (2.5) we get
H3O+ · [Ac–]Ka =[HAc]=⇒[HAc] ≈ 0.664 − H3O+=⇒ 
H3O+ · H3O+
Ka ≈0.664 − H3O+

Plugging in for Ka and letting x = H3O+ we getx·x
1.77 × 10−5 ==⇒x = 3.4195 × 10−3
0.664 − xThis numerical results here matches the previous ones to within several sig figs.
In fact, normally we even make a further approximation to simplify the quadratic equation
px·xx·x≈=⇒x ≈ 0.664 · 1.77 × 10−5 = 3.4282 × 10−3
1.77 × 10−5 =0.664 − x0.664
This second approximation gives H3O+ that is only about 0.3% off from the exact result from the previous section.
The point here is not that the normal approximations save us time. That is obviously always true. The point is that we aremaking approximations not just in the mathematics but to the chemistry. We should be aware that these approximations
may not always work.
If either Ka or the initial concentration of HAc were smaller in this problem, then the usual approximations would notwork.
4. Method A: Dilution Using Pure Water
We need to dilute the solution in the flask. The initial concentration is 0.664 m with an initial volume of 0.975 L. We canuse the dilute formula M1 · V1 = M2 · V2 to the initial HAc concentration[HAc]◦ · V = [HAc]◦ · V| {z } | {z }original

newand get
0.664 m · 0.975 L
0.6474=VV
where V is the unknown final volume which is made up of the original 0.975 L and the unknown added water.
(0.664 m) · (0.975 L) = [HAc]◦ · V=⇒[HAc]◦ =We know that the final pH must be 6.420. Therefore, the final equilibrium concentration of H3O+ will be
H3O+ = 10−6.420 = 3.80 × 10−7 m2(4.1)

Using these changes, the four governing equations eqs. (2.3) to (2.6) would become
Conservation of Mass:

0.6474V
3.80 × 10−7 = [OH–] + [Ac–][HAc] + [Ac–] =Conservation of Charge:
Equilibrium WA:Ka =3.80 × 10−7 · [Ac–][HAc]

Kw = 3.80 × 10−7 · [OH–]Equilibrium Water:where the unknowns are nowV

and

[OH–]

and

[HAc]

[Ac–]

andSolving them we get[HAc] = 7.601 354 636 98 × 10−9≈ 7.60 × 10−9 m[Ac–] = 3.538 867 164 02 × 10−7

≈ 3.54 × 10−7 m

V = 1.790 930 467 33 × 106≈ 1.79 × 106 L[OH–] = 2.630 267 991 × 10−8

≈ 2.63 × 10−8 m

The final volume is insanely large
1.79 million Liters
This would fill more than two-thirds of an Olympic swimming pool.
In practice, to make the buffer just through dilution we would only a tiny sample from the original flask and then dilutethat to achieve the desired pH. For example, we could pipette out one micro-liter (1.0 × 10−6 L) from the original flask
and then enough water to that to make 1.84 L.

5. Why Does The “Dilution Formula” Not Work?
The so-called dilution formula is given in textbooks as
M1 · V1 = M2 · V2

(5.1)

One can apply this to the initial HAc charged into the flask. This is exactly what we did above in eq. (4.1). However, wewould be using it only for the initial salt placed in the flask.
This formula can not be applied to the final equilibrium concentrations. That is, the original concentration of thehydronium ion ispH = 2.466=⇒H3O+ old = 10−2.466
and the desired concentration of the hydronium ion is
pH = 6.420=⇒
H3O+ new = 10−6.420
We can NOT use the dilution formula to write
10−2.466 · (0.975 L) = 10−6.420 · V=⇒

V = 8770 LThis volume is wrong. It is off by a factor of ∼ 200 because we have ignored the equilibrium reaction eq. (2.1) for the weakacid HAc. The decrease in H3O+ concentration from dilute will cause more HAc to dissociate to produce more H3O+ but
this requires more dilution.
The dilution formula eq. (5.1) really only works for initial amounts charged to flasks. It does work for equilibrium
concentrations for salts and strong acids and bases because those completely dissociate. Those reactions are completely
shifted to the right.
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6. Method B: Using A Strong Base
Next, we need to add an unknown amount of the 2.50 m KOH to this flask so that the final pH = 6.420. Since KOH is astrong base its dissociation reaction is completely shifted to the right
KOH(aq) GGA K+(aq) + OH–(aq)Kb ≈ ∞Let V represent the final volume. This is made up of the original volume 0.975 L and the volume of the KOH added. The
previous result eq. (4.1) can be used for the new initial HAc concentration[HAc]◦ =0.6474
0.664 m · 0.975 L=VV

The above will be used in the conservation of mass equation below. The final equilibrium concentration of H3O+ will be
H3O+ = 10−6.420 = 3.80 × 10−7 mThis will be used in the conservation of charge equation below. Note also that K+ is another positive ion in this problem
and contributed to the conservation of charge. The final equilibrium concentration of the potassium ion is found using
the dilution formula  +
K · V = K+ · V| {z } | {z }addedat end

The added volume of the KOH is the final volume V minus the volume in the original flask 0.975 L 
 +
2.4375
V − 0.975 L= 2.50 −(2.50 m) · (V − 0.975 L) = K+ · V=⇒K = 2.50 m ·VV
The conservation of charge equation below includes this.
0.6474Conservation of Mass:[HAc] + [Ac–] =V2.4375−7
Conservation of Charge:
3.80 × 10 + 2.50 −= [OH–] + [Ac–]V
3.80 × 10−7 · [Ac–]
Equilibrium WA:
Ka =
[HAc]
Kw = 3.80 × 10−7 · [OH–]Equilibrium Water:

(6.1)
(6.2)
(6.3)
(6.4)

Solving numerically for the results we get (an unrealistic number of sig figs are shown and then rounded)[HAc] = 1.108 126 800 35 × 10−2−1–≈ 0.0111 m[Ac ] = 5.158 966 703 45 × 10

≈ 0.516 m

V = 1.228 514 427 05≈ 1.23 L–[OH ] = 2.630 267 991 9 × 10−8≈ 2.63 × 10−8 m

The amount of KOH that needs to be added is1.2285 L − 0.975 L = 0.253 514 427 053 L254 mL=⇒(6.5)

7. Quicker Approximation
There is a quicker way to solve this problem. First, substitute the HAc from the conservation of mass equation eq. (6.1)
directly into the equilibrium expression for the weak acid eq. (6.3)
Ka =3.80 × 10−7 · [Ac–]0.6474− [Ac–]
Vor

Ka =3.80 × 10−7 · V · [Ac–]0.6474 − V · [Ac–]

Solve this for the product V · [Ac–]V · [Ac–] = 0.6337865023888

(7.1)
–

Note that the above is not an approximation. It represents the exact number of moles of Ac that remain at equilibrium.
Next, since Kw is small compared with Ka we can set [OH–] ≈ 0 in the conservation of charge equation[Ac–] ≈ 3.80 × 10−7 + 2.50 −2.4375
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Multiply through by V and replace 2.4375 with 2.50 · 0.975 so we can factor out the 2.50
V · [Ac–] ≈ 2.50 · (V − 0.975) + V · 3.80 × 10−7The second term above is much smaller than the first term. We ignore it and get
V · [Ac–]
V · [Ac–] ≈ 2.50 · (V − 0.975)=⇒(V − 0.975) ≈2.50
where V − 0.975 represents the volume of KOH solution added.
Using the previous exact result from eq. (7.1) on the right-hand side we get
0.6337865023888(V − 0.975) ≈= 0.253 514 600 955 L
2.50
This matches the exact result from eq. (6.5) to six sig figs.

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