## 1  Single Acids

### 1.1  The Usual Way

We will first review the simplest method for calculating the pH of a solution for a generic weak acid, hydrofluoric acid, HF, at a fairly high concentration.

What is the [H3O+]eq in a 0.100 M solution of HF (aq) at 25C?
Using the acid ionization reaction with its associated Ka = 6.76 × 10−4:

 R HF (aq) + H2O (ℓ) ⇋ F− (aq) + H3O+ (aq) I 0.100 M 0.0 M 1.00 × 10−7 ≈ 0.0 M C −x +x +x E 0.100 − x x x
6.76 × 10−4 =
 x2 0.100−x

 6.76 × 10−5 − 6.76 × 10−4 x = x2
 x2 − 6.76 × 10−5 + 6.76 × 10−4 x = 0
 7.891 × 10−3
or  −8.567 × 10−3

Instead of solving this quadratic equation with the quadratic formula, you can iterate. x, how far the reaction needs to go to get to equilibrium, is going to be small because Ka is small compared to 0.100 M. If x is very small, x ≈ 0.0, then 0.100 - x ≈ 0.100.

6.76 × 10−4 =
 x2 0.100

 x = 8.22 × 10−3

Now that we have a good guess for x, plug it in instead of 0.0 for the second iteration:

6.76 × 10−4 =
 x2 0.100−8.22 × 10−3

 x = 7.88 × 10−3

Now that we have a better guess for x, plug it in for the third iteration:

6.76 × 10−4 =
 x2 0.100−7.88 × 10−3

 x = 7.89 × 10−3

Fourth iteration:

6.76 × 10−4 =
 x2 0.100−7.89 × 10−3

 x = 7.89 × 10−3

Now we stop because the x we put in is the same x that was returned; the iterations have converged.

Regardless of the mathematical method used to solve the equation:

 [H3O+]eq = x = 7.89 × 10−3 M

This leads to a pH = −log(7.89 × 10−3) = 2.103.

### 1.2  Correction to the usual way

We’ll first address the approximation that [H3O+]0 = 0.0 M.

We know to assume that [H3O+] is zero at initial ”time" because the weak acids we usually test over in general chemistry are strong enough AND in high enough concentrations to make enough H3O+ from acid ionization to dwarf the amount of H3O+ from the auto-ionization of water.

The generic equation used to solve weak acid equilibrium problems is an approximation because we assume [H3O+]0 = 0 M:

 x2 [HA]0 − x
= Ka

This is only valid when:

• Ka is large enough (i.e. the weak acid is strong enough) so that x = [H3O+]eq from the weak acid’s ionization is way larger than that made from water auto-ionization (i.e. [H3O+]0 = 1.00 × 10−7 M).
• [HA]0 is large enough that the amount of H3O+ produced is larger than that made from water auto-ionization.

Let’s work a problem where Ka is very small and with a low acid concentration to see where this approximate equation fails.

What is the pH of 1.00 × 10−5 M HCN with Ka = 6.2 × 10−10?

Assuming [H3O+]0 = 0 M:

 x2 1.00 × 10−5 − x
= 6.2 × 10−10

The solution is:

 x = 7.84 × 10−8
 pH = −log7.84 × 10−8 = 7.10

This pH makes no sense because HCN is an acid, but at 25C, 7.10 is a basic pH.

The amount of H3O+ produced from HCN was less than what is produced by the auto-ionization of water.

A simple fix would be to add the amount of H3O+ from the auto-ionization of water, 1 × 10−7 M H3O+ to get an approximation:

 1.00 × 10−7 + 7.84 × 10−8 = 1.784 × 10−7
 pH = −log1.784 × 10−7 = 6.75

This pH makes more sense, but it is wrong which we can prove by solving for the Ka of HCN.

Ka =

 [H3O+][CN−] [HCN]

=
 (1.784 × 10−7)(7.84 × 10−8) 1 × 10−5 − 7.84 × 10−8
= 1.41 × 10−9 ≠ 6.2 × 10−10

We really need to not assume [H3O+]0 = 0 M in our RICE table.

 R HCN (aq) + H2O (ℓ) ⇋ CN− (aq) + H3O+ (aq) I 1 × 10−5 0 1 × 10−7 C −x +x +x E 1 × 10−5 − x x 1 × 10−7 + x

So the equation to solve is:

 (x)(1 × 10−7 + x) 1 × 10−5 − x
= 6.2 × 10−10

Written generically:

 (x)(1 × 10−7 + x) [HA]0 − x
= Ka

If solved by iteration:

x =
 6.2 × 10−10(1 × 10−5 − x) 1 × 10−7 + x

Starting with x = 0 on the right hand side:

 x = 0 x = 6.20 × 10−8 x = 3.80 × 10−8 x = 4.47 × 10−8 x = 4.26 × 10−8 x = 4.33 × 10−8 x = 4.31 × 10−8 x = 4.31 × 10−8

Having converged:

 [H3O+] = 1 × 10−7 + 4.31 × 10−8 = 1.43 × 10−7 M
 pH = −log1.43 × 10−7 = 6.845

Is this correct? Calculate Ka!

Ka =

 [H3O+][CN−] [HCN]

=
 (1.43 × 10−7)(4.31 × 10−8) 1 × 10−5 − 4.31 × 10−8
= 6.21 × 10−10 ≈ 6.2 × 10−10 = Ka

But this is still wrong because there must be electric charge balance in the solution. The concentration of positive things (H3O+) must be equal to the concentration of negative things (OH and CN).

[OH] + [CN] =
 Kw [H3O+]eq
+ [CN]eq =
 1 × 10−14 1.43 × 10−7
+ 4.31 × 10−8 = 1.13 × 10−7 ≠ 1.43 × 10−7 = [H3O+

As you can see, the concentration of anions is not equal to the concentration of cations.

This issue of charge balance can actually be addressed in how the original problem is solved by treating the two equilibrium problems in tandem (water auto-ionization and acid ionization).

However, this method is long and often requires the solution of a cubic equation (x3 instead of quadratic, x2) which is not trivial, i.e. hard to do.

A quick fix that usually works is to simply average the concentrations of positive with negative things to figure out how they need to meet in the middle; the geometric average seems to align better with the exact solution where the two equilibria are solved simultaneously.

The geometric average is the square root of the product of the two numbers instead of the arithmetic average which is the sum of the two numbers divided by 2.

 ([OH−]+[CN−])([H3O+])
=
 (1.13 × 10−7)(1.431 × 10−7)
= 1.27 × 10−7

For what it’s worth, the arithmetic average is 1.28 × 10−7.

 [H3O+] = [OH−] + [CN−] = 1.27 × 10−7 M
 pH = −log(1.27 × 10−7) = 6.896
[OH] =
 1 × 10−14 1.27 × 10−7
= 7.87 × 10−8 M

 [CN−] = 1.27 × 10−7 − [OH−] = 1.27 × 10−7 − 7.87 × 10−8 = 4.83 × 10−8 M
 [HCN] = 1 × 10−5 − [CN−] = 1 × 10−5 − 4.83 × 10−8 = 9.95 × 10−6 M

We can also realize how much H3O+ came from HCN and how much came from water’s auto-ionization.

 [H3O+]HCN = [CN−] = 4.83 × 10−8 M
 [H3O+]H2O = [OH−] = 7.87 × 10−8 M

Curiously, water made more H3O+ that HCN.

These relationships are true because of the demands of charge balance!

We can see that the auto-ionization of water was suppressed by the acid’s generation of H3O+ because 7.87 × 10−8 < 1 × 10−7.

Let’s check our Ka to make sure everything is still okay.

Ka =

 [H3O+][CN−] [HCN]

=
 (1.27 × 10−7)(4.83 × 10−8) 9.95 × 10−6
= 6.17 × 10−10 ≈ 6.2 × 10−10 = Ka

Looks like we are good to go.

It is important to note that you cannot use the simpler equilibrium expression (i.e. ignore auto-ionization), add the auto-ionization after the fact (+1 × 10−7), and then simply make the charge balance correction as this will lead to an H3O+ concentration that is too high and you would not recover the correct Ka value.

Were you to do this bad thing, [H3O+] = 1.549 × 10−7 M and Ka = 1.41 × 10−09 ≠ 6.2 × 10−10 = Ka, so it’s obviously wrong.

Lastly, there is even more work to be done because this work ignores thermodynamic activities. Oh well.

### 1.3  Iterative Solution

There are 2 equilibrium reactions to consider that generate H3O+:

1. 2 H2O ⇌ OH + H3O+
2. HY + H2O ⇌ Y + H3O+

This gives 2 interdependent RICE tables because both reactions generate H3O+:

 R 2 H2O (ℓ) ⇋ OH− (aq) + H3O+ (aq) I n/a 0.0 M y C n/a +x +x E n/a x y+x
Kw = (x)(x+y)              x =
 Kw (x+y)

 R HY (aq) + H2O (ℓ) ⇋ Y− (aq) + H3O+ (aq) I [HY]0 0.0 M x C −y +y +y E [HY]0 − y y x+y
Ka =
 (y)(x+y) ([HY]0 − y)
(x+y) =
 Ka([HY]0 − y) y

Substituting the solution for (x+y) from Ka into the equation for x gives x in terms of y:

x =
 y Kw Ka([HY]0 − y)
(1)

Substituting these into Ka gives:

Ka =
(y)

 y Kw Ka([HY]0 − y)
+ y

([HY]0 − y)

Collecting y:

Ka =
(y2)

 Kw Ka([HY]0 − y)
+ 1

([HY]0 − y)

Giving the numerators all the same denominator:

Ka =
(y2)

 Kw Ka([HY]0 − y)
+
 Ka([HY]0 − y) Ka([HY]0 − y)

([HY]0 − y)

Combining numerators:

Ka =
(y2)

 Kw + Ka([HY]0 − y) Ka([HY]0 − y)

([HY]0 − y)

Moving the bottom denominator over:

Ka([HY]0 − y) = (y2)

 Kw + Ka([HY]0 − y) Ka([HY]0 − y)

Moving the new denominator over since it’s the same as the left-hand side:

 [Ka([HY]0 − y)]2 = (y2) ⎛ ⎝ Kw + Ka([HY]0 − y) ⎞ ⎠

Rearranging for y2:

[Ka([HY]0 − y)]2
 ⎛ ⎝ Kw + Ka([HY]0 − y) ⎞ ⎠
= y2

Solving for y (taking the positive solution only):

 √

[Ka([HY]0 − y)]2
 ⎛ ⎝ Kw + Ka([HY]0 − y) ⎞ ⎠

= y     (2)
1. Solve for y in Eq. 2 the first time assuming all y inside √ are zero.
2. Solve for x in Eq. 1 using y.
3. Solve for y...
4. Solve for x........until they converge.

[H3O+]eq = x + y

#### Example

[HY]0 = 1.00 × 10−5 M HCN with Ka = 6.2 × 10−10 at 25C where Kw = 1.00 × 10−14.

y =
 √

[Ka([HY]0 − 0.0)]2
 ⎛ ⎝ Kw + Ka([HY]0 − 0.0) ⎞ ⎠

= 4.87 × 10−8  M H3O+ from HCN
x =
 y Kw Ka([HY]0 − y)
= 7.90 × 10−8  M H3O+ from H2

Based on our first iteration, most of the H3O+ will be coming from the auto-ionization of water.

y =
 √

[Ka([HY]0 − y)]2
 ⎛ ⎝ Kw + Ka([HY]0 − y) ⎞ ⎠

= 4.85 × 10−8  M H3O+ from HCN
x =
 y Kw Ka([HY]0 − y)
= 7.86 × 10−8  M H3O+ from H2

We are already very close to a solution; iterating again gives the same values to 3 significant figures.

y =
 √

[Ka([HY]0 − y)]2
 ⎛ ⎝ Kw + Ka([HY]0 − y) ⎞ ⎠

= 4.85 × 10−8  M H3O+ from HCN
x =
 y Kw Ka([HY]0 − y)
= 7.86 × 10−8  M H3O+ from H2

[H3O+]eq = x + y = 1.27 × 10−7 M (pH = 6.896).

Notice we achieved the correct solution without having to worry about charge balance.

———————————————————————–

## 2  Mixture of Two Acids

There are 3 equilibrium reactions to consider that generate H3O+:

1. 2 H2O ⇌ OH + H3O+
2. HY + H2O ⇌ Y + H3O+
3. HZ + H2O ⇌ Z + H3O+

This gives 3 interdependent RICE tables because all 3 reactions generate H3O+:

 R 2 H2O (ℓ) ⇋ OH− (aq) + H3O+ (aq) I n/a 0.0 M y+z C n/a +x +x E n/a x x+y+z
Kw = (x)(x+y+z)              x =
 Kw (x+y+z)

 R HY (aq) + H2O (ℓ) ⇋ Y− (aq) + H3O+ (aq) I [HY]0 0.0 M x+z C −y +y +y E [HY]0 − y y x+y+z
Kay =
 (y)(x+y+z) ([HY]0 − y)
(x+y+z) =
 Kay([HY]0 − y) y

 R HZ (aq) + H2O (ℓ) ⇋ Y− (aq) + H3O+ (aq) I [HZ]0 0.0 M x+y C −z +z +z E [HZ]0 − z z x+y+z
Kaz =
 (z)(x+y+z) ([HZ]0 − z)
z =
 Kaz([HZ]0 − z) (x+y+z)

Substituting the solution for (x+y+z) from Kay into the equations for x and z give them in terms of y:

x =
 y Kw Kay([HY]0 − y)
(3)
z =
 y Kaz([HZ]0 − z) Kay([HY]0 − y)
(4)

Substituting these into Kay gives:

Kay =
(y)

 y Kw Kay([HY]0 − y)
+ y +
 y Kaz([HZ]0 − z) Kay([HY]0 − y)

([HY]0 − y)

Collecting y:

Kay =
(y2)

 Kw Kay([HY]0 − y)
+ 1 +
 Kaz([HZ]0 − z) Kay([HY]0 − y)

([HY]0 − y)

Giving the numerators all the same denominator:

Kay =
(y2)

 Kw Kay([HY]0 − y)
+
 Kay([HY]0 − y) Kay([HY]0 − y)
+
 Kaz([HZ]0 − z) Kay([HY]0 − y)

([HY]0 − y)

Combining numerators:

Kay =
(y2)

 Kw + Kay([HY]0 − y) + Kaz([HZ]0 − z) Kay([HY]0 − y)

([HY]0 − y)

Moving the bottom denominator over:

Kay([HY]0 − y) = (y2)

 Kw + Kay([HY]0 − y) + Kaz([HZ]0 − z) Kay([HY]0 − y)

Moving the new denominator over since it’s the same as the left-hand side:

 [Kay([HY]0 − y)]2 = (y2) ⎛ ⎝ Kw + Kay([HY]0 − y) + Kaz([HZ]0 − z) ⎞ ⎠

Rearranging for y2:

[Kay([HY]0 − y)]2
 ⎛ ⎝ Kw + Kay([HY]0 − y) + Kaz([HZ]0 − z) ⎞ ⎠
= y2

Solving for y (taking the positive solution only):

 √
[Kay([HY]0 − y)]2
 ⎛ ⎝ Kw + Kay([HY]0 − y) + Kaz([HZ]0 − z) ⎞ ⎠
= y     (5)
1. Solve for y in Eq. 5 the first time assuming all y and z inside √ are zero.
2. Solve for x in Eq. 3 using y.
3. Solve for z in Eq. 4 using y and assuming z=0 on the right-hand side the first time.
4. Solve for y...
5. Solve for x...
6. Solve for z........until they converge.

[H3O+]eq = x + y + z

#### Example

[HY]0 = 0.200 M HCOOH with Kay = 1.70 × 10−4 and [HZ]0 = 0.400 M HClO with Kaz = 2.95 × 10−8 at 25C where Kw = 1.00 × 10−14.

y =
 √
[Kay([HY]0 − 0.0)]2
 ⎛ ⎝ Kw + Kay([HY]0 − 0.0) + Kaz([HZ]0 − 0.0) ⎞ ⎠
= 5.83 × 10−3  M H3O+ from HCOOH
x =
 y Kw Kay([HY]0 − y)
= 1.77 × 10−12  M H3O+ from H2
z =
 y Kaz([HZ]0 − 0.0) Kay([HY]0 − y)
= 2.08 × 10−6  M H3O+ from HClO

Based on our first iteration, most of the H3O+ will be coming from the strongest acid present, HCOOH, as expected.

y =
 √
[Kay([HY]0 − y)]2
 ⎛ ⎝ Kw + Kay([HY]0 − y) + Kaz([HZ]0 − z) ⎞ ⎠
= 5.74 × 10−3  M H3O+ from HCOOH
x =
 y Kw Kay([HY]0 − y)
= 1.74 × 10−12  M H3O+ from H2
z =
 y Kaz([HZ]0 − z) Kay([HY]0 − y)
= 2.05 × 10−6  M H3O+ from HClO

We are already very close to a solution; iterating again gives almost the same values to 3 significant figures.

y =
 √
[Kay([HY]0 − y)]2
 ⎛ ⎝ Kw + Kay([HY]0 − y) + Kaz([HZ]0 − z) ⎞ ⎠
= 5.75 × 10−3  M H3O+ from HCOOH
x =
 y Kw Kay([HY]0 − y)
= 1.74 × 10−12  M H3O+ from H2
z =
 y Kaz([HZ]0 − z) Kay([HY]0 − y)
= 2.05 × 10−6  M H3O+ from HClO

Once more for good measure:

y =
 √
[Kay([HY]0 − y)]2
 ⎛ ⎝ Kw + Kay([HY]0 − y) + Kaz([HZ]0 − z) ⎞ ⎠
= 5.75 × 10−3  M H3O+ from HCOOH
x =
 y Kw Kay([HY]0 − y)
= 1.74 × 10−12  M H3O+ from H2
z =
 y Kaz([HZ]0 − z) Kay([HY]0 − y)
= 2.05 × 10−6  M H3O+ from HClO

[H3O+]eq = x + y + z = 5.75 × 10−3 M (pH = 2.242) which is the same value had we simply ignored water’s ionization and HClO entirely.

However, [ClO]eq = z = 2.05 × 10−6 M which is not something obtainable from treating only the acidity of HCOOH.

———————————————————————–

This document was translated from LATEX by HEVEA.