Advanced pH Calculations

Patrick McLaurin

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 25^∘C?
Using the acid ionization reaction with its associated K_a = 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

quadratic formula: x =

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 K_a 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

= K_a

This is only valid when:

K_a 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 K_a 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 K_a = 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 25^∘C, 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 K_a of HCN.

K_a =

[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

= K_a

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

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 K_a!

K_a =

[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 = K_a

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⁻] =

K_w

[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 K_a to make sure everything is still okay.

K_a =

[H₃O⁺][CN⁻]

[HCN]

(1.27 × 10⁻⁷)(4.83 × 10⁻⁸)

9.95 × 10⁻⁶

= 6.17 × 10⁻¹⁰ ≈ 6.2 × 10⁻¹⁰ = K_a

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 K_a value.

Were you to do this bad thing, [H₃O⁺] = 1.549 × 10⁻⁷ M and K_a = 1.41 × 10⁻⁰⁹ ≠ 6.2 × 10⁻¹⁰ = K_a, 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⁺:

2 H₂O ⇌ OH⁻ + H₃O⁺
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

K_w = (x)(x+y) x =

K_w

(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

K_a =

(y)(x+y)

([HY]₀ − y)

(x+y) =

K_a([HY]₀ − y)

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

x =

y K_w

K_a([HY]₀ − y)

(1)

Substituting these into K_a gives:

K_a =

(y)

⎛
⎜
⎜
⎝

y K_w

K_a([HY]₀ − y)

+ y

⎞
⎟
⎟
⎠

([HY]₀ − y)

Collecting y:

K_a =

(y²)

⎛
⎜
⎜
⎝

K_w

K_a([HY]₀ − y)

+ 1

⎞
⎟
⎟
⎠

([HY]₀ − y)

Giving the numerators all the same denominator:

K_a =

(y²)

⎛
⎜
⎜
⎝

K_w

K_a([HY]₀ − y)

⎞
⎟
⎟
⎠

([HY]₀ − y)

Combining numerators:

K_a =

(y²)

⎛
⎜
⎜
⎝

K_w + K_a([HY]₀ − y)

K_a([HY]₀ − y)

⎞
⎟
⎟
⎠

([HY]₀ − y)

Moving the bottom denominator over:

K_a([HY]₀ − y) = (y²)

⎛
⎜
⎜
⎝

K_w + K_a([HY]₀ − y)

K_a([HY]₀ − y)

⎞
⎟
⎟
⎠

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

[K_a([HY]₀ − y)]² = (y²)

⎛
⎝

K_w + K_a([HY]₀ − y)

⎞
⎠

Rearranging for y²:

[K_a([HY]₀ − y)]²

⎛
⎝

K_w + K_a([HY]₀ − y)

⎞
⎠

= y²

Solving for y (taking the positive solution only):

√

[K_a([HY]₀ − y)]²

⎛
⎝

K_w + K_a([HY]₀ − y)

⎞
⎠

= y (2)

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

[H₃O⁺]_eq = x + y

Example

[HY]₀ = 1.00 × 10⁻⁵ M HCN with K_a = 6.2 × 10⁻¹⁰ at 25^∘C where K_w = 1.00 × 10⁻¹⁴.

y =

√

[K_a([HY]₀ − 0.0)]²

⎛
⎝

K_w + K_a([HY]₀ − 0.0)

⎞
⎠

= 4.87 × 10⁻⁸ M H₃O⁺ from HCN

x =

y K_w

K_a([HY]₀ − y)

= 7.90 × 10⁻⁸ M H₃O⁺ from H₂O

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

y =

√

[K_a([HY]₀ − y)]²

⎛
⎝

K_w + K_a([HY]₀ − y)

⎞
⎠

= 4.85 × 10⁻⁸ M H₃O⁺ from HCN

x =

y K_w

K_a([HY]₀ − y)

= 7.86 × 10⁻⁸ M H₃O⁺ from H₂O

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

y =

√

[K_a([HY]₀ − y)]²

⎛
⎝

K_w + K_a([HY]₀ − y)

⎞
⎠

= 4.85 × 10⁻⁸ M H₃O⁺ from HCN

x =

y K_w

K_a([HY]₀ − y)

= 7.86 × 10⁻⁸ M H₃O⁺ from H₂O

[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⁺:

2 H₂O ⇌ OH⁻ + H₃O⁺
HY + H₂O ⇌ Y⁻ + H₃O⁺
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

K_w = (x)(x+y+z) x =

K_w

(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

K_{a_y} =

(y)(x+y+z)

([HY]₀ − y)

(x+y+z) =

K_{a_y}([HY]₀ − 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

K_{a_z} =

(z)(x+y+z)

([HZ]₀ − z)

z =

K_{a_z}([HZ]₀ − z)

(x+y+z)

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

x =

y K_w

K_{a_y}([HY]₀ − y)

(3)

z =

y K_{a_z}([HZ]₀ − z)

K_{a_y}([HY]₀ − y)

(4)

Substituting these into K_{a_y} gives:

K_{a_y} =

(y)

⎛
⎜
⎜
⎝

y K_w

K_{a_y}([HY]₀ − y)

+ y +

y K_{a_z}([HZ]₀ − z)

K_{a_y}([HY]₀ − y)

⎞
⎟
⎟
⎠

([HY]₀ − y)

Collecting y:

K_{a_y} =

(y²)

⎛
⎜
⎜
⎝

K_w

K_{a_y}([HY]₀ − y)

+ 1 +

K_{a_z}([HZ]₀ − z)

K_{a_y}([HY]₀ − y)

⎞
⎟
⎟
⎠

([HY]₀ − y)

Giving the numerators all the same denominator:

K_{a_y} =

(y²)

⎛
⎜
⎜
⎝

K_w

K_{a_y}([HY]₀ − y)

K_{a_z}([HZ]₀ − z)

K_{a_y}([HY]₀ − y)

⎞
⎟
⎟
⎠

([HY]₀ − y)

Combining numerators:

K_{a_y} =

(y²)

⎛
⎜
⎜
⎝

K_w + K_{a_y}([HY]₀ − y) + K_{a_z}([HZ]₀ − z)

K_{a_y}([HY]₀ − y)

⎞
⎟
⎟
⎠

([HY]₀ − y)

Moving the bottom denominator over:

K_{a_y}([HY]₀ − y) = (y²)

⎛
⎜
⎜
⎝

K_w + K_{a_y}([HY]₀ − y) + K_{a_z}([HZ]₀ − z)

K_{a_y}([HY]₀ − y)

⎞
⎟
⎟
⎠

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

[K_{a_y}([HY]₀ − y)]² = (y²)

⎛
⎝

K_w + K_{a_y}([HY]₀ − y) + K_{a_z}([HZ]₀ − z)

⎞
⎠

Rearranging for y²:

[K_{a_y}([HY]₀ − y)]²

⎛
⎝

K_w + K_{a_y}([HY]₀ − y) + K_{a_z}([HZ]₀ − z)

⎞
⎠

= y²

Solving for y (taking the positive solution only):

√

[K_{a_y}([HY]₀ − y)]²

⎛
⎝

K_w + K_{a_y}([HY]₀ − y) + K_{a_z}([HZ]₀ − z)

⎞
⎠

= y (5)

Solve for y in Eq. 5 the first time assuming all y and z inside √ are zero.
Solve for x in Eq. 3 using y.
Solve for z in Eq. 4 using y and assuming z=0 on the right-hand side the first time.
Solve for y...
Solve for x...
Solve for z........until they converge.

[H₃O⁺]_eq = x + y + z

Example

[HY]₀ = 0.200 M HCOOH with K_{a_y} = 1.70 × 10⁻⁴ and [HZ]₀ = 0.400 M HClO with K_{a_z} = 2.95 × 10⁻⁸ at 25^∘C where K_w = 1.00 × 10⁻¹⁴.

y =

√

[K_{a_y}([HY]₀ − 0.0)]²

⎛
⎝

K_w + K_{a_y}([HY]₀ − 0.0) + K_{a_z}([HZ]₀ − 0.0)

⎞
⎠

= 5.83 × 10⁻³ M H₃O⁺ from HCOOH

x =

y K_w

K_{a_y}([HY]₀ − y)

= 1.77 × 10⁻¹² M H₃O⁺ from H₂O

z =

y K_{a_z}([HZ]₀ − 0.0)

K_{a_y}([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 =

√

[K_{a_y}([HY]₀ − y)]²

⎛
⎝

K_w + K_{a_y}([HY]₀ − y) + K_{a_z}([HZ]₀ − z)

⎞
⎠

= 5.74 × 10⁻³ M H₃O⁺ from HCOOH

x =

y K_w

K_{a_y}([HY]₀ − y)

= 1.74 × 10⁻¹² M H₃O⁺ from H₂O

z =

y K_{a_z}([HZ]₀ − z)

K_{a_y}([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 =

√

[K_{a_y}([HY]₀ − y)]²

⎛
⎝

K_w + K_{a_y}([HY]₀ − y) + K_{a_z}([HZ]₀ − z)

⎞
⎠

= 5.75 × 10⁻³ M H₃O⁺ from HCOOH

x =

y K_w

K_{a_y}([HY]₀ − y)

= 1.74 × 10⁻¹² M H₃O⁺ from H₂O

z =

y K_{a_z}([HZ]₀ − z)

K_{a_y}([HY]₀ − y)

= 2.05 × 10⁻⁶ M H₃O⁺ from HClO

Once more for good measure:

y =

√

[K_{a_y}([HY]₀ − y)]²

⎛
⎝

K_w + K_{a_y}([HY]₀ − y) + K_{a_z}([HZ]₀ − z)

⎞
⎠

= 5.75 × 10⁻³ M H₃O⁺ from HCOOH

x =

y K_w

K_{a_y}([HY]₀ − y)

= 1.74 × 10⁻¹² M H₃O⁺ from H₂O

z =

y K_{a_z}([HZ]₀ − z)

K_{a_y}([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 L^AT_EX by H^EV^EA.