 Calculating pH of a Polyprotic Strong Acid (Sulfuric Acid) Chemistry Tutorial

Key Concepts

• A polyprotic acid is a Brønsted-Lowry acid that is able to dontate more than one proton.

A diprotic acid is able to donate two protons
A triprotic acid is able to donate three protons
• A strong acid is an acid that fully dissociates (ionises).

A weak acid is an acid that does not fully dissociate (ionise).

• Sulfuric acid is a strong diprotic acid.
• There are two stages to the dissociation of H2SO4 each with its own acid dissociation constant, Ka:

 Stage 1:     Stage 2: H2SO4 → H+(aq) + HSO4- Ka1 HSO4-(aq) H+(aq) + SO42- Ka2

• Ka1 is much, much, larger than Ka2
Ka1 >> Ka2
• Ka1 is large enough for us to assume that the first stage of the dissociation goes to completion.
• If we assume Ka2 is large enough for the second stage dissociation reaction to also go to completion, we can quickly calculate a very approximate value for the pH of a sulfuric acid solution.
• If we want a more accurate value for the pH of a sulfuric acid solution, we must use the value of Ka2 to calculate the final concentration of H+(aq) in solution.
This involves :

(i) setting up a R.I.C.E. Table

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Concept: Approximate pH of Sulfuric Acid

Sulfuric acid, H2SO4(aq), is a diprotic acid, it dissociates (ionises) in 2 stages by losing a proton, H+, at each stage1:

Stage 1: Stage 2: H2SO4 donates a proton H2SO4(aq) → H+(aq) + HSO4-(aq) HSO4- donates a proton HSO4-(aq) → H+(aq) + SO42-(aq)

If we assume that both H2SO4(aq) and HSO4-(aq) are both strong acids that dissociate fully then all the HSO4-(aq) produced in the first stage will dissociate into H+(aq) and SO42-(aq):

 H2SO4(aq) → H+(aq) + HSO4-(aq) HSO4-(aq) → H+(aq) + SO42-(aq) H2SO4(aq) → 2H+(aq) + SO42-(aq)

A solution of sulfuric acid will contain H+(aq) and SO42-(aq) but will NOT contain H2SO4(aq) nor HSO4-(aq).

From the balanced chemical equation above, we can use the mole ratio (or stoichiometric ratio) to see that 1 mole of H2SO4(aq) will produce 2 moles of H+(aq)
So, n moles of H2SO4(aq) will produce 2n moles of H+(aq)

Since the volume of solution is the same for both H2SO4(aq) and H+(aq), we can say that
the concentration of H+(aq), [H+(aq)] is 2 times the concentration of H2SO4(aq), that is 2[H2SO4(aq)],:

[H+(aq)] = 2 [H2SO4(aq)]

Therefore we can calculate an approximate pH for the sulfuric acid solution:

pH = -log10[H+(aq)]

As we shall see below, the pH we calculate using these assumptions is only very approximate!

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Example: Calculating Approximate pH of Sulfuric Acid

Calculate the approximate pH of a 0.5 mol L-1 aqueous solution of sulfuric acid.

1. What is the question asking you to do?
Calculate the approximate pH
pH = ?
2. What information (data) has been given in the question?
[H2SO4(aq)] = 0.5 mol L-1
3. How will you calculate the pH?
pH = -log10[H+(aq)]
4. What will you need to find out before you can calculate the pH?
concentration of H+(aq), [H+(aq)]
5. How will you calculate [H+(aq)]?

• Assume that both H2SO4(aq) and HSO4-(aq) are strong acids that dissociate fully.
• Write the balanced chemical equation based on this assumption:
H2SO4(aq) → 2H+(aq) + SO42-(aq)
• Use the stoichiometric (mole) ratio to calculate the concentration of H+(aq):

 H+(aq) : H2SO4(aq) 2 : 1 [H+(aq)] = 2[H2SO4(aq)] = 2 × 0.5 = 1.0 mol L-1

6. Calculate the pH of the sulfuric acid solution:
pH = -log10[H+(aq)]
[H+(aq)] = 1.0 mol L-1 (calculated in the step above)
pH = -log10[1.0] = 0

The approximate pH of a 0.5 mol L-1 aqueous solution of sulfuric acid is 0

Quick Question

What is the approximate pH of a 0.0639 mol L-1 aqueous solution of sulfuric acid?

pH =

Concept: Calculating pH of Sulfuric Acid using R.I.C.E. Table and Quadratic Equation

Sulfuric acid is a strong acid, it fully dissociates in the first stage, that is, its first acid dissociation constant, Ka1, is very large so that the equilibrium position lies almost completely to the right:

H2SO4(aq) → H+(aq) + HSO4-(aq)     Ka1 is very large

After this first stage the species in solution are H+(aq) and HSO4-(aq), but H2SO4(aq) is NOT present in solution.
From the balanced chemical equation above we see that :

• 1 mol H2SO4(aq) dissociates to produce 1 mole of H+(aq) and 1 mole of HSO4-(aq)
• So n moles of H2SO4(aq) dissociate to produce n moles of H+(aq) and n moles of HSO4-(aq)
• Therefore [H+(aq)] = [HSO4-(aq)] = stated concentration of sulfuric acid

In the second stage, the HSO4-(aq) partially dissociates:

HSO4-(aq) H+(aq) + SO42-(aq)     Ka2 = 1.2 × 10-2 (at 25°C)

 Ka2 = [H+(aq)][SO42-(aq)][HSO4-(aq)] = 1.2 × 10-2

After this second stage the species in solution are:

• HSO4-(aq) that has not dissociated
• SO42-(aq) from the HSO4- that has dissociated
• H+(aq) from 2 sources, the initial full dissociation of H2SO4(aq) AND from the partial dissociation of HSO4-(aq)

For the first stage of the dissociation, H2SO4(aq) → H+(aq) + HSO4-(aq) :

• Let c = stated concentration of sulfuric acid
• Then c = [H+(aq)] = [HSO4-(aq)]

For the second stage of the dissociation, HSO4-(aq) H+(aq) + SO42-(aq)
Set up a R.I.C.E. Table as shown below:

R.I.C.E. Table
Reaction: HSO4-(aq) H+(aq) + SO42-(aq)
Initial concentration: c   c   0
Change in concentration: -x   +x   +x
Equilibrium concentration: c - x   c + x   0 + x

Explanation of entries in R.I.C.E. Table:

• initial concentrations:
[HSO4-(aq)] = c (from the first stage of dissociation)
[H+(aq)] = c (from the first stage of dissociation)
[SO42-(aq)] = 0 (no SO42-(aq) is present until some HSO4- dissociates)
• change in concentration, x, as a result of the dissociation of HSO4-(aq):
[HSO4-(aq)] will decrease by an amount, x
[H+(aq)] will increase by the same amount, x
[SO42-(aq)] will increase by the same amount, x
• final concentrations:
[HSO4-(aq)] = c - x
[H+(aq)] = c + x
[SO42-(aq)] = 0 + x = x

So, the pH of the sulfuric acid solution can then be calculated:

• pH = -log10[H+(aq)]
• pH = -log10[c + x]

We can use the expression for the acid dissociation constant, Ka2, to find the value of x

 Ka2 = [H+(aq)][SO42-(aq)][HSO4-(aq)] 1.2 × 10-2 = [c + x][x][c - x] 1.2 × 10-2 [c - x] = [c + x][x] 1.2 × 10-2c - 1.2 × 10-2x = [c + x][x] 1.2 × 10-2c - 1.2 × 10-2x = cx + x2 x2 + cx + 1.2 × 10-2x - 1.2 × 10-2c = 0 x2 + (c + 1.2 × 10-2)x - 1.2 × 10-2c = 0

Because we will know the value of c, that is, the stated concentration of the sulfuric acid, (c + 1.2 x 10-2) is a known number and 1.2 x 10-2c is also a known number.
This means we have a quadratic equation of the form:

ax2 + bx + constant = 0

which we can solve for X:
 x = -b ± √(b2 -4 × a × constant)2a x = -[c + 1.2 × 10-2] ± √( [c + 1.2 × 10-2]2 - [4 × 1 × 1.2 × 10-2c] )2 × 1 x = -[c + 1.2 × 10-2] ± √( [c + 1.2 × 10-2]2 - [0.048 × c] )2 (We only use positive values of x, ignore negative values)

We can then use this value of x to calculate the final concentration of H+(aq) in solution:
[H+(aq)] = c + x
(remember than n was the concentration of H+(aq) after the first dissociation)

Then we can use this concentration of H+(aq) to calculate the pH of the sulfuric acid solution:
pH = -log10[H+(aq)] = -log10[c + x]

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Example: Calculating pH of Sulfuric Acid using R.I.C.E. Table and Quadratic Equation

Calculate the pH of a 0.5 mol L-1 aqueous solution of sulfuric acid at 25°C given Ka2 = 1.2 × 10-2.

1. What is the question asking you to do?
Calculate pH
pH = ?
2. What information (data) has been given in the question?
[H2SO42-(aq)] = 0.5 mol L-1
Ka2 = 1.2 × 10-2 (at 25°C)
3. How will you calculate pH?
pH = -log10[H+(aq)]
4. What will you need to find before you can calculate the pH?
concentration of H+(aq) in solution, that is, [H+(aq)]
5. How will you calculate [H+(aq)]?

• First stage of dissociation: Assume complete dissociation of H2SO4(aq):
H2SO4 → H+(aq) + HSO4-(aq)

Calculate intial concentrations of H+(aq) and HSO4-(aq)
stoichiometric ratio H+(aq) : H2SO4 : HSO4-(aq) is 1 : 1 : 1
0.5 mol L-1 H2SO4 fully dissociates to produce
0.5 mol L-1 H+(aq) and 0.5 mol L-1 HSO4-(aq)
[H+(aq)] = 0.5 mol L-1
[HSO4-(aq)] = 0.5 mol L-1
• Second stage of dissociation, HSO4- partially dissociates:
Set up a R.I.C.E. Table as shown below:
R.I.C.E. Table
Reaction: HSO4-(aq) H+(aq) + SO42-(aq)
Initial concentration: 0.5   0.5   0
Change in concentration: -x   +x   +x
Equilibrium concentration: 0.5 - x   0.5 + x   0 + x

Use the equilibrium expression to write an expression for the value of x:
 Ka2 = [H+(aq)][SO42-(aq)][HSO4-(aq)] 1.2 × 10-2 = [0.5 + x][x][0.5 - x] 1.2 × 10-2 [0.5 - x] = [0.5 + x][x] 6.0 × 10-3 - 1.2 × 10-2x = 0.5x + x2 x2 + 0.5x + 1.2 × 10-2x - 6.0 × 10-3 = 0 x2 + 0.512x - 6.0 × 10-3 = 0

Calculate only the positive values of x by solving the quadratic equation:
 x = -b ± √(b2 -4 × a × constant)2a x = -[0.512] ± √( [0.512]2 - [4 × 1 × -6.0 × 10-3] )2 × 1 x = -[0.512] ± √( [0.262] - [-0.024] )2 x = -0.512 ± √0.2862 x = -0.512 ± 0.5352 (i) x = (-0.512 + 0.535)/2 = 0.012 (ii) x = (-0.512 - 0.535)/2 = -0.52 We only use positive values of x, ignore negative values, so x = 0.012

Substitute the value of x back into the expression for [H+(aq)]:
[H+(aq)] = 0.5 + x = 0.5 + 0.12 = 0.512 mol L-1
6. Calculate the pH of the sulfuric acid solution:
pH = -log10[H+(aq)] = -log10[0.512] = 0.29

The pH of 0.5 mol L-1 aqueous sulfuric acid at 25°C is 0.29

Quick Question

What is the pH of a 0.0824 mol L-1 aqueous solution of sulfuric acid given Ka2 = 1.2 × 10-2 ?

pH =

1. If the dissociation occurs in water, then the proton is hydrated so we can write H+(aq) or H3O+