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, K_a:

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

• K_a1 is much, much, larger than K_a2

  K_a1 >> K_a2

• K_a1 is large enough for us to assume that the first stage of the dissociation goes to completion.
• If we assume K_a2 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 K_a2 to calculate the final concentration of H⁺(aq) in solution.

  This involves :

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

  (ii) solving a quadratic equation.

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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:

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 SO₄^2-(aq):

A solution of sulfuric acid will contain H⁺(aq) and SO₄^2-(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!

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) + SO₄^2-(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, K_a1, is very large so that the equilibrium position lies almost completely to the right:

H₂SO₄(aq) → H⁺(aq) + HSO₄^-(aq)     K_a1 is very large

• 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) + SO₄^2-(aq)     K_a2 = 1.2 × 10^-2 (at 25°C)

• HSO₄^-(aq) that has not dissociated
• SO₄^2-(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) + SO₄^2-(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)
  [SO₄^2-(aq)] = 0 (no SO₄^2-(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
  [SO₄^2-(aq)] will increase by the same amount, x

• final concentrations:

  [HSO₄^-(aq)] = c - x
  [H⁺(aq)] = c + x
  [SO₄^2-(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, K_a2, to find the value of x

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

[H⁺(aq)] = c + x
(remember than n was the concentration of H⁺(aq) after the first dissociation)

pH = -log₁₀[H⁺(aq)] = -log₁₀[c + x]

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 K_a2 = 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?

   [H₂SO₄^2-(aq)] = 0.5 mol L^-1
   K_a2 = 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 ± √0.286 2
     x =	-0.512 ± 0.535 2
     (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 K_a2 = 1.2 × 10^-2 ?

pH =