pH of a Polyprotic Strong Acid (Sulfuric Acid)
Key Concepts
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Concept: Approximate pH of Sulfuric Acid
Sulfuric acid, H_{2}SO_{4}(aq), is a diprotic acid, it dissociates (ionises) in 2 stages by losing a proton, H^{+}, at each stage^{1}:
Stage 1: 
H_{2}SO_{4} donates a proton 

H_{2}SO_{4}(aq) → H^{+}(aq) + HSO_{4}^{}(aq) 
Stage 2: 
HSO_{4}^{} donates a proton 

HSO_{4}^{}(aq) → H^{+}(aq) + SO_{4}^{2}(aq) 
If we assume that both H_{2}SO_{4}(aq) and HSO_{4}^{}(aq) are both strong acids that dissociate fully then all the HSO_{4}^{}(aq) produced in the first stage will dissociate into H^{+}(aq) and SO_{4}^{2}(aq):
H_{2}SO_{4}(aq) 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{}(aq) 
HSO_{4}^{}(aq) 
→ 
H^{+}(aq) 
+ 
SO_{4}^{2}(aq)


H_{2}SO_{4}(aq) 
→ 
2H^{+}(aq) 
+ 
SO_{4}^{2}(aq) 
A solution of sulfuric acid will contain H^{+}(aq) and SO_{4}^{2}(aq) but will NOT contain H_{2}SO_{4}(aq) nor HSO_{4}^{}(aq).
From the balanced chemical equation above, we can use the mole ratio (or stoichiometric ratio) to see that 1 mole of H_{2}SO_{4}(aq) will produce 2 moles of H^{+}(aq)
So, n moles of H_{2}SO_{4}(aq) will produce 2n moles of H^{+}(aq)
Since the volume of solution is the same for both H_{2}SO_{4}(aq) and H^{+}(aq), we can say that
the concentration of H^{+}(aq), [H^{+}(aq)] is 2 times the concentration of H_{2}SO_{4}(aq), that is 2[H_{2}SO_{4}(aq)],:
[H^{+}(aq)] = 2 [H_{2}SO_{4}(aq)]
Therefore we can calculate an approximate pH for the sulfuric acid solution:
pH = log_{10}[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.
 What is the question asking you to do?
Calculate the approximate pH
pH = ?
 What information (data) has been given in the question?
[H_{2}SO_{4}(aq)] = 0.5 mol L^{1}
 How will you calculate the pH?
pH = log_{10}[H^{+}(aq)]
 What will you need to find out before you can calculate the pH?
concentration of H^{+}(aq), [H^{+}(aq)]
 How will you calculate [H^{+}(aq)]?
 Assume that both H_{2}SO_{4}(aq) and HSO_{4}^{}(aq) are strong acids that dissociate fully.
 Write the balanced chemical equation based on this assumption:
H_{2}SO_{4}(aq) → 2H^{+}(aq) + SO_{4}^{2}(aq)
 Use the stoichiometric (mole) ratio to calculate the concentration of H^{+}(aq):
H^{+}(aq)  :  H_{2}SO_{4}(aq) 
2  :  1 
[H^{+}(aq)]  =  2[H_{2}SO_{4}(aq)] 
 =  2 × 0.5 
 =  1.0 mol L^{1} 
 Calculate the pH of the sulfuric acid solution:
pH = log_{10}[H^{+}(aq)]
[H^{+}(aq)] = 1.0 mol L^{1} (calculated in the step above)
pH = log_{10}[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?
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:
H_{2}SO_{4}(aq) → H^{+}(aq) + HSO_{4}^{}(aq) K_{a1} is very large
After this first stage the species in solution are H^{+}(aq) and HSO_{4}^{}(aq), but H_{2}SO_{4}(aq) is NOT present in solution.
From the balanced chemical equation above we see that :
 1 mol H_{2}SO_{4}(aq) dissociates to produce 1 mole of H^{+}(aq) and 1 mole of HSO_{4}^{}(aq)
 So n moles of H_{2}SO_{4}(aq) dissociate to produce n moles of H^{+}(aq) and n moles of HSO_{4}^{}(aq)
 Therefore [H^{+}(aq)] = [HSO_{4}^{}(aq)] = stated concentration of sulfuric acid
In the second stage, the HSO_{4}^{}(aq) partially dissociates:
HSO_{4}^{}(aq) H^{+}(aq) + SO_{4}^{2}(aq) K_{a2} = 1.2 × 10^{2} (at 25°C)
K_{a2} 
= 
[H^{+}(aq)][SO_{4}^{2}(aq)]
[HSO_{4}^{}(aq)] 
= 
1.2 × 10^{2} 
After this second stage the species in solution are:
 HSO_{4}^{}(aq) that has not dissociated
 SO_{4}^{2}(aq) from the HSO_{4}^{} that has dissociated
 H^{+}(aq) from 2 sources, the initial full dissociation of H_{2}SO_{4}(aq) AND from the partial dissociation of HSO_{4}^{}(aq)
For the first stage of the dissociation, H_{2}SO_{4}(aq) → H^{+}(aq) + HSO_{4}^{}(aq) :
 Let c = stated concentration of sulfuric acid
 Then c = [H^{+}(aq)] = [HSO_{4}^{}(aq)]
For the second stage of the dissociation, HSO_{4}^{}(aq) H^{+}(aq) + SO_{4}^{2}(aq)
Set up a R.I.C.E. Table as shown below:
R.I.C.E. Table 
Reaction: 
HSO_{4}^{}_{(aq)} 

H^{+}_{(aq)} 
+ 
SO_{4}^{2}_{(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:
[HSO_{4}^{}(aq)] = c (from the first stage of dissociation)
[H^{+}(aq)] = c (from the first stage of dissociation)
[SO_{4}^{2}(aq)] = 0 (no SO_{4}^{2}(aq) is present until some HSO_{4}^{} dissociates)
 change in concentration, x, as a result of the dissociation of HSO_{4}^{}(aq):
[HSO_{4}^{}(aq)] will decrease by an amount, x
[H^{+}(aq)] will increase by the same amount, x
[SO_{4}^{2}(aq)] will increase by the same amount, x
 final concentrations:
[HSO_{4}^{}(aq)] = c  x
[H^{+}(aq)] = c + x
[SO_{4}^{2}(aq)] = 0 + x = x
So, the pH of the sulfuric acid solution can then be calculated:
 pH = log_{10}[H^{+}(aq)]
 pH = log_{10}[c + x]
We can use the expression for the acid dissociation constant, K_{a2}, to find the value of x
K_{a2} 
= 
[H^{+}(aq)][SO_{4}^{2}(aq)]
[HSO_{4}^{}(aq)] 
1.2 × 10^{2} 
= 
[c + x][x]
[c  x] 
1.2 × 10^{2} [c  x] 
= 
[c + x][x] 
1.2 × 10^{2}c  1.2 × 10^{2}x 
= 
[c + x][x] 
1.2 × 10^{2}c  1.2 × 10^{2}x 
= 
cx + x^{2} 
x^{2} + cx + 1.2 × 10^{2}x  1.2 × 10^{2}c 
= 
0 
x^{2} + (c + 1.2 × 10^{2})x  1.2 × 10^{2}c 
= 
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^{2}c is also a known number.
This means we have a quadratic equation of the form:
ax^{2} + bx + constant = 0
which we can solve for X:
x = 
b ± √(b^{2} 4 × a × constant)
2a 
x = 
[c + 1.2 × 10^{2}] ± √( [c + 1.2 × 10^{2}]^{2}  [4 × 1 × 1.2 × 10^{2}c] )
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 = log_{10}[H^{+}(aq)] = log_{10}[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}.
 What is the question asking you to do?
Calculate pH
pH = ?
 What information (data) has been given in the question?
[H_{2}SO_{4}^{2}(aq)] = 0.5 mol L^{1}
K_{a2} = 1.2 × 10^{2} (at 25°C)
 How will you calculate pH?
pH = log_{10}[H^{+}(aq)]
 What will you need to find before you can calculate the pH?
concentration of H^{+}(aq) in solution, that is, [H^{+}(aq)]
 How will you calculate [H^{+}(aq)]?
 First stage of dissociation: Assume complete dissociation of H_{2}SO_{4}(aq):
H_{2}SO_{4} → H^{+}(aq) + HSO_{4}^{}(aq)
Calculate intial concentrations of H^{+}(aq) and HSO_{4}^{}(aq)
stoichiometric ratio H^{+}(aq) : H_{2}SO_{4} : HSO_{4}^{}(aq) is 1 : 1 : 1
0.5 mol L^{1} H_{2}SO_{4} fully dissociates to produce
0.5 mol L^{1} H^{+}(aq) and 0.5 mol L^{1} HSO_{4}^{}(aq)
[H^{+}(aq)] = 0.5 mol L^{1}
[HSO_{4}^{}(aq)] = 0.5 mol L^{1}
 Second stage of dissociation, HSO_{4}^{} partially dissociates:
Set up a R.I.C.E. Table as shown below:
R.I.C.E. Table 
Reaction: 
HSO_{4}^{}_{(aq)} 

H^{+}_{(aq)} 
+ 
SO_{4}^{2}_{(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:
K_{a2} 
= 
[H^{+}(aq)][SO_{4}^{2}(aq)]
[HSO_{4}^{}(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^{2}x 
= 
0.5x + x^{2} 
x^{2} + 0.5x + 1.2 × 10^{2}x  6.0 × 10^{3} 
= 
0 
x^{2} + 0.512x  6.0 × 10^{3} 
= 
0 
Calculate only the positive values of x by solving the quadratic equation:
x = 
b ± √(b^{2} 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}
 Calculate the pH of the sulfuric acid solution:
pH = log_{10}[H^{+}(aq)] = log_{10}[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} ?
1. If the dissociation occurs in water, then the proton is hydrated so we can write H^{+}(aq) or H_{3}O^{+}