pH of a strong acid – Examples
In a previous post entitled “Strong Acids and Bases – Ionic Equilibrium – A general relation for the pH of a strong acid” the following equation was derived for the hydronium ion concentration [H+] for strong acids:
For strong acids at equilibrium:
[H+] = C where C is the initial concentration of the acid (1)
Let us see the following examples:
Example #1
Calculate the pH of a 0.1 M solution of HCl.
HCl is considered to be a strong acid (Table I.1). It dissociates completely in water to produce 0.1 M of H+and 0.1 M of Cl-.
HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)
STEPS | RESULTS | NOTES | ||||||||||||
I |
| Write down the data given and the unknown. | ||||||||||||
II | pH = -log [H+] (1) HCl → H+ + Cl- (2)
| Write down equations relating data given and unknowns. Write down the relevant chemical equation. [H+]andpH’s are unknowns but they can be calculated. | ||||||||||||
III | From Step II above at equilibrium: [H+] = 0.1 M (3) |
IV | From (1) at Step ΙΙand (3): pH = -log [H+] = -log (0.1) = -log (10-1) = -(-1) = 1 pH = 1 |
Mathematical approach:
From equation (1) above and for strong acids:
[H+] = C
Since the concentration of a strong acid at equilibrium is given as:
[H+] = C = [HCl] = 0.1 M (2)
The pH of the 0.1 M HCl solution would be:
pH = -log([H+]) = -log (0.1) = -log (10-1) = -(-1) = 1
Example #2
Calculate the pH of a 10-7 M solution of HCl
Since HCl is a strong acid dissociates completely in water to produce at equilibrium 10-7 M H+and 10-7 M Cl- :
HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)
-10-7M +10-7M +10-7 M
From the above reaction at equilibrium:
H3O+= 10-7 M and pH = -log [H+] = -log (10-7) = -(-7) = 7 (2)
However, we do know from theory that an acid solution has to have a pH less than 7. This erroneous result was obtained because we do not take into account the autoionazation of H2O. This autoionaziation of H2O becomes important when the initial concentration of an acid is lower or equal to 10-7.
STEP | RESULTS | NOTES | |||||
I |
| Write down the data given and the unknown. | |||||
II | Write down equations relating data given and unknowns. Write down the relevant chemical equation. [H+]andpH’s are unknowns but they can be calculated. pH = -log([H3O+]) (1) HCldissociates in water according to: ΗCl + H2O→H3O+ + Cl- (2) 10-7mol/ℓ 10-7mol/ℓ From (1): pH = -log([H3O+]) = -log([10-7]) = 7 However, it is known from theory that the pH of an acid must be pH<7. This erroneous result was obtained because the autoionization of water becomes important and must be taken into consideration where the concentration of an acid is [acid] ≤ 10-7Μ. Suppose that χmol/ℓ water react and χmol/ℓ H3O+ is produced: H2O + H2O → H3O+ + OH- (3) -xmol/ℓ react +xmol/ℓ is produced From(2)+(3) : [H3O+] = (10-7 + x) M και [OH-] = xM (4) Forkwat 25 °C: kw = [H3O+].[OH-] = 10-14 (5) | ||||||
III | By substituting in equation (5) the concentrations derived in (4) at step ΙΙ χcan be calculated. From (5) and (4): kw = [H3O+].[OH-] = (10-7 + x). x = 10-14 x2+(10-7).x-10-14 = 0 and x = (6.2)*10-8 M (6) | The quadratic equation has to be solved since x is appreciable at low concentrations of acids. | |||||
IV | From (4) and (6): x = [H3O+] = (10-7 + x) = 10-7 + (6.2)*10-8 = (1.6)*10-7M (7) From (1) and (7): pH = -log((1.6)*10-7) = -log(1.6)-log(10-7) = -0.2-(-7) ≈ 6.8 | ||||||
References
J-L. Burgot “Ionic Equilibria in Analytical Chemistry”, Springer Science & Business Media, 2012
J.N. Butler “Ionic Equilibrium – Solubility and pH calculations”, Wiley – Interscience, 1998
Clayden, Greeves, Waren and Wothers “Organic Chemistry”, Oxford,
D. Harvey, “Modern Analytical Chemistry”, McGraw-Hill Companies Inc., 2000
Toratane Munegumi, World J. of Chem. Education, 1.1, 12-16 (2013)
J.N. Spencer et al., “Chemistry structure and dynamics”, 5th Edition, John Wiley & Sons, Inc., 2012
L. Cardellini, Chem. Educ. Res. Pract. Eur., 1, 151-160, (2000)