pH of a strong acid – Examples
pH of a strong acid - Examples
pH of a strong acid - Examples
For strong acids at equilibrium :
[H+] = C where C is the initial concentration of the strong 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 |
Data given
| [HCl] = 0.1 M | Asked for | pH = ? |
|
Write down the data given and the unknown. |
II |
pH = -log [H+] (1)
HCl → H+ + Cl- (2)
Initially | 0.1 Μ | 0 Μ | 0 Μ | Change | -0.1 M | 0.1 M | 0.1 M | Final (at equilibrium | 0 | 0.1 M | 0.1 M |
|
Write down equations relating data given and unknowns. Write down the relevant chemical equation. [H+] and pH’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 ΙΙ and (3): pH = -log [H+] = -log (0.1) = -log (10-1) = -(-1) = 1
pH = 1 |
|
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 |
Data given
|
HCl solution [HCl] = 10-7 M | Asked for | pH = ? |
|
Write down the data given and the unknown. |
II | Write down equations relating data given and unknowns. Write down the relevant chemical equation. [H+] and pH’s are unknowns but they can be calculated.
pH = -log([H3O+]) (1)
HCl dissociates in water according to:
ΗCl + H2O → H3O+ + Cl- (2) 10-7 mol/ℓ 10-7 mol/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] ≤ 10-7 Μ. Suppose that χ mol/ℓ water react and χ mol/ℓ H3O+ is produced: H2O + H2O → H3O+ + OH- (3) -x mol/ℓ react +x mol/ℓ is produced
From (2)+(3) : [H3O+] = (10-7 + x) M και [OH-] = x M (4)
For kw at 25 °C:
kw = [H3O+].[OH-] = 10-14 (5)
|
III |
By substituting in equation (5) the concentrations derived in (4) at step ΙΙ χ can 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)
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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)*10-7 M (7) From (1) and (7): pH = -log((1.6)*10-7) = -log(1.6)-log(10-7) = -0.2-(-7) ≈ 6.8 |
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Relevant Posts
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)
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