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
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RESULTS
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NOTES
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I
|
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Write down the data given and the unknown.
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II
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pH = -log [H+] (1)
HCl → H+ +
Cl- (2)
|
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.
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III
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From Step II above at equilibrium: [H+] = 0.1 M (3)
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IV
|
From (1) at Step ΙΙ and (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)
-10-7
M +10-7
M +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+] 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)
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III
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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)
|
The quadratic equation has to be solved since x
is appreciable at low concentrations of acids.
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IV
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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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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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