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Brown's Advanced A Level Chemistry Revision Notes  theoretical–Physical
Advanced Level
Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 5
Part
5.3 Definition of a strong acid, theory, examples and pH calculations
of strong acids
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What is a strong acid? How to calculate the pH of a strong acid
solution given its concentration. What is the real pH of sulfuric acid solutions?
Subindex for
Part 5
Equilibria:
Lewis and BronstedLowry acidbase theories
Selfionisation of water and pH scale
Strong acids  examples and pH calculations
Weak acids  examples & pH, K_{a} and pK_{a} calculations
Strong bases  examples and pH calculations
Weak bases  examples and pH, K_{b} and pK_{b} calculations
Basic notes and equations on acids, bases, salts,
uses of
acidbase titrations  upgrade from GCSE!
5.3 Definition, examples and pH calculations
of strong acids
Note: H^{+}(aq) = aqueous hydrogen
ion = aqueous proton = oxonium ion = hydroxonium ion
Appendix 1.
What is the real aqueous hydrogen ion concentration in dilute sulfuric
acid?
To fully understand this
calculation its handy to have studied the
weak acid calculations page.
e.g. take 0.500 molar H_{2}SO_{4}
(aq), if fully ionised, you would expect ...
the [H+(aq)]
concentration to be 2 x 0.500 = 1.000 mol dm^{–3}
and the pH to be
–log_{10}(1.000) = 0.00
BUT, what is the reality?
The 1st dissociation is
virtually complete: H_{2}SO_{4(aq)} ==>
H^{+}_{(aq)}
+ HSO_{4}^{–}_{(aq)}
K_{a1} =

[H^{+}_{(aq)}]
[HSO_{4}^{–}_{(aq)}] 
–––––––––––––––
= VERY LARGE 
[H_{2}SO_{4(aq)}] 
and will provide an
initial concentration of 0.500 mol dm^{–3} of hydrogen ions.
The 2nd ionisation
HSO_{4}^{–}_{(aq)}
H^{+}_{(aq)}
+ SO_{4}^{2–}_{(aq)} is not complete,
The hydrogensulfate ion is a
weak acid, but will still provide
further hydrogen ions, which can be calculated via a weak acid
calculation using the equilibrium expression below.
K_{a2} =

[H^{+}_{(aq)}]
[SO_{4}^{2}^{–}_{(aq)}] 
–––––––––––––––
= 1.20 x 10^{–2}
mol dm^{–3} 
[HSO_{4}^{–}_{(aq)}] 
Prior to the 2nd
ionisation, theoretically, the initial concentrations of
hydrogensulfate ions and hydrogen ions will be equal, and both 0.500
mol dm^{–3}.
On the 2nd ionisation, the
hydrogensulfate ion will provide the extra hydrogen ions and the only
sulfate ions, but in doing so, the hydrogensulfate ion is reduced.
If we call the 'equal' extra
hydrogen ion concentration and the final sulfate ion concentration x,
then the total hydrogen ion concentration is (0.5 + x) and the
hydrogensulfate ion concentration is reduced to (0.5 – x)
1.20 x 10^{–2} =

(0.5 + x) x 
––––––––––––––– 
(0.5 – x) 
This gives the quadratic
equation 0 = x^{2} + 0.512x – 0.006
On solving this using the
quadratic equation formula, gives roots of –0.523 and +0.0114
Therefore x must be 0.0114
(the 'extra' H^{+}), which then gives (0.5 + 0.0114) ...
a total hydrogen ion
concentration [H^{+}(aq)] of 0.511 mol dm^{–3},
not 1.000 mol dm^{–3}
and the real calculated
pH = –log_{10}(0.511) =
0.29, not pH 0.00 and significantly
higher.
You may think the extra
hydrogen ion concentration is very low, but this is because the HSO_{4}^{–}
ion is a weak acid AND the 2nd ionisation is actually heavily suppressed
by the 1st ionisation – think Le Chatelier's equilibrium principle as
regards the concentration effect.
To fully understand this
calculation its handy to have studied the
weak acid calculations page.
WHAT NEXT?
Advanced Equilibrium Chemistry Notes Part 1. Equilibrium,
Le Chatelier's Principle–rules
* Part 2. K_{c} and K_{p} equilibrium expressions and
calculations * Part 3.
Equilibria and industrial processes * Part 4
Partition between two
phases, solubility product K_{sp}, common ion effect,
ion–exchange systems *
Part 5. pH, weak–strong acid–base theory and
calculations * Part 6. Salt hydrolysis,
acid–base titrations–indicators, pH curves and buffers * Part 7.
Redox equilibria, half–cell electrode potentials,
electrolysis and electrochemical series
*
Part 8.
Phase equilibria–vapour
pressure, boiling point and intermolecular forces watch out for subindexes
to multiple sections or pages
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