Doc 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

email doc brown - comments - query?

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?

Sub-index for Part 5

5.1 Equilibria:  Lewis and Bronsted-Lowry acid-base theories

5.2 Self-ionisation of water and pH scale

5.3 Strong acids - examples and pH calculations (this page)

5.4 Weak acids - examples & pH, K_a and pK_a calculations

5.5 Strong bases - examples and pH calculations

5.6 Weak bases - examples and pH, K_b and pK_b calculations

5.0 Basic notes and equations on acids, bases, salts, uses of acid-base 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

• 5.3.1 Definition and examples of STRONG ACIDS

  • Strong acids are highly ionised in water, in many cases approaching 100% dissociation into ions.

  • This means for the general reaction of an acid HX interacting in an acid–base manner with water ...

    • HX_(aq) + H₂O_(l) ==> H₃O⁺_(aq) + X^–_(aq)

    • K_a = [H₃O⁺_(aq)] [X^–_(aq)]/[HX_(aq)] (units are mol dm^–3)

    • K_a the equilibrium constant for this reaction is called the acid dissociation/ionisation constant.

    • [H₂O_(l)] is considered constant and incorporated into K_a.

    • The high % of ionisation gives the maximum concentration of hydrogen ions and therefore the most acidic solution of lowest possible pH.

    • Note that [H₂O_(l)] is considered constant because water is the solvent and on ionisation the concentration of water is only slightly reduced, and is not quoted in the equilibrium expression.

    • Argument:

      • 1000 g of water ~1000 cm³ ~ 1 dm³, M_r(H₂O) = 18.

      • Molarity of water in pure water = 1000/18 = 55.6 mol dm^-3

      • Theoretically this is only reduced to 54.6 if a 1 molar solution of HX completely ionises.

  • In 'strong acid' pH calculations, ionic dissociation is assumed to be ~100% and in reality they have very large K_a values and very negative pK_a values (pKa = –log(K_a/mol dm^–3), compare K_a for weak acids e.g.

    • HCl K_a  ~10⁷ pK_a ~–7, HBr K_a  ~10⁹ pK_a ~–9, HI K_a  ~10¹⁰ pK_a ~–10

    • The Ka is so high it is virtually 100% ionised and so the equilibrium sign is pointless and omitted.

    • Note the acid gets stronger down group 7 as the H–X bond enthalpy/strength decreases as the halogen atom gets bigger and the bond length increases.

    • These are all monobasic/monoprotic acids, meaning only one proton is available for transfer to a base, see HCl below, or the conjugate base of the acid can only accept one proton).

    • -

  • 5.3.1a: HCl_(g) + H₂O_(l) ==> H₃O⁺_(aq) + Cl^–_(aq)

    • dissolving hydrogen chloride in water to form hydrochloric acid,

    • or more simply HCl_(aq) ==> H⁺_(aq) + Cl^–_(aq)

    • This has very high equilibrium constant K_a i.e. virtually 100% to the right.

    • The other gases, hydrogen bromide and hydrogen iodide similarly dissolve to form the very strong hydrobromic acid and hydriodic acid respectively.

    • However, hydrogen fluoride gas dissolves in water to form the relatively weak hydrofluoric acid (see 5.4.2d) in dilute solution.

    • -

  • 5.3.1b: HNO_3(aq) + H₂O_(l) ===> H₃O⁺_(aq) + NO₃^–_(aq)

    • or the dissociation of dilute nitric acid (monobasic/monoprotic) can be simply shown as

    • HNO_3(aq) ==> H⁺_(aq) + NO₃^–_(aq)

    • K_a = 40, pK_a = –1.4

    • -

  • 5.3.1c: H₂SO_4(l) + 2H₂O_(l) ===> 2H₃O⁺_(aq) + SO₄^2–_(aq)

    • dissolving concentrated sulfuric acid in water to make dilute sulfuric acid (dibasic).

    • or more simply: H₂SO_4(aq) ===> 2H⁺_(aq) + SO₄^2–_(aq)

    • Strictly speaking the ionization occurs in two stages since it is a dibasic/diprotic^* acid

      1. H₂SO_4(aq) ===> H⁺_(aq) + HSO₄^–_(aq)

         • K_a1 is very high, pK_a1 is very negative, when the 1st conjugate base, HSO₄^-, is  formed, this ionisation is ~100%, but the 2nd is a weak acid ionisation.

         • So the molecule of sulfuric acid is a VERY strong acid.

      2. HSO₄^–_(aq) H⁺_(aq) + SO₄^2–_(aq)

         • K_a2 =  1.20 x 10^–2 mol dm^–3, pK_a2 = 1.92, positive, when 2nd conjugate base formed.

         • Strictly speaking, ionisation 2. is incomplete, but is often ignored in AS–A2 level calculations.

         • So, H₂SO₄ is a strong acid but the first conjugate based formed, the hydrogensulfate ion, HSO₄^– is a weak acid!

    • ^*Dibasic/diprotic means a maximum of two protons per molecule are available for transfer to a base, or the 2nd conjugate base of the acid can accept two protons. (See section 5.1.3 for more examples)

  • See 5.3 for a brief comparison of selected weak/strong acid properties

• 5.3.2 Calculating the pH of a strong acid

  • Calculation example 5.3.2a

    • (a) Calculate the hydrogen ion concentration and pH of a 0.25 mol dm^–3 solution of hydrochloric acid.

      • HCl is monobasic/monoprotic acid, so [H⁺_(aq)] = 0.25 mol dm^–3

      • pH = –log(0.25) = 0.602

  • Calculation example 5.3.2b

    • (a) Calculate the hydrogen ion concentration and pH of a 1.5 mol dm^–3 solution of sulfuric acid.

      • H₂SO₄ is dibasic/diprotic acid, so [H⁺_(aq)] = 2 x 1.5 = 3.0 mol dm^–3

        • This isn't strictly true, the 1st ionisation is 100%, but the ionisation of the hydrogensulfate ion to release the 2nd proton is not complete, but 100% ionisation is assumed at this academic level.

      • pH = –log(3.0) = –0.477

      • but in reality the ph will be higher (see above discussion on the double ionisation of sulfuric acid and the calculation in appendix 1 at the bottom of the page).

  • Calculation example 5.3.2c

    • Calculate the hydrogen ion concentration solution of hydrochloric acid of pH 1.2

    • [H⁺_(aq)] = 10^–pH = 10^–1.2 = 0.0631 mol dm^–3

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 H₂SO₄ (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 –log₁₀(1.000) = 0.00

BUT, what is the reality?

The 1st dissociation is virtually complete: H₂SO_4(aq) ==> H⁺_(aq) + HSO₄^–_(aq)

Ka1 =	[H+(aq)] [HSO4–(aq)]
	–––––––––––––––    =   VERY LARGE
	[H2SO4(aq)]

and will provide an initial concentration of 0.500 mol dm^–3 of hydrogen ions.

The 2nd ionisation HSO₄^–_(aq) H⁺_(aq) + SO₄^2–_(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.

Ka2 =	[H+(aq)] [SO42–(aq)]
	–––––––––––––––    =   1.20 x 10–2 mol dm–3
	[HSO4–(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 = x² + 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 = –log₁₀(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 HSO₄^– 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 sub-indexes to multiple sections or pages

TOP OF PAGE