The pH of Weak Acid and Bases

The hydrogen ion concentration of a solution of a weak acid is much weaker (by orders of magnitude) than a solution of a strong acid of the same concentration, and the pH of the former is, therefore, greater.

Equilibrium equation:

CH₃COOH (aq) + H₂O (l) <=====> H₃O⁺ (aq) + CH₃COO^- (aq)

Species CH₃COOH H₃O⁺ CH₃OO^-
Step 1 - Initial concentration 0.10 0.00 0.00
Step 2 - Change in
concentration -x +x +x
Step 3 - Equilibrium
concentration 0.10-x x x
Step 4 --- --- ---

Neglecting x in the denominator (it is small), it follows that x²is approximately = 0.1x1.8x10⁵. Hence x >> 1.3x10mol L^-1 and pH >> 2.9. One can also calculate % of molecules ionised = [HA ionised] / [HA original] x 100% = 1.3% here.

In general, the neglect of the small factor in the denominator in step 4 is justified if the degree of ionisation is less than 5%.

Species	CH₃COOH	H₃O⁺	CH₃OO^-
Step 1 - Initial concentration	0.10	0.00	0.00
Step 2 - Change in concentration	-x	+x	+x
Step 3 - Equilibrium concentration	0.10-x	x	x
Step 4	---	---	---

Polyprotic acids and bases

The pK_A constants of an acid such as phosphoric acid are discussed, especially the trend with increasing ionisation.

H₃A (aq) + H₂0 (l) <====> H₃O⁺ (aq) + H₂A^- (aq) This yields the equilibrium constant of :

H₂A^- (aq) + H₂0 (l) <====> H₃O⁺ (aq) + HA^2- (aq) This yields the equilibrium constant of :

HA^2-(aq) + H₂0 (l) <====> H₃O⁺ (aq) + A^3- (aq) This yields the equilibrium constant of :

Buffer Solutions - the Henderson Hasselbach equation

The shape of the titration curves for weak acids and strong bases (or vice versa) show very slow variation in the region around half way to the stoichiometric point (near pH = pK). Hence, when the concentrations of an ion and its parent weak acid (or base) are roughly equal one can add small amounts of acid or base with very little effect on the pH. Solutions which show this property are called buffers. Cell fluids are buffered solutions (blood has pH 7.4) and even a small deviation in pH can cause death or illness.

Buffer solutions can be analysed by the following sequence of operations:

HA (aq) + H₂O (l) <====> H₃O⁺ (aq) + A^-(aq)

The above equation is exact. For a weak acid the percentage ionisation of HA, and the protonation of Aare very low in a mixed solution of acid and salt. For this reason we can say that to a good approximation,

which is called the Henderson- Hasselbach equation.

Calculating the pH of a buffer solution

Calculate the pH of 100mL of a buffer solution comprising 0.10M ethanoic acid and 0.10M sodium acetate, given that the pK_aacetic acid is 4.74. This is achieved as follows.

State the equilibrium

CH₃COOH (aq) + H₂O (l) <=====> H₃O⁺ (aq) + CH₃COO^- (aq)

Using the Henderson Hasselbach equation,

Acids, Bases and Salts

Students are expected to know all of material from Atkins and Beran Chapter 14. Students will be expected to read this chapter and to be able to do basic pH calculations. They will also be expected to know the material from Chapter 15 up to and including section 15.5 (not including section 15.6 onwards on solubility equilibria).