Page 75 - Vogel's TEXTBOOK OF QUANTITATIVE CHEMICAL ANALYSIS

BUFFER SOLUTIOWS 2.20

comparison with c, and cs. Equation (19) then reduces to:

Ca [Acid]
CH'] =-.Ka, or CH+] =-
CS [sait] Ka
[Salt]
or pH = pK, +log----
[ Acid]
The equations can be readily expressed in a somewhat more general form
when applied to a Br~nsted-Lowry acid A and its conjugate base B:

(e.g. CH3COOH and CH3CO0 -, etc.). The expression for pH is:

where Ka = [H+][B]/[A].
Similarly for a mixture of a weak base of dissociation constant Kb and its
Salt with a strong acid:

[Base]
[OH-] = ---- x Kb
[Salt]
[Salt]
or pOH = pK, + log ---
[Base]
Confining attention to the case in which the concentrations of the acid and
its Salt are equal, i.e. of a half-neutralised acid then pH = pK,. Thus the pH of
a half-neutralised solution of a weak acid is equal to the negative logarithm
of the dissociation constant of the acid. For acetic (ethanoic) acid, Ka =
1.75 x mol L-', pK, = 4.76; a half-neutralised solution of, Say 0.1M acetic
acid wiil have a pH of 4.76. If we add a small concentration of H + ions to such
a solution, the former will combine with acetate ions to form undissociated
acetic acid:
H+ + CH3COO- = CH3COOH
Similarly, if a small concentration of hydroxide ions be added, the latter will
combine with the hydrogen ions arising from the dissociation of the acetic acid
and form water; the equilibrium wiil be disturbed, and more acetic acid will
dissociate to replace the hydrogen ions removed in this way. In either case, the
concentration of the acetic acid and acetate ion (or salt) will not be appreciably
changed. It follows from equation (21) that the pH of the solution will not be
materially affected.
Example 12. Calculate the pH of the solution produced by adding 10 mL of
1 M hydrochloric acid to 1 L of a solution which is 0.1 M in acetic (ethanoic)
acid and 0.1 M in sodium acetate (Ka = 1.75 x 10-'mol L-').
The pH of the acetic acid-sodium acetate buffer solution is given by the
equation:
[Salt]
pH = pK, + log - = 4.76 + 0.0 = 4.76
[ Acid]