Page 305 - Physical chemistry understanding our chemical world

272 ACIDS AND BASES

Strategy: we first calculate the number of moles of hydrochloric
The acetate buffer is acid added. Second, we calculate the new concentrations of ethanoic
an extremely popu- acid and ethanoate. And third, we employ the Henderson–Hasselbach
lar choice in the food
equation once more.
industry. The buffer (1) 10 cm 3 represents one-hundredth of a litre. From
might be described on
Equation (6.42), the number of moles is 0.01 mol.
a food packet as an (2) Before adding the hydrochloric acid, the concentrations
acidity regulator. −3
of ethanoate and ethanoic acid are constant at 0.1 mol dm .The
hydrochloric acid added reacts with the conjugate base in the buffer
−
(the ethanoate anion) to form ethanoic acid. Accordingly, the concentration [CH 3 COO ]
decreases and the concentration [CH 3 COOH] increases. (We assume the reaction is
quantitative.) Therefore, the concentration of ethanoate is (0.1 − 0.01) mol dm −3 =
−3
0.09 mol dm . The concentration of ethanoic acid is (0.1 + 0.01) mol dm −3 =
−3
0.11 mol dm .
(3) Inserting values into Equation (6.50):

0.09
pH = 4.70 + log 10
0.11
pH = 4.70 + log (0.818)
pH = 4.70 + (−0.09)

so
pH = 4.61

So, we see how the pH shifts by less than one tenth of a pH unit after adding quite
a lot of acid. Adding this same amount of HCl to distilled water would change the
pH from 7 to 2, a shift of five pH units.

SAQ 6.14 Consider the ammonia–ammonium buffer in Worked Exam-
3
ple 6.12. Starting with 1 dm of buffer solution containing 0.05 mol dm −3
each of NH 3 and NH 4 Cl, calculate the pH after adding 8 cm 3 of NaOH
solution of concentration 0.1 mol dm −3 .

Justification Box 6.5
We start by writing the equilibrium constant for a weak acid HA dissociating in water,
+
HA + H 2 O → H 3 O + A , where each ion is solvated. The dissociation constant for
−
the acid K a is given by Equation (6.35):
+
−
[H 3 O ][A ]
K a =
[HA]
where, as usual, we ignore the water term. Taking logarithms of Equation (6.35) yields
+
−
[H 3 O ][A ]
log K a = log (6.51)
10 10
[HA]