Page 285 - Physical chemistry understanding our chemical world

252 ACIDS AND BASES

Table 6.4 The pHs of naturally occurring substances, listed in order
of decreasing acidity
Human body Common foodstuffs
Gastric juices 1.0–3.0 Limes 1.8–2.0
Faeces 4.6–8.4 Rhubarb 3.1–3.2
Duodenal contents 4.8–8.2 Apricots 3.6–4.0
Urine 4.8–8.4 Tomatoes 4.0–4.4
Saliva 6.5–7.5 Spinach 5.1–5.7
Milk 6.6–7.6 Salmon 6.1–6.3
Bile 6.8–7.0 Maple syrup 6.5–7.0
Spinal fluid 7.3–7.5 Tap water 6.5–8.0
Blood plasma 7.3–7.5 Egg white (fresh) 7.6–8.0
Source: Handbook of Chemistry and Physics (66th Edition), R. C. Weast (ed.),
CRC Press, Boca Raton, Florida, 1985, page D-146.

substances is higher than 7, so we cannot call them either ‘acidic’ or ‘neutral’. The
pHs range from 1.8 for limes (which explains why they taste so sour) to 7.8 or so
for fresh egg white (albumen).
Blood ‘plasma’ is that part of the blood remaining after removal
The word ‘product’ of the haemoglobin cells that impart a characteristic ‘blood-red’
is used here in its colour. According to Table 6.4, most people’s plasma has a pH
mathematical sense of in the range 7.3–7.5. So, what is the concentration of solvated
‘multiplied by’. protons in such plasma? We met the autoprotolysis constant K w
in Equation (6.4). Although we discussed it in terms of super-pure
water, curiously the relationship still applies to any aqueous system. The product of
the concentrations of solvated protons and hydroxide ions is always 10 −14 at 298 K.
If Equation (6.4) applies although the water contains dissolved solute, then we can
calculate the concentration of solvated protons and the concentration of hydroxide
ions, and hence ascertain what a pH of more than ‘7’ actually means.

−
Worked Example 6.8 What is the concentration [OH (aq) ] in blood plasma of pH 7.4?

Answer strategy. (1) We first calculate the concentration of solvated protons from the pH,
+
via Equation (6.20). (2) Second, we compare the concentration [H 3 O (aq) ] with that of a
neutral solution, via Equation (6.4).
+
(1) The concentration [H 3 O (aq) ] is obtained from Equation (6.20). Inserting values:
+
[H 3 O (aq) ] = 10 −7.4
so
+
[H 3 O (aq) ] = 4 × 10 −8 mol dm −3
(2) We see how the concentration obtained in part (1) is in fact less than the 1 ×
10 −7 mol dm −3 we saw for pure water, as calculated in Worked Example 6.1.