Page 36 - Modern Analytical Chemistry

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Chapter 2 Basic Tools of Analytical Chemistry 19

EXAMPLE 2.3
The maximum allowed concentration of chloride in a municipal drinking
–
2
water supply is 2.50 ´10 ppm Cl . When the supply of water exceeds this
limit, it often has a distinctive salty taste. What is this concentration in moles
–
Cl /liter?
SOLUTION
2
.
250 ´ 10 mg Cl – 1 g 1 mole Cl – 3 –
10 M
.
´ ´ = 705 ´
m
L 1 000 g 35 453 gCl –
.

2B.6 p-Functions
Sometimes it is inconvenient to use the concentration units in Table 2.4. For exam-
ple, during a reaction a reactant’s concentration may change by many orders of mag-
nitude. If we are interested in viewing the progress of the reaction graphically, we
might wish to plot the reactant’s concentration as a function of time or as a function
of the volume of a reagent being added to the reaction. Such is the case in Figure 2.1,
+
where the molar concentration of H is plotted (y-axis on left side of figure) as a
+
function of the volume of NaOH added to a solution of HCl. The initial [H ] is 0.10
M, and its concentration after adding 75 mL of NaOH is 5.0 ´10 –13 M. We can easily
+
follow changes in the [H ] over the first 14 additions of NaOH. For the last ten addi-
+
tions of NaOH, however, changes in the [H ] are too small to be seen.
When working with concentrations that span many orders of magnitude, it is
often more convenient to express the concentration as a p-function. The p-func- p-function
tion of a number X is written as pX and is defined as A function of the form pX, where
pX =-log(X).
pX = –log(X)
+
Thus, the pH of a solution that is 0.10 M H is
+
pH =–log[H ] =–log(0.10) =1.00
+
and the pH of 5.0 ´10 –13 M H is
+
pH =–log[H ] =–log(5.0 ´10 –13 ) =12.30
+
Figure 2.1 shows how plotting pH in place of [H ] provides more detail about how
+
the concentration of H changes following the addition of NaOH.
EXAMPLE 2. 4

–3
What is pNa for a solution of 1.76 ´10 M Na 3 PO 4 ?
SOLUTION
+
Since each mole of Na 3 PO 4 contains three moles of Na , the concentration of
+
Na is
3 mol Na + –3 –3
+
[Na ] = ´1 .76 ´10 M =5 .28 ´ 10 M

mol Na PO 4
3
and pNa is
+
–3
pNa =–log[Na ] =–log(5.28 ´10 ) =2.277