346
Hydrogen-Ion Concentration and pH
(-5 60) must first be changed to a positive number To do this, rewrite it as
10 o 40 x 10 » The antilog of 0 40 (from the log table) is 2 51 This gives
[OH-] = 2 51 x 10-" M
PROBLEMHOW many grams of NaOH must be added to 200 ml of water to give a solution
of pH 11 5
9
SOLUTION:
From the pH we find the pOH, and from that the [OH ] Since NaOH is completely ionized, this [OH~] will also be the NaOH concentration needed
pOH = 14 00 - pH = 14 00 - 1 1 5 = 2 50
Proceeding as in the previous problem, [OH~] = 3 16 x 10~
J M We need 200 ml
of 3 16 x 10~
3 M NaOH, so
wt of NaOH needed = (0 200 liter) (3 16 x 10
3 IH2l£i) LQ 0
\
liter / \
mole
= 0 0253 g NaOH
PROBLEMS A
1 Calculate the pH of solutions with the following H
+ concentrations (in moles/
liter)
(a) 104
(f) 8 9 x 102
(b) 106
^
(g) 3 7 x 10 '
(c) 10
8
(h) 6 5 x 10
H
(d) 10
(i) 3 5
(e) 0012
0) 0 5
2 Calculate the pH of solutions with the following OH concentrations (in
moles/liter)
(a) 10-"
(f) 791 x 10'
2
(b) 10-"
(g) 4 65 x 10 ••
(c) 108
(h) 2 56 x 10
8
(d) 10
(i) 6 5
(e) 0 025
0) 0 72
3 Calculate the H
+ concentration for each of the solutions with the following
values for pH
(a) 361
(f) 8 96
(b) 7 52
(g) 0
(c) 13 43
(h) 2 80
(d) 0 77
(i) -0 6
(e) 6 45
0) 14 8