Sample Problem 19.3
(continued)
= 4.3 x 10-6 M
pH = 5.37
(c) When 40.00 mL of NaOH are added, all of the HPr will be reacted and the [Pr-] will be:
Ka x Kb = Kw
Kb = Kw/Ka = 1.0 x 10-14/1.3 x 10-5 = 7.7 x 10-10
pH = 8.80
(d) 50.00 mL of NaOH will produce an excess of OH-.
mol excess OH- = (0.1000 M)(0.05000 L - 0.04000 L) = 0.00100 mol
M = 0.01111
[H3O+] = 1.0 x 10-14/0.01111 = 9.0 x 10-13 M
pH = 12.05
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