Secobarbital [C_12 H_15N_2 O_2] is usually delivered as the sodium salt (NaC_12,
ID: 513044 • Letter: S
Question
Explanation / Answer
pKa = 7.8
pKb = 14 - 7.8 = 6.2
pKb = -log(Kb)
Kb = 6.31 x 10^-7
initial [NaC12H17N2O3] = 0.1 g/238.3 g/mol x 0.4 L = 1.05 x 10^-3 M
NaC12H17N2O3 --> Na+ + C12H17N2O3-
C12H17N2O3- + H2O <==> C12H18N2O3 + OH-
let x amount has reacted
Kb = [C12H18N2O3][OH-]/[C12H17N2O3-]
6.31 x 10^-7 = x^2/1.05 x 10^-3
x = [OH-] = 2.57 x 10^-5 M
pOH = -log[OH-] = 4.59
thus,
pH = 14 - pOH = 9.41
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