Kb = 1.7*10^-9
pKb = - log (Kb)
= - log(1.7*10^-9)
= 8.77
POH = 14 - pH
= 14 - 5.59
= 8.41
use formula for buffer
pOH = pKb + log ([C5H5NHCl]/[C5H5N])
8.41 = 8.7696 + log ([C5H5NHCl]/[C5H5N])
log ([C5H5NHCl]/[C5H5N]) = -0.3596
[C5H5NHCl]/[C5H5N] = 0.437
[C5H5NHCl]/0.9 = 0.437
[C5H5NHCl] = 0.3933
volume , V = 4.5*10^2 mL
= 0.45 L
use:
number of mol,
n = Molarity * Volume
= 0.3933*0.45
= 0.177 mol
Molar mass of C5H5NHCl,
MM = 5*MM(C) + 6*MM(H) + 1*MM(N) + 1*MM(Cl)
= 5*12.01 + 6*1.008 + 1*14.01 + 1*35.45
= 115.558 g/mol
use:
mass of C5H5NHCl,
m = number of mol * molar mass
= 0.177 mol * 1.156*10^2 g/mol
= 20.45 g
Answer: 20. g
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