The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pK b.
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Answered by
48
Concentration of codeine= C=0.005 MpH=9.95pOH=-log[OH ]
=14 - 9.95
=4.05
[OH⁻] = 8.92x 10⁻⁵ M
Kᵇ= [H⁺][OH⁻]
= (8.92 x 10⁻⁵)²/ 0.005
=1.59 x 10⁻⁶
pKᵇ= - log Kᵇ= log 1.59 x 10⁻⁶
= 6 log1.59
=5.8
=14 - 9.95
=4.05
[OH⁻] = 8.92x 10⁻⁵ M
Kᵇ= [H⁺][OH⁻]
= (8.92 x 10⁻⁵)²/ 0.005
=1.59 x 10⁻⁶
pKᵇ= - log Kᵇ= log 1.59 x 10⁻⁶
= 6 log1.59
=5.8
Answered by
12
For log 1.588
Look at 15 at 8 & at mean difference 8
Add all these
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