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प्रश्न
The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb.
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उत्तर
c = 0.005
pH = 9.95
pOH = 4.05
pH = – log (4.105)
`4.05 = -log["OH"^-]`
`["OH"^-] = 8.91 xx 10^(-5)`
`"c" alpha = 8.91 xx 10^(-5)`
`alpha = (8.91 xx 10^(-5))/(5xx10^(-3)) = 1.782 xx 10^(-2)`
Thus `"K"_"b" = "c" alpha^2`
`= 0.005 xx (1.782)^2 xx 10^(-4)`
`= 0.005 xx 3.1755 xx 10^(-4)`
`= 0.0158 xx 10^(-4)`
`"K"_"b" = 1.58 xx 10^(-6)`
`"Pk"_"b" = - log "K"_"b"`
`= - log (1.58 xx 10^(-6))`
= 5.80
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