-->

Equilibrium

Question
CBSEENCH11006560

The pH of 0·005M codeine (C18H21NO3) solution is 9·95. Calculate its ionization constant and pKb.

Solution

We know,
   negative log space open square brackets straight H to the power of plus close square brackets space equals space pH
or space space log space open square brackets straight H to the power of plus close square brackets space equals negative pH space equals space minus 9.95 space equals space minus 9 minus 0.95 space minus 1
space space space space space space space space space space space space space space space space space equals 10 with bar on top.05
or space space space space space open square brackets straight H to the power of plus close square brackets space equals space antilog space left parenthesis 10 with bar on top.05 right parenthesis
space space space space space space space space space space space space space space space equals 1.12 space cross times space 10 to the power of negative 10 end exponent straight M
Also comma space space straight K subscript straight w space equals open square brackets straight H to the power of plus close square brackets space open square brackets straight O with bar on top straight H close square brackets
or space space space space open square brackets OH to the power of minus close square brackets space equals space fraction numerator straight K subscript straight w over denominator open square brackets straight H to the power of plus close square brackets end fraction
space space space space space space space space space space space space space space space space equals fraction numerator 10 to the power of negative 14 end exponent over denominator 1.12 space cross times 10 to the power of 10 end fraction
                 equals space 8.93 space cross times space 10 to the power of negative 5 end exponent straight M
The concentration of codeine ion is also same as that of hydroxyl ion. The concentration of the ions is very small and hence the concentration of undissociated base can be taken as equal to 0·005 M.
Thus comma space space space straight K subscript straight b space equals space fraction numerator open square brackets straight M to the power of plus close square brackets space open square brackets OH to the power of minus close square brackets over denominator open square brackets MOH close square brackets end fraction space equals fraction numerator left parenthesis 8.93 space cross times space 10 to the power of negative 5 end exponent right parenthesis squared over denominator 0.005 end fraction
space space space space space space space space space space space space space space equals space 1.6 space cross times space 10 to the power of negative 6 end exponent
pK subscript straight b space equals space minus log space straight K subscript straight b space equals space minus log space left parenthesis 1.6 space cross times space 10 to the power of negative 6 end exponent right parenthesis space equals space 5.8