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The pH of 0.005 M codeine (C_18H_21NO_3) solution is 9.95 . Calculate its ionization constant and pK_b. |
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Answer» Solution :Calculation of concentrationof `OH^-` : pH=-log `[H^+]`= 9.95 So, pOH=(14.0-9.95)=4.05 pOH=-log `[OH^-]` =4.05 = `bar5.95` `therefore [OH^-]`=-Antilog `(bar5.95)` `[OH^-]=10^(-4.05) = 8.9125xx10^(-5)` pH = 9.95 `gt` 7.0, So codeine is weak BASE. So, ionic EQUILIBRIUM is as under : `{:(,C_18H_21NO_(3(aq))+H_2O_((L)) hArr, C_18H_21NO_3H_((aq))^(+) + , OH_((aq))^(-)),("INITIAL concen.", 0.005 M,0.0 M, 0.0M),("Change in equilibrium:",0.005 M ,0.0 M ,0.0 M),("Molarity at equili.",underset(approx0.005 M)(0.005-x),underset(=8.91xx10^(-5))(xM),underset(=8.91xx10^(-5))(x M)):}` Calculation of `K_a` : `K_b=([C_18 H_21 NO_3H^+][OH^-])/([C_18H_21NO_3])` `K_b=((8.9125)^2(10^(-5))^2)/0.005=1.5886xx10^(-6)` `pK_b=-log (K_b)=-log (1.5886xx10^(-6))` =-(0.2010-6.0)=-(-5.799)=+ 5.799 `APPROX` 5.8 |
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