codeine is a drug used in painkillers and is a weak base with kb 1.6X10-6. calculate the ph of a 2.2X10-3M solution of codeine. calculate % dissociation of codeine.
The Pkb of codeine = -Log(Kb) = -Log(1.6*10-6) = 5.8
a) pOH = 1/2 {pKb - Log[codeine]}
= 1/2 {5.8 - Log(2.2*10-3)}
= 4.23
Now, pH + pOH = 14
i.e. pH + 4.23 = 14
i.e. pH = 14 - 4.23 = 9.77
b) The percent dissociation of codeine = (Kb/[codeine])1/2 *100
= {(1.6*10-6)/(2.2*10-3)}*100
= 0.073%
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