What is the pH of a solution of 3.22g Codeine (C18H21NO3, kb = 1.6*10-6) and 2.44g of codeine hydrochloride (C18H22NO3Cl ) in 100mL solution with water, H2O?
I thought I knew how to tackle the problem, but I'm now thinking I've done it incorrectly.
Mass of codeine = 3,22 g
molar mass of codeine = 299 g/mol
So,moles of codeine = 3.22 g / 299 g/mol == 0.01077 mol
Mass of codeine hydrochloride =2.44 g
Molar mass of codeine hydrochloride =335.5 g/mol
Moles of of codeine hydrochloride =2.44 g / 335.5 g/mol == 0.00727 mol
Volume of solution = 0.100 L
Given, Kb =1.6 x 10-6
Ka = 10-14 / 1.6 x 10-6 == 0.625 x 10-8
pKa = - log (0.625 x 10-8 ) == 8.204
pH = pKa + log (Salt / acid )
= 8.204 + log ( 0.00727 mol / 0.01077 mol )
= 8.03
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