Question

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.

Answer #1

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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