Codeine (C18H12NO3) is a weak base, (pKb= 5.79). A solution of 5*10^-2M solution has a pH...

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Codeine (C18H12NO3) is a weak base, (pKb= 5.79). A solution of5*10^-2M solution has a pH of 10. Calculate the [C18H22NO3],[C18H21NO3] and [OH-] at equilibruim.

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4.4 Ratings (607 Votes)

first find pOH from pH

pH + pOH = 14

pOH = 14 - pHÂ Â Â = 14 - 10 = 4

from the pOH we can calculate the [HO-]

[HO-] = 10^--4 = 0.0001 M

Â Â Â Â Â C18H12NO3 (aq) + H2O (l) ----> C18H13NO3- (aq) + HO- (aq) construct ICE table

IÂ Â Â Â Â Â 0.05 MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

CÂ Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +x

EÂ Â Â Â Â Â Â 0.05-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +x

from the ICE table concentration of C18H13NO3- of concetration HO-

and we have alredy concentration of HO- = 0.001 M = [C18H13NO3-]

conceration of C18H12NO3 = 0.05 - x = 0.05 - 0.0001 = 0.0499 M

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