Q. 514.7( 7 Votes )

The pH of 0.005M

Answer :

Given:

pH of Codeine(C18H21NO3) solution = 9.95


[Cod] = 0.005M


Ionisation of Cod:


Cod + H2O CodH+ + OH-1


As we know that pH + pOH = 14


9.95 + pOH =14


pOH = 14- 9.95


pOH = 4.05


As we know that, pOH = -log[OH-]


4.05 = -log[OH]


By taking antilog of both the sides, we get


Antilog 4.05 = -[OH-]


[OH-] = 8.913 × 10-5


As we know that, Kb(ionization constant)


Kb=


Hence, for the reaction, Cod + H2O CodH+ + OH-1


Kb =


As the reaction is completely ionized,


[CodH+] = [OH-]


Hence, Kb=


As [OH-] = 8.913 × 10-5 (calculated)


[Cod] = 0.005M


Kb=


Kb = 1.588 × 10-6


Thus, the ionization constant of codeine is 1.588 × 10-6


To calculate Pkb we apply the formula:


Pka = -logKb


As Kb = 1.74 × 10-3


pkb = -log (1.588 × 10-6)


Pkb = -log 1.588 – (-6) log 10


Pkb = 5.80


Thus, the Pkb of codeine is 5.80


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