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  • The pH of 0.005M codeine (C_18H_21NO_3) solution is 9.95. Calculate its ionization constant and pKb.

7.51     The pH of 0.005M codeine \left ( C_{18}H_{21}NO_{3} \right ) solution is 9.95. Calculate its ionization constant and pKb.

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best_answer

We have,
C = 0.005 M
P^H = 9.95 and P^{OH} = 14 - 9.95 = 4.05 

we know that P^{OH} = -\log[OH^-]
                       4.05 = -\log [OH^-]

By taking antilog on both sides we get.

concentration of [OH^-] = 8.91 \times 10^{-5}

C.a = 8.91 \times 10^{-5}
So, a = 1.782\times 10^{-2}

We know that,
K_b = C.a^2
        \\= 0.005 \times (1.782\times 10^{-2})^2\\ =0.0158\times 10^{-4}

Thus 
pK_b = -\log (K_b)
            \\= -\log (0.0158\times 10^{-4})\\ =5.80

Posted by

manish

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