Question 50 of 69intermediate🔧 ApplyNumerical3 marks

The pH of 0.005 M codeine (C18H21NO3C_{18}H_{21}NO_3) solution is 9.95. Calculate its ionization constant and pKbpK_b.

Correct Answer

The ionization constant KbK_b of codeine is 1.59×10−61.59 \times 10^{-6} and its pKbpK_b is 5.80.

Exercise: EXERCISES | Q: 6.51 | (Page 46)
For More Understanding

Explanation

Codeine is a weak base. Given the pH of the solution, we can first find the hydrogen ion concentration. Using the ionic product of water (KwK_w), we then determine the hydroxyl ion concentration. Since the base ionizes to produce an equal amount of conjugate acid and hydroxyl ions, we can use these values along with the initial concentration of the base to calculate the ionization constant KbK_b. Finally, pKbpK_b is calculated by taking the negative logarithm of KbK_b.

Solution Steps

  1. Step 1: Calculate [H+][H^+] from the given pH [H+]=antilog(−pH)=antilog(−9.95)=1.12×10−10[H^+] = \text{antilog}(-pH) = \text{antilog}(-9.95) = 1.12 \times 10^{-10} M

  2. Step 2: Calculate [OH−][OH^-] using KwK_w [OH−]=Kw[H+]=1×10−141.12×10−10=8.91×10−5[OH^-] = \frac{K_w}{[H^+]} = \frac{1 \times 10^{-14}}{1.12 \times 10^{-10}} = 8.91 \times 10^{-5} M

  3. Step 3: Set up the ionization equilibrium for codeine Let codeine be represented by B. The reaction is: B+H2O⇌BH++OH−B + H_2O \rightleftharpoons BH^+ + OH^- At equilibrium, [BH+]=[OH−]=8.91×10−5[BH^+] = [OH^-] = 8.91 \times 10^{-5} M. Since ionization is very small, the concentration of undissociated base [B]≈0.005[B] \approx 0.005 M.

  4. Step 4: Calculate the ionization constant KbK_b

    Kb=[BH+][OH−][B]=(8.91×10−5)20.005=79.38×10−100.005=1.59×10−6K_b = \frac{[BH^+][OH^-]}{[B]} = \frac{(8.91 \times 10^{-5})^2}{0.005} = \frac{79.38 \times 10^{-10}}{0.005} = 1.59 \times 10^{-6}

  5. Step 5: Calculate pKbpK_b

    pKb=−log⁡(Kb)=−log⁡(1.59×10−6)=5.80pK_b = -\log(K_b) = -\log(1.59 \times 10^{-6}) = 5.80
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