The $pH$ of $0.005 \, M$ codeine $(C_{18}H_{21}NO_{3})$ solution is $9.95$. Calculate its ionization constant and $pK_{b}$.
Given: Concentration $C = 0.005 \, M$,$pH = 9.95$.
Step $1$: Calculate $pOH$.
$pOH = 14 - pH = 14 - 9.95 = 4.05$.
Step $2$: Calculate $[OH^-]$.
$[OH^-] = 10^{-pOH} = 10^{-4.05} = 8.91 \times 10^{-5} \, M$.
Step $3$: Calculate the ionization constant $K_b$.
For a weak base,$[OH^-] = \sqrt{K_b \times C}$.
$K_b = \frac{[OH^-]^2}{C} = \frac{(8.91 \times 10^{-5})^2}{0.005} = \frac{7.94 \times 10^{-9}}{0.005} = 1.588 \times 10^{-6}$.
Step $4$: Calculate $pK_b$.
$pK_b = -\log(K_b) = -\log(1.588 \times 10^{-6}) = 6 - \log(1.588) = 6 - 0.20 = 5.80$.
Step $1$: Calculate $pOH$.
$pOH = 14 - pH = 14 - 9.95 = 4.05$.
Step $2$: Calculate $[OH^-]$.
$[OH^-] = 10^{-pOH} = 10^{-4.05} = 8.91 \times 10^{-5} \, M$.
Step $3$: Calculate the ionization constant $K_b$.
For a weak base,$[OH^-] = \sqrt{K_b \times C}$.
$K_b = \frac{[OH^-]^2}{C} = \frac{(8.91 \times 10^{-5})^2}{0.005} = \frac{7.94 \times 10^{-9}}{0.005} = 1.588 \times 10^{-6}$.
Step $4$: Calculate $pK_b$.
$pK_b = -\log(K_b) = -\log(1.588 \times 10^{-6}) = 6 - \log(1.588) = 6 - 0.20 = 5.80$.
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