Bio Chem English gmail Google Docs Netflix Disney+ spotify Hulu Adids end Bases Calculating the pH of a weak base solution The base protonation constant \( K_{b} \) of azetidine \( \left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{NH}\right) \) is \( 1.5 \times 10^{-8} \). Calculate the pH of a 1.2 M solution of azetidine at \( 25^{\circ} \mathrm{C} \). Round your answer to 1 decimal place. \[ \mathrm{pH}= \]

To calculate the pH of a 1.2 M solution of azetidine, we first need to find the concentrations of hydroxide ions \([\text{OH}^-]\) produced when azetidine acts as a weak base. The equilibrium expression can be set up using the \(K_b\) value: \[ K_b = \frac{[C_3H_6NH_2^+][OH^-]}{[C_3H_6NH]} = 1.5 \times 10^{-8} \] Since azetidine essentially dissociates into its protonated form and hydroxide ions, we set \([C_3H_6NH_2^+]\) and \([\text{OH}^-] = x\) (from the dissociation) and the equilibrium concentration of azetidine will be \(1.2 - x\). For very weak bases like azetidine, we can often approximate \(1.2 - x \approx 1.2\): \[ 1.5 \times 10^{-8} = \frac{x^2}{1.2} \] Rearranging gives us: \[ x^2 = 1.5 \times 10^{-8} \times 1.2 = 1.8 \times 10^{-8} \] Taking the square root gives: \[ x \approx 4.24 \times 10^{-5} \] This \(x\) value corresponds to \([\text{OH}^-]\), and we can find the pOH: \[ \text{pOH} = -\log(4.24 \times 10^{-5}) \approx 4.37 \] Finally, using the relationship \(\text{pH} + \text{pOH} = 14\): \[ \text{pH} = 14 - 4.37 = 9.63 \Rightarrow \text{pH} \approx 9.6 \]