Calculate pKb of Weak Base from Titration Curve

To determine the $\text{p}K_b$ of the weak base $\text{B}$ from the given titration curve, we follow these steps:

1. Equivalence Point Determination: From the titration curve, the sharp inflection drop in $\text{pH}$ occurs at a volume of added acid: $V_{\text{eq}} = 6\text{ mL}$

2. Half-Equivalence Point: The half-equivalence point occurs when exactly half of the volume of strong acid needed for complete neutralization is added: $V_{\text{half}} = \frac{V_{\text{eq}}}{2} = \frac{6\text{ mL}}{2} = 3\text{ mL}$

3. Reading $\text{pH}$ from the Graph: At $V_{\text{HA}} = 3\text{ mL}$, we look at the corresponding value on the vertical $\text{pH}$ axis of the grid: $\text{pH} = 11$

4. Calculating $\text{p}K_b$: At the half-equivalence point, half of the weak base $\text{B}$ has reacted to form its conjugate acid $\text{BH}^+$. Therefore, the concentrations of the weak base and its conjugate acid are equal: $[\text{B}] = [\text{BH}^+]$

   Applying the Henderson-Hasselbalch equation for a basic buffer solution: $\text{pOH} = \text{p}K_b + \log_{10}\left(\frac{[\text{BH}^+]}{[\text{B}]}\right)$

   Since $[\text{B}] = [\text{BH}^+]$, the logarithmic term becomes $\log_{10}(1) = 0$: $\text{pOH} = \text{p}K_b$

   Knowing that $\text{pH} + \text{pOH} = 14$ (at $25^\circ\text{C}$): $\text{pOH} = 14 - \text{pH} = 14 - 11 = 3$

   Therefore: $\text{p}K_b = 3$