Question

A solution of $0.1\mathrm{M}$ weak base $\left(\mathrm{B}\right)$ is titrated with $0.1\mathrm{M}$ strong acid $\left(\mathrm{HA}\right)$. The variation of $\mathrm{pH}$ of the solution with the volume of $\mathrm{HA}$ added is shown in the figure below. What is the ${\mathrm{pK}}_{\mathrm{b}}$ of the base? The neutralization reaction is given by-

$\mathrm{B}+\mathrm{HA}\to {\mathrm{BH}}^{+}+{\mathrm{A}}^{-}$

Answer 1

Step 1: The Henderson-Hasselbalch equation for the buffer solution is given below-

$\mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]}$

$\mathrm{pOH}=14-\mathrm{pH}$

$\left[\mathrm{salt}\right]=$concentration of salt

$\left[\mathrm{base}\right]=$concentration of base

Step 2: Calculation of $\mathrm{pOH}$-

In the graph, at the equivalence point $(\mathrm{V}=6\mathrm{mL})$- complete neutralisation takes place.

At half neutralisation point $(\mathrm{V}=3\mathrm{mL})$- buffer formation occurs $(\mathrm{pH}=11)$.

$\mathrm{pOH}=14-11$

$\mathrm{pOH}=3$

Step 3: Calculation of $\left[\mathrm{salt}\right]$ and $\left[\mathrm{base}\right]$ -

$$\begin{gathered} \mathrm{B}+\mathrm{HA}\to {\mathrm{BH}}^{+}+{\mathrm{A}}^{-} \\ \mathrm{At}\mathrm{pH}=11[0.1\times 3\mathrm{mmol}]\left[0.1\times 3\mathrm{mmol}\right][0.3\mathrm{mmol}] \\ \mathrm{total}\mathrm{volume}=6\mathrm{mL} \end{gathered}$$

$\left[\mathrm{salt}\right]=\frac{0.3}{6}$ $\mathrm{mol}/\mathrm{L}$

$\left[\mathrm{base}\right]=\frac{0.3}{6}$$\mathrm{mol}/\mathrm{L}$

Step 3: Calculation of ${\mathrm{pK}}_{\mathrm{b}}$ for the given reaction -

$\mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]}$

$3={\mathrm{pK}}_{\mathrm{b}}+\log \frac{{\displaystyle \frac{0.3}{6}}}{{\displaystyle \frac{0.3}{6}}}$

$3={\mathrm{pK}}_{\mathrm{b}}+\log 1$

$3={\mathrm{pK}}_{\mathrm{b}}$

So, the ${\mathrm{pK}}_{\mathrm{b}}$ of the base is $3$.