Question

In a Bowler's solution containing the same concentrations of B^- and HB, the ratio of K_B to B^- is 10^-10. The pH of the Bowler's solution will be:

• A. 4
• B. 6
• C. 7
• D. 10

Hint:

For a buffer solution containing a weak acid (HA) and its conjugate base ($\text{A}^-$), if $[\text{HA}] = [\text{A}^-]$, then $\text{pH} = \text{p}\text{K}_{\text{a}}$. Similarly, for a buffer containing a weak base (B) and its conjugate acid ($\text{BH}^+$), if $[\text{B}] = [\text{BH}^+]$, then $\text{pOH} = \text{p}\text{K}_{\text{b}}$. Always remember the relationship $\text{pH} + \text{pOH} = 14$ and $\text{p}\text{K}_{\text{a}} + \text{p}\text{K}_{\text{b}} = 14$.

Answer 1

The Correct Option is A

Step 1: Identify the equilibrium.
The Bowler's solution contains a base (B\(^-\)) and its conjugate acid (HB) at equal concentrations, forming a buffer. The pOH of this buffer is: \[ \text{pOH} = \text{p}K_\text{b} + \log \left( \frac{[\text{HB}]}{[\text{B}^-]} \right) \] Step 2: Apply the given data.
Since \([B^-] = [HB]\), \[ \log \left( \frac{[\text{HB}]}{[\text{B}^-]} \right) = \log(1) = 0 \] Therefore, \[ \text{pOH} = \text{p}K_\text{b} \] We know: \[ \frac{K_\text{B}}{[\text{B}^-]} = 10^{-10} \Rightarrow K_\text{B} = 10^{-10} \times [\text{B}^-] \] As \([B^-]\) cancels out when \([B^-] = [HB]\), we have: \[ K_b = 10^{-10} \Rightarrow \text{p}K_b = 10 \] Thus, \[ \text{pOH} = 10 \Rightarrow \text{pH} = 14 - \text{pOH} = 14 - 10 = 4 \]