Question

In Section 8.3 an equation was derived for the exact treatment of HA/
NA-type buffers. What would be the expression for B/BHCl-type buffers stated in terms of Kb, [OH-], [B], and [BH+]? Would it be necessary to use this exact
expression to solve for the pH of a solution containing 1.0 x 10-4M HONH2and 1.0 x 10-5M HONH3CI?
Explain.

Short answer

Answer

The expression for B/BHCl types can be written as ${\mathit{K}}_{\mathbf{b}}\mathbf{,}\left[O{H}^{-}\right]\mathbf{,}\left[B\right]\mathbf{\&}\left[B{H}^{+}\right]$

Step-by-step solution

Given Information

The solution has $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{H}\mathit{O}\mathit{N}{\mathit{H}}_{\mathbf{2}}\mathbf{}\mathbf{\&}\mathbf{}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathit{M}\mathit{H}\mathit{O}\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{l}\mathbf{.}$The equation can be denoted as,

$$\begin{gathered} \mathit{B}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{}\mathbf{}\boxed{\xcancel{}}\mathbf{}\mathbf{}\mathbf{}\mathit{B}{\mathit{H}}^{\mathbf{+}}\mathbf{}\mathbf{+}\mathbf{}\mathit{O}{\mathit{H}}^{\mathbf{-}} \\ {\mathit{K}}_{\mathbf{b}}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{BH}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}{\left[B\right]} \end{gathered}$$

The equation for buffer same as the equation of HA/A- type buffers

It can be denoted as,

$$\begin{gathered} {\mathit{K}}_{\mathbf{b}}\mathbf{=}\frac{\left[O{H}^{-}\right]\left({\left[B{H}^{+}\right]}_0+{\displaystyle \frac{{\left[O{H}^{-}\right]}^2-K_w}{\left[O{H}^{-}\right]}}\right)}{{\left[B\right]}_{\mathbf{0}}\mathbf{+}{\displaystyle \frac{{\left[O{H}^{-}\right]}^{\mathbf{2}}\mathbf{-}{\mathbf{K}}_{\mathbf{w}}}{\left[O{H}^{-}\right]}}} \\ {\mathit{K}}_{\mathbf{b}}\mathbf{=}\frac{\left[HON{H}_3^{-}\right]\left[O{H}^{-}\right]}{\left[HON{H}_2\right]} \\ \mathbf{1}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{=}\frac{\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{x}}{\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}} \\ \mathit{x}\mathbf{=}\left[O{H}^{-}\right]\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}} \end{gathered}$$

The reduction of Kb equation

It can be reduced as,

$$\begin{gathered} {\mathit{K}}_{\mathbf{b}}\mathbf{=}\frac{{\left[HON{H}_3^{+}\right]}_{\mathbf{0}}\left[O{H}^{-}\right]}{{\left[HON{H}_2\right]}_{\mathbf{0}}} \\ \frac{{\left[O{H}^{-}\right]}^{\mathbf{2}}\mathbf{-}{\mathbf{K}}_{\mathbf{w}}}{\left[O{H}^{-}\right]}\mathbf{=}\frac{{\left(1.1\times {10}^{-7}\right)}^{\mathbf{2}}\mathbf{-}\left(1\times {10}^{-14}\right)}{\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}} \\ \frac{{\left[O{H}^{-}\right]}^{\mathbf{2}}\mathbf{-}{\mathbf{K}}_{\mathbf{w}}}{\left[O{H}^{-}\right]}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}} \end{gathered}$$

The value can be smaller thanrole="math" localid="1657799350323" $\mathbf{\left[}\mathit{H}\mathit{O}\mathit{N}{{\mathit{H}}_{\mathbf{3}}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\&}\mathbf{}\mathbf{\left[}\mathit{H}\mathit{O}\mathit{N}{\mathit{H}}_{\mathbf{2}}\mathbf{\right]}$. The calculation of hydroxide concentration using the usual procedure.