Question

What is the pK of the weak acid HA if a solution containing 0.1 M HA and 0.2 M A⁻ has a pH of 6.5?

Short answer

The pK value of the solution is 6.2.

Step-by-step solution

Introduction of Hasselbach-Henderson equation.

The Hasselbach-Henderson equation states that when the molar concentration of an acid (HA) and its conjugate base (A^-) are equal, then

$\mathbf{log}\frac{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}{\mathbf{\left[}\mathbf{HA}\mathbf{\right]}}\mathbf{=}\mathbf{log}\mathbf{1}\mathbf{=}\mathbf{0}$and hence, the pH value of the solution is equal to the pK value.

Mathematically,

$\mathbf{pH}\mathbf{=}\mathbf{pK}\mathbf{+}\mathbf{log}\frac{\mathbf{\left[}{\mathbf{A}}^{\mathbf{-}}\mathbf{\right]}}{\mathbf{\left[}\mathbf{HA}\mathbf{\right]}}$

(i)

Calculation of the pK value of the solution.

Substitute $\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{M}$for $\mathbf{\left[}{\mathbf{A}}^{\mathbf{-}}\mathbf{\right]},\mathbf{}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{M}\mathrm{for}\mathbf{}\mathbf{\left[}\mathbf{HA}\mathbf{\right]}$, and 6.5 for pH.

$$\begin{gathered} \mathbf{6}\mathbf{.}\mathbf{5}\mathbf{=}\mathbf{pK}\mathbf{+}\mathbf{log}\left(\frac{0.2M}{0.1M}\right) \\ \mathbf{6}\mathbf{.}\mathbf{5}\mathbf{=}\mathbf{pK}\mathbf{+}\mathbf{log}\mathbf{2} \\ \mathbf{pK}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{5}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{3} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{2} \end{gathered}$$

Therefore, the pK value of the final solution is 6.2.