Question

A solution is made by mixing 50 mL of 2.0 M K₂HPO₄ and 25 mL of 2.0 M KH₂PO₄. The solution is diluted to a final volume of 200 mL. What is the pH of the final solution?

Short answer

The pH value of the final solution is 7.15.

Step-by-step solution

Concentration of K2HPO4 and KH2PO4 in the solution

The final volume of the solution is 200 mL.

Concentration of K₂HPO₄in the solution is calculated as:.

$$\begin{gathered} \mathbf{\left[}{\mathbf{K}}_{\mathbf{2}}{\mathbf{HPO}}_{\mathbf{4}}\mathbf{\right]}\mathbf{=}\frac{\mathbf{50}\mathbf{mL}\mathbf{\times }\mathbf{2}\mathbf{M}}{\mathbf{200}\mathbf{mL}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{M} \end{gathered}$$

Concentration of KH₂PO₄ in the solution is calculated as:

$$\begin{gathered} \mathbf{\left[}{\mathbf{KH}}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right]}\mathbf{=}\frac{\mathbf{25}\mathbf{mL}\mathbf{\times }\mathbf{2}\mathbf{M}}{\mathbf{200}\mathbf{mL}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{M} \end{gathered}$$

Calculation of pH value of the final solution

According to Hasselbach-Henderson equation,

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

Here, $\mathbf{\left[}{\mathbf{A}}^{\mathbf{-}}\mathbf{\right]}$is the concentration of base (K₂HPO₄) and is the concentration of acid (KH₂PO₄).

The pK value ofKH₂PO₄ is 6.85.

Substitute $\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{M}\mathbf{}$for $\mathbf{\left[}{\mathbf{A}}^{\mathbf{-}}\mathbf{\right]},\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{M}\mathrm{for}\mathbf{\left[}\mathbf{HA}\mathbf{\right]},$and 6.85 for pK.

$$\begin{gathered} \mathbf{pH}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{85}\mathbf{+}\mathbf{log}\left(\frac{0.5M}{0.25M}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{85}\mathbf{+}\mathbf{log}\mathbf{2} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{85}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{3} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{15} \end{gathered}$$

Therefore, the pH value of final solution is 7.15.