Question

Calculate the pH of a 200 mL solution of pure water to which has been added 50 mL of 1 mM HCl.

Short answer

The pH value of total solution is 3.7.

Step-by-step solution

Concentration of H+ ions in the total solution

Total amount of solution=250 mL

Pure water has a pH of 7. So, its hydrogen ion concentration is:

$\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{M}$

Here,H⁺ is hydrogen ion concentration.

The added 50 ml of solution had H⁺ concentration of 1x10^-3 M (1mM HCl).

Thus, total concentration of H⁺ in the solution is calculated as:.

[H⁺]= (50mLx1mM)/250mL

$$\begin{gathered} \mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}\mathbf{=}\frac{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{L}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{001}\mathbf{M}\mathbf{)}}{\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{L}}\mathbf{+}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{M} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{+}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{)}\mathbf{}\mathbf{M}\mathbf{} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{M}\mathbf{.} \end{gathered}$$

Calculation of pH

$$\begin{gathered} \mathbf{pH}\mathbf{=}\mathbf{log}\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]} \\ \mathit{N}\mathit{o}\mathit{w}\mathbf{,}\mathbf{}{\mathbf{\left[}\mathbf{H}\mathbf{\right]}}^{\mathbf{+}}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathbf{M} \\ \mathbf{pH}\mathbf{=}\mathbf{-}\mathbf{log}\mathbf{(}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathbf{M}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{7} \end{gathered}$$

Therefore, the pH value of the total solution is 3.7.