Chapter U4.20, Problem 4E

Interpretation Introduction

Interpretation:

How much water is needed to be added to 10 mL of a solution of HCl with a pH of 3 to change the pH to 6 must be explained.

Concept introduction:

The pH of solution is related to hydrogen ion concentration is as follows:

  $\text{pH=-log}\left[{\text{H}}^{\text{+}}\right]$

Here, pH is changed by 1 unit when [H⁺] changes 10 fold as pH is a logarithmic scale.

990 mL of water must be added to 10 mL of a solution of HCl with a pH of 3 to change the pH to 6.

When pH is increased from 3 to 6 then it is an increase of pH by 3 units. This means the concentration of H+ is decreased by 10³ fold. Initially, $\text{pH=3}\text{.}$ Thus, initial concentration of ${\text{H}}^{\text{+}}{\text{=10}}^{\text{-3}}\text{mol/L}\text{.}$

  ${\text{H}}^{\text{+}}{\text{=10}}^{\text{-3}}\text{mol/L}\text{.}$

Final $\text{pH=6}$ thus, concentration of ${\text{H}}^{\text{+}}{\text{=10}}^{\text{-6}}$

  ${\text{H}}^{\text{+}}{\text{=10}}^{\text{-6}}$ .

So, new volume of the solution must be 1000 times as compared to the initial volume.

So extra water which needs to be added $$

  $\text{=}\left(\text{1000-10}\right)\text{ mL=990 mL}\text{.}$

Thus, 990 mL of water must be added.

Answer to Problem 4E

990 mL of water must be added to 10 mL of a solution of HCl with a pH of 3 to change the pH to 6.

Explanation of Solution

When pH is increased from 3 to 6 then it is an increase of pH by 3 units. This means the concentration of H+ is decreased by 10³ fold. Initially, $\text{pH=3}\text{.}$ Thus, initial concentration of ${\text{H}}^{\text{+}}{\text{=10}}^{\text{-3}}\text{mol/L}\text{.}$

  ${\text{H}}^{\text{+}}{\text{=10}}^{\text{-3}}\text{mol/L}\text{.}$

Final $\text{pH=6}$ thus, concentration of ${\text{H}}^{\text{+}}{\text{=10}}^{\text{-6}}$

  ${\text{H}}^{\text{+}}{\text{=10}}^{\text{-6}}$ .

So, new volume of the solution must be 1000 times as compared to the initial volume.

So extra water which needs to be added $$

  $\text{=}\left(\text{1000-10}\right)\text{ mL=990 mL}\text{.}$

Conclusion

Thus, 990 mL of water must be added.