Chapter U4.21, Problem 5E

Interpretation Introduction

Interpretation: The reason for the pH of a solution of H₂SO₄ to remain below 7 when 1 mole of NaOH is added to the solution needs to be explained.

Concept Introduction: The pH of an aqueous solution is defined as

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

Here, $\left[{\text{H}}^{\text{+}}\right]$ is the concentration of hydrogen ion solution.

The pOH of an aqueous solution is the negative logarithm of the hydroxide, ${\text{OH}}^{-}$ ion concentration, i.e.,

  $\text{pOH }=-\text{log }\left[{\text{OH}}^{-}\right]$

Here, $\left[{\text{OH}}^{-}\right]$ is hydrogen ion concentration in the solution.

An acid-base neutralization reaction is a reaction of ${\text{H}}^{+}$ with ${\text{OH}}^{-}$ ions to produce undissociated water, H₂O molecules as per the equation below:

  ${\text{H}}^{\text{+}}\left(aq\right){\text{ + OH}}^{\text{-}}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)$

The reaction above is quantitative and consumes equal amounts of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions.

It is known that for solutions rich in ${\text{H}}^{+}$ (i.e, solutions containing excess ${\text{H}}^{+}$ ), the pH is less than 7. Such solutions are known as acidic solutions. Similarly, for solutions containing an excess of ${\text{OH}}^{-}$ , the pH is greater than 7 and such solutions are known as basic solutions.

Answer to Problem 5E

The pH of the solution of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ remains below 7 upon the addition of 1 mole of NaOH due to the presence of excess ${\text{H}}^{+}$ ions in the solution. The added NaOH cannot completely neutralize the ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ present in the solution and thus, the solution contains unreacted ${\text{H}}^{+}$ .

Explanation of Solution

Sulfuric acid, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ ionizes in an aqueous solution as below:

  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to 2{\text{ H}}^{+}\left(aq\right)+{\text{ SO}}_{\text{4}}{}^{2-}\left(aq\right)$

The ionization of 1 mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ furnishes 2 moles of protons, ${\text{H}}^{+}$ .

Sodium hydroxide, NaOH ionizes in an aqueous solution as follows:

  $\text{NaOH }\left(aq\right)\to {\text{Na}}^{+}\left(aq\right)+{\text{ OH}}^{-}\left(aq\right)$

The ionization of 1 mole of NaOH gives 1 mole of ${\text{OH}}^{-}$ ions.

Now, 1 mole of the added ${\text{OH}}^{-}$ ions consumes 1 mole of ${\text{H}}^{+}$ ions as per the reaction given in the introduction section. However, the solution still contains one mole of unreacted ${\text{H}}^{+}$ ions and these excess ${\text{H}}^{+}$ ions result in the pH of the solution being lower than 7.

Conclusion

The addition of NaOH to the solution of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ cannot completely neutralize the acid present and thus, the pH of the solution remains below 7.