Question

A neutral solution:
(A) lacks all forms of ions
(B) lacks ${{\text{H}}_{3}}{{\text{O}}^{+}}$ ion only
(C) has a pH of 7
(D) has equal concentrations of ${{\text{H}}_{3}}{{\text{O}}^{+}}$ and $\text{O}{{\text{H}}^{-}}$ ions

Answer 1

Hint: By the definition of neutral solution, neutral solution is a type of solution whose pH is 7, which is neither acidic solution (pH <7) nor basic solution (pH >7). Write the equation and expression of the ionic product of water and its value. Calculate the concentration of ions in the solution. To find pH apply the formula, pH = $-\text{log}\left[ {{\text{H}}^{+}} \right]$.

Complete step by step solution:
Let us discuss more information related to neutral solutions and how its pH is 7.
-The reaction of dissociation of water is ${{\text{H}}_{2}}\text{O}\to {{\text{H}}^{+}}+\text{O}{{\text{H}}^{-}}$. Water dissociates into hydroxide ions $\left( \text{O}{{\text{H}}^{-}} \right)$ and hydrogen ions $\left( {{\text{H}}^{+}} \right)$.
-The ionic product of water is ${{\text{K}}_{\text{w}}}$, whose value is ${{10}^{-14}}$. The expression of ${{\text{K}}_{\text{w}}}$ of the reaction is $\left[ {{\text{H}}^{+}} \right]\left[ \text{O}{{\text{H}}^{-}} \right]={{10}^{-14}}$.
-In neutral solution or pure water, the concentration of hydrogen ions $\left( {{\text{H}}^{+}} \right)$ is equal to hydroxide ions $\left( \text{O}{{\text{H}}^{-}} \right)$. So, $\left[ {{\text{H}}^{+}} \right]=\left[ \text{O}{{\text{H}}^{-}} \right]$.
-The concentration of hydrogen ions $\left( {{\text{H}}^{+}} \right)$ or the concentration of hydroxide ions $\left( \text{O}{{\text{H}}^{-}} \right)$ will be equal to $\left[ {{\text{H}}^{+}} \right]\left[ {{\text{H}}^{+}} \right]={{10}^{-14}}$ or ${{\left[ {{\text{H}}^{+}} \right]}^{2}}={{10}^{-14}}$.
-The concentration of hydrogen ions will be $\left[ {{\text{H}}^{+}} \right]=\sqrt{{{10}^{-14}}}$ or $\left[ {{\text{H}}^{+}} \right]={{10}^{-7}}\text{M}$.
The concentration of hydrogen ions and hydroxide ions will be $\left[ {{\text{H}}^{+}} \right]={{10}^{-7}}\text{M}$ and $\left[ \text{O}{{\text{H}}^{-}} \right]={{10}^{-7}}\text{M}$.
- The pH of the ions will be 7, as the formula of pH is pH = $-\text{log}\left[ {{\text{H}}^{+}} \right]$.
The pH is $-\text{log}\left[ {{10}^{-7}} \right]$ or $-\left( -7 \right)\text{log}\left[ 10 \right]$ or +7.

A neutral solution has a pH of 7 and has equal concentrations of ${{\text{H}}_{3}}{{\text{O}}^{+}}$ and $\text{O}{{\text{H}}^{-}}$ ions, which is option (C) and (D).

Note: There is a relationship between $\text{p}{{\text{K}}_{\text{w}}},\text{ pH and pOH}$. We know that the value of ${{\text{K}}_{\text{w}}}$ is ${{10}^{-14}}$.
The expression of ${{\text{K}}_{\text{w}}}$ of the reaction is $\left[ {{\text{H}}^{+}} \right]\left[ \text{O}{{\text{H}}^{-}} \right]={{10}^{-14}}$.
Taking log both sides and multiplying negative sign, we get $-\text{log}\left[ {{\text{H}}^{+}} \right]-\text{log}\left[ \text{O}{{\text{H}}^{-}} \right]=-\left( -14 \right)$.
We know that pOH = $-\text{log}\left[ \text{O}{{\text{H}}^{-}} \right]$ and pH = $-\text{log}\left[ {{\text{H}}^{+}} \right]$.
The net equation or relation is pH + pOH = 14.