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Question: A neutral solution: (A) lacks all forms of ions (B) lacks \({{\text{H}}_{3}}{{\text{O}}^{+}}\) i...

A neutral solution:
(A) lacks all forms of ions
(B) lacks H3O+{{\text{H}}_{3}}{{\text{O}}^{+}} ion only
(C) has a pH of 7
(D) has equal concentrations of H3O+{{\text{H}}_{3}}{{\text{O}}^{+}} and OH−\text{O}{{\text{H}}^{-}} ions

Explanation

Solution

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 = −log[H+]-\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 H2O→H++OH−{{\text{H}}_{2}}\text{O}\to {{\text{H}}^{+}}+\text{O}{{\text{H}}^{-}}. Water dissociates into hydroxide ions (OH−)\left( \text{O}{{\text{H}}^{-}} \right) and hydrogen ions (H+)\left( {{\text{H}}^{+}} \right).
-The ionic product of water is Kw{{\text{K}}_{\text{w}}}, whose value is 10−14{{10}^{-14}}. The expression of Kw{{\text{K}}_{\text{w}}} of the reaction is [H+][OH−]=10−14\left[ {{\text{H}}^{+}} \right]\left[ \text{O}{{\text{H}}^{-}} \right]={{10}^{-14}}.
-In neutral solution or pure water, the concentration of hydrogen ions (H+)\left( {{\text{H}}^{+}} \right) is equal to hydroxide ions (OH−)\left( \text{O}{{\text{H}}^{-}} \right). So, [H+]=[OH−]\left[ {{\text{H}}^{+}} \right]=\left[ \text{O}{{\text{H}}^{-}} \right].
-The concentration of hydrogen ions (H+)\left( {{\text{H}}^{+}} \right) or the concentration of hydroxide ions (OH−)\left( \text{O}{{\text{H}}^{-}} \right) will be equal to [H+][H+]=10−14\left[ {{\text{H}}^{+}} \right]\left[ {{\text{H}}^{+}} \right]={{10}^{-14}} or [H+]2=10−14{{\left[ {{\text{H}}^{+}} \right]}^{2}}={{10}^{-14}}.
-The concentration of hydrogen ions will be [H+]=10−14\left[ {{\text{H}}^{+}} \right]=\sqrt{{{10}^{-14}}} or [H+]=10−7M\left[ {{\text{H}}^{+}} \right]={{10}^{-7}}\text{M}.
The concentration of hydrogen ions and hydroxide ions will be [H+]=10−7M\left[ {{\text{H}}^{+}} \right]={{10}^{-7}}\text{M} and [OH−]=10−7M\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 = −log[H+]-\text{log}\left[ {{\text{H}}^{+}} \right].
The pH is −log[10−7]-\text{log}\left[ {{10}^{-7}} \right] or −(−7)log[10]-\left( -7 \right)\text{log}\left[ 10 \right] or +7.

A neutral solution has a pH of 7 and has equal concentrations of H3O+{{\text{H}}_{3}}{{\text{O}}^{+}} and OH−\text{O}{{\text{H}}^{-}} ions, which is option (C) and (D).

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