Ionic Product of Water: Water is essential for the normal functioning of all living organisms. In pure form, it is made up of two molecules of hydrogen and one molecule of oxygen. The molecular formula of water is \({{{\rm{H}}_{\rm{2}}}{\rm{O}}}.\) Pure water is a poor conductor of electricity. It shows that water is a weak electrolyte, and undergoes a self-ionisation process and splits into two molecular ion parts. They are hydrogen ions and hydroxyl ions.

What is the Ionic Product of Water?

Pure water is a weak electrolyte. It ionises itself to a minimal extent producing protons and hydroxyl ions. The self-ionisation of water may be represented as:

\(\begin{array}{*{20}{c}}{{{\rm{H}}_2}{\rm{O}}\left( {\rm{l}} \right){\rm{\;}}}\\{{\rm{acid}}}\end{array} + \begin{array}{*{20}{c}}{{{\rm{H}}_2}{\rm{O}}\left( {\rm{l}} \right)}\\{{\rm{base}}}\end{array}\;\; \leftrightarrow \begin{array}{*{20}{c}}{{{\rm{H}}_3}{{\rm{O}}^ + }}\\{{\rm{conjugate\;acid}}}\end{array} + \begin{array}{*{20}{c}}{{\rm{O}}{{\rm{H}}^ – }}\\{{\rm{conjugate\;base}}}\end{array}\)

It shows that water has the dual nature of proton donor and proton acceptor.

Since water is ionised to a minimal extent, out of millions of water molecules, only a few are dissociated into \({{\rm{H}}^ + }\) and \({{\rm{O}}{{\rm{H}}^{\rm{ – }}}}\) Ions. Thus, the concentration of unionised water molecules, i.e.,\(\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]\) remains almost constant (being equal to \({\rm{1000 / 18 = 55}}{\rm{.55}}\) moles per litre, because \(1\) litre of water \( = 1000\,{\rm{cc}} = 1000\;{\rm{g}}\) and molar mass of \({{\rm{H}}_2}{\rm{O}} = 18\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ – 1}}\)), i.e., \(\left[ {{{\rm{H}}_2}{\rm{O}}} \right] = \) constant.

Then the equilibrium constant, \({\rm{K}}\), is given by;

\({\rm{K}} = \frac{{\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]{\rm{\;}}\left[ {{\rm{O}}{{\rm{H}}^ – }} \right]}}{{\left[ {{{\rm{H}}_2}} \right.{{\left. {\rm{O}} \right]}^2}}}\)

\({\rm{K}}{\left[ {{{\rm{H}}_2}{\rm{O}}} \right]^2} = \left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\)

Water is a pure liquid, and its concentration remains constant.

Therefore, \({\rm{K}}{\left[ {{{\rm{H}}_2}{\rm{O}}} \right]^2} = \) constant, which can take as \({{\rm{K}}_{\rm{w}}}\)

\({{\rm{K}}_{\rm{w}}} = \left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\) (or) simply, \({{\rm{K}}_{\rm{w}}} = \left[ {{{\rm{H}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\)

Here, the constant \({{\rm{K}}_{\rm{w}}}\) is known as an ionic product of water at a given temperature.

The new constant, a product of the concentration of water and equilibrium constant, is called an ionic product of water \(\left( {{{\rm{K}}_{\rm{w}}}} \right)\). It is also called the auto protolysis constant of water. The ionic product of water \(\left( {{{\rm{K}}_{\rm{w}}}} \right)\) at a given temperature is defined as the concentration of proton \(\left[ {{{\rm{H}}^ + }} \right]\) and hydroxyl ions \({\left( {{\rm{O}}{{\rm{H}}^ – }} \right)}\) in water or aqueous solutions.

At \(298\;{\rm{K}},\;{{\rm{K}}_{\rm{w}}}\) is \(1 \times {10^{ – 14}}\;{\rm{mo}}{{\rm{l}}^2}/{{\rm{L}}^2}\). As the temperature increases, the degree of ionisation of water increases and \({{\rm{K}}_{\rm{w}}}\) increases due to increased \({{{\rm{H}}^ + }}\) Ion concentration. The values of the dissociation constant of water and the ionic product of water at different temperatures are listed below.

Variation of dissociation constant and ionic product of water with temperature.

Temperature \(\left( {^{\rm{o}}{\rm{C}}} \right)\)	lonic product of water \({{\rm{K}}_{\rm{w}}}\) in \({\rm{mo}}{{\rm{l}}^{\rm{2}}}{\rm{/}}{{\rm{L}}^{\rm{2}}}\)
\(0\)	\(0.114 \times {10^{ – 14}}\)
\(10\)	\(0.31 \times {10^{ – 14}}\)
\(25\)	\(1.008 \times {10^{ – 14}}\)
\(50\)	\(5.500 \times {10^{ – 14}}\)
\(100\)	\(7.50 \times {10^{ – 14}}\)

Pure water is a neutral liquid. But it has both proton and hydroxyl ions in equal proportions, though less in quantity. Conductivity measurements support this.

In pure water, \(\left[ {{{\rm{H}}^ + }} \right] = \left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\)

\({{\rm{K}}_{\rm{w}}} = \left[ {{{\rm{H}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^ – }} \right] = {\left[ {{{\rm{H}}^ + }} \right]^2} = {\left[ {{\rm{O}}{{\rm{H}}^ – }} \right]^2} = 1 \times {10^{ – 14}}\;{\rm{mo}}{{\rm{l}}^2}/{{\rm{L}}^2}\) (at \(\left. {{\rm{2}}{{\rm{5}}^{\rm{o}}}{\rm{C}}} \right)\) )

\(\left[ {{{\rm{H}}^ + }} \right] = \left[ {{\rm{O}}{{\rm{H}}^ – }} \right] = \sqrt {{{\rm{K}}_{\rm{w}}}} \)

\(\sqrt {{{\rm{K}}_{\rm{w}}}} = \sqrt {1 \times {{10}^{ – 14}}} \)

\( = 1 \times {10^{ – 7}}\;{\rm{mol}}/{\rm{L}}\)

At \({\rm{2}}{{\rm{5}}^{\rm{o}}}{\rm{C}}\), the proton concentration of water is \(1 \times {10^{ – 7}}\;{\rm{mol}}/{\rm{L}}\), which is also the hydroxyl ion concentration of water.

The ionic product of water is also applicable for all aqueous solution. Therefore, knowing either \(\left[ {{{\rm{H}}^ + }} \right]\) or  \(\left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\) the other value can be calculated using the value of the ionic product of water.

\({{\rm{K}}_{\rm{w}}} = \left[ {{{\rm{H}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\)

\(\left[ {{{\rm{H}}^ + }} \right] = \frac{{{{\rm{K}}_{\rm{w}}}}}{{\left[ {{\rm{O}}{{\rm{H}}^ – }} \right]}}\)

At any temperature for water and neutral solutions \(\left[ {{{\rm{H}}^ + }} \right] = \left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\).

Aqueous solutions of various substances are divided into neutral, acidic or basic solutions. In neutral solutions, proton and hydroxyl ion concentration are equal, solution in which proton concentration dominates are called acidic solutions and solution in which hydroxyl ion dominates are called basic solutions. These three types of solutions are summarised below:

Types of Solution	Condition of Concentration	Concentration of \({{\rm{H}}^ + }\)$ ion at \({\rm{2}}{{\rm{5}}^{\rm{o}}}{\rm{C}}\)	Concentration of \({\rm{O}}{{\rm{H}}^ – }\) ion at \({\rm{2}}{{\rm{5}}^{\rm{o}}}{\rm{C}}\)
Acidic solution	\(\left[ {{{\rm{H}}^ + }} \right] > \left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\)	\(\left[ {{{\rm{H}}^ + }} \right] > {10^{ – 7}}{\rm{M}}\)	\(\left[ {{\rm{O}}{{\rm{H}}^ – }} \right] < {10^{ – 7}}{\rm{M}}\)
Neutral solution (or) water	\(\left[ {{{\rm{H}}^ + }} \right] = \left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\)	\(\left[ {{{\rm{H}}^ + }} \right] = {10^{ – 7}}{\rm{M}}\)	\(\left[ {{\rm{O}}{{\rm{H}}^ – }} \right] = {10^{ – 7}}{\rm{M}}\)
Basic solution	\(\left[ {{{\rm{H}}^ + }} \right] < \left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\)	\(\left[ {{{\rm{H}}^ + }} \right] < {10^{ – 7}}{\rm{M}}\)	\(\left[ {{\rm{O}}{{\rm{H}}^ – }} \right] > {10^{ – 7}}{\rm{M}}\)

Note carefully that the dissociation/ionisation constant of water \(\left( {\rm{K}} \right)\) is different from the ionic product of \({\rm{K}}\) water \(\left( {{{\rm{K}}_{\rm{w}}}} \right)\). The two are related as \({\rm{K = }}\frac{{{{\rm{K}}_{\rm{w}}}}}{{{\rm{55}}.{\rm{55}}}}.\)

By substituting \(\left[ {{{\rm{H}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^ – }} \right] = {{\rm{K}}_{\rm{w}}}\) and \(\left[ {{{\rm{H}}_2}{\rm{O}}} \right] = 55.55{\rm{M}}\)

We can write, \({\rm{K}} = \frac{{{{10}^{ – 14}}}}{{55.55}} = 1.8 \times {10^{ – 16}}\)

Further, in pure water, \(\left[ {{{\rm{H}}^ + }} \right] = \left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\).

But at \(298\;{\rm{K}},\,\,\left[ {{{\rm{H}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^ – }} \right] = {{\rm{K}}_{\rm{w}}} = {10^{ – 14}}\).

Therefore, \(\left[ {{{\rm{H}}^ + }} \right] = \left[ {{\rm{O}}{{\rm{H}}^ – }} \right] = {10^{ – 7}}\;{\rm{mol}}/{\rm{L}}\).

Hence, the ratio of dissociated water to the undissociated water \( = {10^{ – 7}}/55.55 = 1.8 \times {10^{ – 9}}\). This shows that only a few molecules of \({{\rm{H}}_2}{\rm{O}}\) out of millions are dissociated.

Effect of Temperature on \({{\rm{K}}_{\rm{w}}}\)

The ionic product of water increases if the increase of temperature is increased. It is because with the rise of temperature, the degree of ionisation of water increases. In other words, more \({{\rm{H}}_2}{\rm{O}}\) molecules dissociate into \({{\rm{H}}^ + }\) (or \({{\rm{H}}_3}{{\rm{O}}^ + }\) ) ions and \({\rm{O}}{{\rm{H}}^ – }\) ions. Thus, the concentration of \({{\rm{H}}^ + }\) (or \({{\rm{H}}_3}{{\rm{O}}^ + }\) ) ions and \({\rm{O}}{{\rm{H}}^ – }\) ions increases, and hence the ionic product also increases.

\({{\rm{H}}_3}{{\rm{O}}^ + }\) Ion & \({\rm{O}}{{\rm{H}}^ – }\) Ion Concentrations in Aqueous Solutions of Acids and Bases

As already explained above, for pure water at \(298\;{\rm{K}},\,\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right] = \left[ {{\rm{O}}{{\rm{H}}^ – }} \right] = {10^{ – 7}}{\rm{M}}\).

Now, if some acid (say \({\rm{HCl}}\)) is added to the pure water, then \(\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right] > {10^{ – 7}}{\rm{M}}\)

However, experiments show that the equation \({{\rm{K}}_{\rm{w}}} = \left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\) is still valid. Thus, the \(\left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\) decreases and may be calculated as: \(\left[ {{\rm{O}}{{\rm{H}}^ – }} \right] = \frac{{{{\rm{K}}_{\rm{W}}}}}{{\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]}}\)

Again, if some base (say \({\rm{NaOH}}\)) is added to pure water, then \(\left[ {{\rm{O}}{{\rm{H}}^ – }} \right] > {10^{ – 7}}{\rm{M}}\)

The equation \({{\rm{K}}_{\rm{W}}} = \left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\) is still found to hold good. Hence, \(\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]\) decreases and may be calculated as: \(\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right] = \frac{{{{\rm{K}}_w}}}{{\left[ {{\rm{O}}{{\rm{H}}^ – }} \right]}}\)

The increase or decrease of the \({{\rm{H}}_3}{{\rm{O}}^ + }\) ion concentration in an aqueous solution of an acid or a base may be explained qualitatively based on Le Chatelier’s principle (or common ion effect) as follows:

\(2{{\rm{H}}_2}{\rm{O}}({\rm{l}}) \leftrightarrow {{\rm{H}}_3}{{\rm{O}}^ + } + {\rm{O}}{{\rm{H}}^ – }\)

If some acid is added to pure water, \({{\rm{H}}_3}{{\rm{O}}^ + }\) ion concentration increases; therefore, the equilibrium shifts in the backward direction (or ionisation of \({{\rm{H}}_2}{\rm{O}}\) is suppressed). Thus, the \({\rm{O}}{{\rm{H}}^ – }\) ion concentration decreases. Similarly, if a base added, \({\rm{O}}{{\rm{H}}^ – }\) ion concentration increases, again, the equilibrium shifts in the backward direction (or the ionisation of \({{\rm{H}}_2}{\rm{O}}\) is suppressed) and hence the  \({{\rm{H}}_3}{{\rm{O}}^ + }\) ion concentration decreases. Hence, it can be concluded that the ionic product of water changes when an acid or a base is added to it.

Solved Examples on Ionic Product of Water

Q.1. Neutral solutions have \({\rm{pH}} = 7\). A sample of pure water is found to have a \({\rm{pH}} < 7\). Does it mean that it is acidic? Explain.
Ans: \({\rm{pH}} < 7\) for pure \({{\rm{H}}_2}{\rm{O}}\) shows that water is at a temperature higher than \({\rm{298 K}}\). It is neutral at all temperatures. At higher Temperature, \({{\rm{H}}_2}{\rm{O}}\) dissociates more to give larger concentrations of \({{\rm{H}}^ + }\) ions and \({\rm{O}}{{\rm{H}}^ – }\) ions. Hence, \({\rm{pH}} < 7\). However, \(\left[ {{{\rm{H}}^ + }} \right] = \left[ {{\rm{O}}{{\rm{H}}^ – }} \right]\) hold at all temperatures.

Summary

In this article, we studied the ionic product of water definition, its formula and properties. In addition, we looked into how the concentration of hydrogen ions increases whenever the concentration of hydroxide ions decreases and vice versa. Conversely, the concentration of hydrogen ions falls when the level of hydroxide ions rises to maintain the constancy of the ionic product for water in aqueous solutions.

In this article, we learnt the importance of the ionic product of water, its formula, and its units. The ionic product of water helps in classifying the aqueous solutions as basic, acidic and neutral.

FAQs on Ionic Product of Water

Q.1. What is the value of the ionic product of water at \(298\;{\rm{K}}\)?
Ans: The value of the ionic product of water at \(298\;{\rm{K}}\) is \(1 \times {10^{ – 14}}\;{\rm{mo}}{{\rm{l}}^2}/{{\rm{L}}^2}\) 

Q.2. How do you calculate the ionic product of water?
Ans: Ionic product of water is applicable for all aqueous solution. Knowing either \(\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\) or \(\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\) the other value can be calculated using the value of the ionic product of water, where the value of \({{\rm{K}}_{\rm{w}}}\) is constant at a specific temperature.

\({{\rm{K}}_{\rm{w}}}{\rm{ = }}\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\)

\(\left[ {{{\rm{H}}^{\rm{ + }}}} \right]{\rm{ = }}\frac{{{{\rm{K}}_{\rm{w}}}}}{{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}\)

Q.3. What is the ionic product of water? Write its value.
Ans: The ionic product of water may be defined as the product of the molar concentration of \({{\rm{H}}^{\rm{ + }}}\) ions and \({\rm{O}}{{\rm{H}}^ – }\) ions. As \({{\rm{H}}^{\rm{ + }}}\) ions in water (or in an aqueous solution) exist as \({{\rm{H}}_3}{{\rm{O}}^ + }\) ions, the ionic product may also be defined as the product of molar concentrations of \({{\rm{H}}_{\rm{3}}}{{\rm{O}}^{\rm{ + }}}\) and \({\rm{O}}{{\rm{H}}^{\rm{ – }}}\) ions, i.e., \({{\rm{K}}_{\rm{w}}}{\rm{ = }}\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\). At \({\rm{298}}\;{\rm{K}},\)
\({{\rm{K}}_{\rm{w}}}{\rm{ = 1}} \times {\rm{1}}{{\rm{0}}^{{\rm{ – 14}}}}\;{\rm{mo}}{{\rm{l}}^{\rm{2}}}{\rm{/}}{{\rm{L}}^{\rm{2}}}.\)

Q.4. What is \({{\rm{K}}_{\rm{w}}}\) ionic product of water?
Ans: The Ionic Product of Water, \({{\rm{K}}_{\rm{w}}}\) is the equilibrium constant for the reaction in which water undergoes an acid-base reaction. That is, water is behaving simultaneously as both a base and an acid.

Q.5. Why ionic product of water increases with temperature?
Ans: The ionic product of water \(\left( {{{\rm{K}}_{\rm{w}}}} \right)\) increases with the increase of temperature. It is obviously because with the rise of temperature, the degree of ionisation of water increases. In other words, more of \({{\rm{H}}_2}{\rm{O}}\) molecules dissociate into \({{\rm{H}}^ + }\) (or \({{\rm{H}}_3}{{\rm{O}}^ + }\) ) ions and \({\rm{O}}{{\rm{H}}^ – }\) ions. Thus, the concentration of \({{\rm{H}}^ + }\) (or \({{\rm{H}}_3}{{\rm{O}}^ + }\) ) ions and \({\rm{O}}{{\rm{H}}^ – }\) ions increases, and hence the ionic product also increases.

Q.6. Why is the ionic product of water constant?
Ans: At any given time, the amount of hydroxide ions and hydronium ions present in water is extremely small. Consequently, the concentration of undissociated water molecules is virtually unchanged by this minute ionisation and may be considered constant. The product is a constant termed the ionic product for water.

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