15.6: Water as an Acid and as a Base

⚙️ Learning Objectives

• Describe the autoionization of water.
• Calculate the concentrations of H₃O⁺ and OH⁻ in aqueous solutions, knowing the other concentration.

We have already seen that H₂O can act as an acid, as with its reaction with NH₃:

\[\color[rgb]{0.0, 0.0, 0.8}{\underbrace{\ce{NH3}}_{\text{base}}} + \color[rgb]{0.8, 0.0, 0.0}{\underbrace{\ce{H2O}}_{\text{acid}}} \color{black} \ce{<=> NH4^{+} + OH^{−}}\]

and that H₂O can act as a base, as with its reaction with HCl:

\[\color[rgb]{0.8, 0.0, 0.0}{\underbrace{\ce{HCl}}_{\text{acid}}} + \color[rgb]{0.0, 0.0, 0.8}{\underbrace{\ce{H2O}}_{\text{base}}} \color{black} \ce{-> H3O^{+} + Cl^{−}}\]

Therefore, it may not surprise you to learn that within any given sample of water, some H₂O molecules act as an acid and some H₂O molecules act as a base. The chemical equation is as follows:

\[\color[rgb]{0.8, 0.0, 0.0}{\underbrace{\ce{H2O}}_{\text{acid}}} + \color[rgb]{0.0, 0.0, 0.8}{\underbrace{\ce{H2O}}_{\text{base}}} \color{black} \ce{<=> H3O^{+} + OH^{−}}\]

However, this process (shown in Figure \(\PageIndex{1}\)), called the autoionization of water, occurs to a very small extent, with only 6 out of every 100,000,000 H₂O molecules participating in autoionization at 25°C.
 

Figure \(\PageIndex{1}\): The autoionization of water, resulting in hydronium ions and hydroxide ions. (Manuel Almagro Rivas via Wikimedia Commons)

This represents a concentration of 1.0 × 10^–7 M for both H₃O⁺ (aq) and OH⁻ (aq) in a sample of pure H₂O (at 25°C). As previously discussed, square brackets, [ ], around a dissolved species represents the molarity of that species. This means that for any sample of pure water:

\[{\color[rgb]{0.8, 0.0, 0.0}\left[{\mathrm H}_3\mathrm O^+\right]}={\color[rgb]{0.0, 0.0, 0.8}\left[\mathrm{OH}^-\right]}=1.0\times10^{-7}\;\mathrm M\]

The product of these two concentrations is 1.0 × 10⁻¹⁴:

\[{\color[rgb]{0.8, 0.0, 0.0}\left[{\mathrm H}_3\mathrm O^+\right]}\times{\color[rgb]{0.0, 0.0, 0.8}\left[\mathrm{OH}^-\right]}=\left(1.0\times10^{-7}\right)\left(1.0\times10^{-7}\right)=1.0\times10^{-14}\]

It turns out that for an aqueous solution at 25°C, the product of the two concentrations, \({\color[rgb]{0.8, 0.0, 0.0}\left[{\mathrm H}_3\mathrm O^+\right]}\times{\color[rgb]{0.0, 0.0, 0.8}\left[\mathrm{OH}^-\right]}\), is always equal to 1.0 × 10⁻¹⁴, regardless of whether the aqueous solution is acidic, basic, or neutral. In other words, 

\[{\color[rgb]{0.8, 0.0, 0.0}\left[{\mathrm H}_3\mathrm O^+\right]}{\color[rgb]{0.0, 0.0, 0.8}\left[\mathrm{OH}^-\right]}=1.0\times10^{-14}\]

The product of these two concentrations is so important for aqueous solutions that it is called the autoionization constant of water and is denoted K_w:

\[K_w=\left[{\mathrm H}_3\mathrm O^+\right]\left[\mathrm{OH}^-\right]=1.0\times10^{-14}\]

This means that if \(\left[{\mathrm H}_3\mathrm O^+\right]\) is known for an aqueous solution, \(\left[\mathrm{OH}^-\right]\) may be calculated, since their product is always 1.0 × 10⁻¹⁴; or if \(\left[\mathrm{OH}^-\right]\) is known, that \(\left[{\mathrm H}_3\mathrm O^+\right]\) may be calculated. This also implies that as one concentration goes up, the other must go down to ensure their product always equals the value of K_w.
 

⚓️ Temperature Matters

The degree of autoionization of water, as shown in Equation \(\PageIndex{3}\) changes with temperature. This means that the value of K_w also changes with temperature. Therefore, Equation \(\PageIndex{4}\) through Equation \(\PageIndex{7}\) are only valid at room temperature (25°C). In the event the temperature is not known, it is assumed to be 25°C.

Since pure water is neutral and \(\left[{\mathrm H}_3\mathrm O^+\right]=\left[\mathrm{OH}^-\right]=1.0\times10^{-7}\;\mathrm M\) in pure water, then any aqueous solution in which \(\left[{\mathrm H}_3\mathrm O^+\right]=\left[\mathrm{OH}^-\right]=1.0\times10^{-7}\;\mathrm M\) must also be neutral (at 25°C). From this, we can conclude the following:

• Since acids produce H₃O⁺ when dissolved in water, \(\left[{\mathrm H}_3\mathrm O^+\right]\;>1.0\times10^{-7}\;\mathrm M\) in acidic solutions (at 25°C).
• Since bases produce OH^– when dissolved in water, \(\left[\mathrm{OH}^-\right]>1.0\times10^{-7}\;\mathrm M\) in basic solutions (at 25°C).
       

✅ Example \(\PageIndex{1}\)

What is the \(\left[\mathrm{OH}^-\right]\) of an aqueous solution if the \(\left[{\mathrm H}_3\mathrm O^+\right]\) is 1.0 × 10⁻⁴ M?

Solution

Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."	Given: \(\left[{\mathrm H}_3\mathrm O^+\right]\) =1.0 × 10^−4 M Find: \(\left[\mathrm{OH}^-\right]\) = ? M
List known relationship(s).	\(\left[{\mathrm H}_3\mathrm O^+\right]\left[\mathrm{OH}^-\right]=1.0\times10^{-14}\)
Plan the problem.	Rearrange the equation to solve for \(\left[\mathrm{OH}^-\right]\). \(\left[\mathrm{OH}^-\right]=\dfrac{\cancel{\left[{\mathrm H}_3\mathrm O^+\right]}\left[\mathrm{OH}^-\right]}{\cancel{\left[{\mathrm H}_3\mathrm O^+\right]}}=\dfrac{1.0\times10^{-14}}{\left[{\mathrm H}_3\mathrm O^+\right]}\)
Calculate the answer.	Now substitute the known quantities into the equation and solve. \(\left[\mathrm{OH}^-\right]=\dfrac{1.0\times10^{-14}}{\left[{\mathrm H}_3\mathrm O^+\right]}=\dfrac{1.0\times10^{-14}}{1.0\times10^{-4}}=\boxed{1.0\times10^{-10}\;\mathrm M}\)
Think about your result.	The \(\left[{\mathrm H}_3\mathrm O^+\right]\) is high (> 1 × 10^-7 M), so \(\left[\mathrm{OH}^-\right]\) should be low. Furthermore, a quick answer check shows: \(\left[{\mathrm H}_3\mathrm O^+\right]\left[\mathrm{OH}^-\right]=\left(1.0\times10^{-4}\right)\left(1.0\times10^{-10}\right)=1.0\times10^{-14}\)

 

✏️ Exercise \(\PageIndex{1}\)

What is \(\left[{\mathrm H}_3\mathrm O^+\right]\) in solution where \(\left[\mathrm{OH}^-\right]\) = 0.00032 M?

Answer

\(3.1\times10^{-11}\,\mathrm M\)

[H₃O⁺] and [OH^–] of Strong Acids and Bases

Finding the \(\left[{\mathrm H}_3\mathrm O^+\right]\) and \(\left[\mathrm{OH}^-\right]\) of strong acids and strong bases is relatively straightforward. Finding the \(\left[{\mathrm H}_3\mathrm O^+\right]\) and \(\left[\mathrm{OH}^-\right]\) of weak acids and weak bases is a bit more complicated and beyond the scope of this text.

As previously discussed, strong acids ionize completely. For monoprotic strong acids (strong acids with only one H),

\[\left[{\mathrm H}_3\mathrm O^+\right]\;=\;\left[\mathrm{acid}\right]\]

This means that \(\left[{\mathrm H}_3\mathrm O^+\right]\) = 0.10 M in a solution of 0.10 M HNO₃, a strong acid, since one H₃O⁺ ion forms for each molecule of HNO₃:

\({\mathrm{HNO}}_3\;(aq)\;+\;{\mathrm H}_2\mathrm O\;(l)\;\xrightarrow{100\%}\;{\mathrm H}_3\mathrm O^+\;(aq)\;+\;\mathrm{NO}_3^-\;(aq)\)

For H₂SO₄, the only strong acid that is diprotic (has two H atoms) , \(\left[{\mathrm H}_3\mathrm O^+\right]\) = [acid] is a close enough approximation, though the \(\left[{\mathrm H}_3\mathrm O^+\right]\) is slightly greater than [acid], since more than one H₃O⁺ ion forms for each molecule of H₂SO₄. However, it is much for difficult for the second proton (H⁺) to be released from H₂SO₄, which is why \(\left[{\mathrm H}_3\mathrm O^+\right]\) = [acid] remains a close enough approximation.

Also previously discussed, strong bases are soluble in water and dissociate completely. For strong bases,

\[\left[\mathrm{OH}^-\right]=\left[\mathrm{base}\right]\times\mathrm{number}\;\mathrm{of}\;\mathrm{OH}^-\;\mathrm{ions}\;\mathrm{in}\;\mathrm{the}\;\mathrm{chemical}\;\mathrm{formula}\]

This means that \(\left[\mathrm{OH}^-\right]\) = 0.10 M in a solution of 0.10 M NaOH, a strong base, since there is only one OH⁻ ion in NaOH:

\(\mathrm{NaOH}\;(aq)\;\xrightarrow{100\%}\;\mathrm{Na}^+\;(aq)\;+\;\mathrm{OH}^-\;(aq)\)

and that \(\left[\mathrm{OH}^-\right]\) = 0.20 M in a solution of 0.10 M Ba(OH)₂, also a strong base, since there are two OH⁻ ions in Ba(OH)₂:

\(\mathrm{Ba}{(\mathrm{OH})}_2\;(aq)\;\xrightarrow{100\%}\;\mathrm{Ba}^{2+}\;(aq)\;+\;2\;\mathrm{OH}^-\;(aq)\)
 

✅ Example \(\PageIndex{2}\)

What is the:

1. \(\left[{\mathrm H}_3\mathrm O^+\right]\) in 0.0023 M HBr?
2. \(\left[{\mathrm H}_3\mathrm O^+\right]\) in 0.0023 M HNO₃?
3. \(\left[\mathrm{OH}^-\right]\) in 0.0023 M KOH?
4. \(\left[\mathrm{OH}^-\right]\) in 0.0023 M Sr(OH)₂?

Solution

1. \(\left[{\mathrm H}_3\mathrm O^+\right]\;=\;\left[\mathrm{acid}\right]\) in a strong acid, so \(\left[{\mathrm H}_3\mathrm O^+\right]\) = 0.0023 M
2. \(\left[{\mathrm H}_3\mathrm O^+\right]\;=\;\left[\mathrm{acid}\right]\) in a strong acid, so \(\left[{\mathrm H}_3\mathrm O^+\right]\) = 0.0023 M
3. \(\left[\mathrm{OH}^-\right]=\left[\mathrm{base}\right]\times\mathrm{number}\;\mathrm{of}\;\mathrm{OH}^-\;\mathrm{ions}\;\mathrm{in}\;\mathrm{the}\;\mathrm{chemical}\;\mathrm{formula}\) for a strong base, so \(\left[\mathrm{OH}^-\right]\) = 0.0023 M, since there is only one OH⁻ ion in KOH.
4. \(\left[\mathrm{OH}^-\right]=\left[\mathrm{base}\right]\times\mathrm{number}\;\mathrm{of}\;\mathrm{OH}^-\;\mathrm{ions}\;\mathrm{in}\;\mathrm{the}\;\mathrm{chemical}\;\mathrm{formula}\) for a strong base, so \(\left[\mathrm{OH}^-\right]\) = 0.0046 M, since there are two OH⁻ ions in Sr(OH)₂. (0.0023 M × 2 = 0.0046 M).

 

✅ Example \(\PageIndex{3}\)

What is \(\left[{\mathrm H}_3\mathrm O^+\right]\) in a 0.0044 M solution of Ca(OH)₂?

Solution

Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."	Given: Ca(OH)2 = 0.0044 M Find: \(\left[{\mathrm H}_3\mathrm O^+\right]\) = ? M
List known relationship(s).	\(\left[\mathrm{OH}^-\right]=\left[\mathrm{base}\right]\times\mathrm{number}\;\mathrm{of}\;\mathrm{OH}^-\;\mathrm{ions}\;\mathrm{in}\;\mathrm{the}\;\mathrm{chemical}\;\mathrm{formula}\) \(\left[{\mathrm H}_3\mathrm O^+\right]\left[\mathrm{OH}^-\right]=1.0\times10^{-14}\)
Plan the problem.	Rearrange the equation to solve for \(\left[{\mathrm H}_3\mathrm O^+\right]\). \(\left[{\mathrm H}_3\mathrm O^+\right]=\dfrac{\left[{\mathrm H}_3\mathrm O^+\right]\cancel{\left[\mathrm{OH}^-\right]}}{\cancel{\left[\mathrm{OH}^-\right]}}=\dfrac{1.0\times10^{-14}}{\left[\mathrm{OH}^-\right]}\)
Calculate.	\(\left[\mathrm{OH}^-\right]=0.0044\;\mathrm M\;\times\;2\;=\;0.0088\;\mathrm M\) \(\left[{\mathrm H}_3\mathrm O^+\right]=\dfrac{1.0\times10^{-14}}{\left[\mathrm{OH}^-\right]}=\dfrac{1.0\times10^{-14}}{0.0088}=\boxed{1.1\times10^{-12}\;\mathrm M}\)
Think about your result.	The \(\left[\mathrm{OH}^-\right]\) is high (> 1 × 10^-7 M), so \(\left[{\mathrm H}_3\mathrm O^+\right]\) should be low (< 1 × 10^-7 M).

 

✏️ Exercise \(\PageIndex{3}\)

Calculate the \(\left[{\mathrm H}_3\mathrm O^+\right]\) and \(\left[\mathrm{OH}^-\right]\) of 6.1×10^–3 M HI.

Answer

\(\left[{\mathrm H}_3\mathrm O^+\right]\) = 6.1×10^–3 M; \(\left[\mathrm{OH}^-\right]\) = 1.6×10^–12 M

 

     

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This page is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Lance S. Lund (Anoka-Ramsey Community College), Marisa Alviar-Agnew, and Henry Agnew.