# 14.8: Water- Acid and Base in One

Learning Objectives

• Describe the autoionization of water.
• Calculate the concentrations of $$\ce{H3O^{+}}$$ and $$\ce{OH^{−}}$$ in aqueous solutions, knowing the other concentration.

We have already seen that $$\ce{H2O}$$ can act as an acid or a base:

$\color{blue}{\underbrace{\ce{NH3}}_{\text{base}}} + \color{red}{\underbrace{\ce{H2O}}_{\text{acid}}} \color{black} \ce{<=> NH4^{+} + OH^{−}}$

where $$\ce{H2O}$$ acts as an $$\color{red}{\text{acid}}$$ (in red).

$\color{red}{\underbrace{\ce{HCl}}_{\text{acid}}} + \color{blue}{\underbrace{\ce{H2O}}_{\text{base}}} \color{black} \ce{-> H3O^{+} + Cl^{−}}$

where $$\ce{H2O}$$ acts as an $$\color{blue}{\text{base}}$$ (in blue).

It may not surprise you to learn, then, that within any given sample of water, some $$\ce{H2O}$$ molecules are acting as acids, and other $$\ce{H2O}$$ molecules are acting as bases. The chemical equation is as follows:

$\color{red}{\underbrace{\ce{H2O}}_{\text{acid}}} + \color{blue}{\underbrace{\ce{H2O}}_{\text{base}}} \color{black} \ce{<=> H3O^{+} + OH^{−}} \label{Auto}$

This occurs only to a very small degree: only about 6 in 108 $$\ce{H2O}$$ molecules are participating in this process, which is called the autoionization of water.

At this level, the concentration of both $$\ce{H3O^{+}(aq)}$$ and $$\ce{OH^{−}(aq)}$$ in a sample of pure $$\ce{H2O}$$ is about $$1.0 \times 10^{−7}\, M$$ (at room temperature). If we use square brackets—[ ]—around a dissolved species to imply the molar concentration of that species, we have

$\color{red}{\ce{[H3O^{+}]}} \color{black}{ = } \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-7} \label{eq5}$

for any sample of pure water because H2O can act as both an acid and a base. The product of these two concentrations is $$1.0\times 10^{−14}$$:

$\color{red}{\ce{[H3O^{+}]}} \color{black}{\times} \color{blue}{\ce{[OH^{-}]}} \color{black} = (1.0 \times 10^{-7})( 1.0 \times 10^{-7}) = 1.0 \times 10^{-14}$

• For acids, the concentration of $$\ce{H3O^{+}(aq)}$$ (i.e., $$\ce{[H3O^{+}]}$$) is greater than $$1.0 \times 10^{−7}\, M$$.
• For bases the concentration of $$\ce{OH^{−}(aq)}$$ (i.e., $$\ce{[OH^{−}]}$$) is greater than $$1.0 \times 10^{−7}\, M$$.

However, the product of the two concentrations—$$\ce{[H3O^{+}][OH^{−}]}$$—is always equal to $$1.0 \times 10^{−14}$$, no matter whether the aqueous solution is an acid, a base, or neutral:

$\color{red}{\ce{[H_3O^+]}} \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-14}$

This value of the product of concentrations is so important for aqueous solutions that it is called the autoionization constant of water and is denoted $$K_w$$:

$K_w = \color{red}{\ce{[H_3O^+]}} \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-14} \label{eq10}$

This means that if you know $$\ce{[H3O^{+}]}$$ for a solution, you can calculate what $$\ce{[OH^{−}]}$$) has to be for the product to equal $$1.0 \times 10^{−14}$$; or if you know $$\ce{[OH^{−}]}$$), you can calculate $$\ce{[H3O^{+}]}$$. This also implies that as one concentration goes up, the other must go down to compensate so that their product always equals the value of $$K_w$$.

Warning: Temperature Matters

The degree of autoionization of water (Equation \ref{Auto})—and hence the value of $$K_w$$—changes with temperature, so Equations \ref{eq5} - \ref{eq10} are accurate only at room temperature.

Example $$\PageIndex{1}$$: Hydroxide Concentration

What is $$\ce{[OH^{−}]}$$) of an aqueous solution if $$\ce{[H3O^{+}]}$$ is $$1.0 \times 10^{−4} M$$?

Solution

Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."

Given: $$\ce{[H3O^{+}]} =1.0 \times 10^{−4}\, M$$

Find: [OH] = ? M

List other known quantities.

none

Plan the problem.

Using the expression for $$K_w$$, (Equation \ref{eq10}), rearrange the equation algebraically to solve for [OH].

$\left [ \ce{OH^{-}} \right ]=\dfrac{1.0\times 10^{-14}}{\left [ H_3O^+ \right ]} \nonumber$

Calculate.

Now substitute the known quantities into the equation and solve.

$\left [\ce{ OH^{-}} \right ]=\dfrac{1.0\times 10^{-14}}{1.0\times 10^{-4}}=1.0\times 10^{-10}M\nonumber$

It is assumed that the concentration unit is molarity, so $$\ce{[OH^{−}]}$$ is 1.0 × 10−10 M.

Think about your result. The concentration of the acid is high (> 1 x 10-7 M), so $$\ce{[OH^{−}]}$$ should be low.

Exercise $$\PageIndex{1}$$

What is $$\ce{[OH^{−}]}$$ in a 0.00032 M solution of H2SO4?

Hint

Assume both protons ionize from the molecule...although this is not the case.

$$3.1 \times 10^{−11}\, M$$

When you have a solution of a particular acid or base, you need to look at the formula of the acid or base to determine the number of H3O+ or OH ions in the formula unit because $$\ce{[H_3O^{+}]}$$ or $$\ce{[OH^{−}]}$$) may not be the same as the concentration of the acid or base itself.

Example $$\PageIndex{2}$$: Hydronium Concentration

What is $$\ce{[H_3O^{+}]}$$ in a 0.0044 M solution of $$\ce{Ca(OH)_2}$$?

Solution

Steps for Problem Solving

Identify the "given" information and what the problem is asking you to "find."

Given: $$[\ce{Ca(OH)_2}] =0.0044 \,M$$

Find: $$\ce{[H_3O^{+}]}$$ = ? M

List other known quantities.

We begin by determining $$\ce{[OH^{−}]}$$. The concentration of the solute is 0.0044 M, but because $$\ce{Ca(OH)_2}$$ is a strong base, there are two OH ions in solution for every formula unit dissolved, so the actual $$\ce{[OH^{−}]}$$ is two times this:

$\ce{[OH^{−}] = 2 \times 0.0044\, M = 0.0088 \,M.} \nonumber$

Plan the problem.

Use the expression for $$K_w$$ (Equation \ref{eq10}) and rearrange the equation algebraically to solve for $$\ce{[H_3O^{+}]}$$.

$\left [ H_3O^{+} \right ]=\dfrac{1.0\times 10^{-14}}{\left [ OH^{-} \right ]} \nonumber$

Calculate.

Now substitute the known quantities into the equation and solve.

$\left [ H_3O^{+} \right ]=\frac{1.0\times 10^{-14}}{(0.0088)}=1.1\times 10^{-12}M \nonumber$

$$\ce{[H_3O^{+}]}$$ has decreased significantly in this basic solution.

Think about your result. The concentration of the base is high (> 1 x 10-7 M) so $$\ce{[H_3O^+}]}$$ should be low.

Exercise $$\PageIndex{2}$$

What is $$\ce{[H_3O^{+}]}$$ of an aqueous solution if $$\ce{[OH^{−}]}$$ is $$1.0 \times 10^{−9}\, M$$?

In any aqueous solution, the product of $$\ce{[H_3O^{+}]}$$ and $$\ce{[OH^{−}]}$$ equals $$1.0 \times 10^{−14}$$ (at room temperature).