4.15: Water - Acid and Base in One
- Page ID
- 333662
Learning Objectives
- Describe the autoionization of water.
- Calculate the concentrations of \(\ce{H^{+}}\) and \(\ce{OH^{−}}\) in aqueous solutions, knowing the other concentration.
In the autoionization of water, one \(\ce{H2O}\) molecule breaks up into an \(\ce{H^{+}}\) and a \(\ce{OH^{−}}\). The chemical equation is as follows:
\(\ce{H2O}\) <=> \(\ce{H^{+}}\) + \(\ce{OH^{−}}\)
This occurs only to a very small degree: only about 6 in 108 \(\ce{H2O}\) molecules are participating in this process.
At this level, the concentration of both \(\ce{H^{+}(aq)}\) 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
\[\color{red}{\ce{[H^{+}]}} \color{black}{ = } \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-7} \label{eq5}\]
for any sample of pure water because H2O can act as both an acid and a base. The product of these two concentrations is \(1.0\times 10^{−14}\):
\[\color{red}{\ce{[H^{+}]}} \color{black}{\times} \color{blue}{\ce{[OH^{-}]}} \color{black} = (1.0 \times 10^{-7})( 1.0 \times 10^{-7}) = 1.0 \times 10^{-14}\]
- For acids, the concentration of \(\ce{H^{+}(aq)}\) (i.e., \(\ce{[H^{+}]}\)) 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—\(\ce{[H^{+}][OH^{−}]}\)—is always equal to \(1.0 \times 10^{−14}\), no matter whether the aqueous solution is an acid, a base, or neutral:
\[\color{red}{\ce{[H^+]}} \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-14} \]
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 \(K_w\):
\[K_w = \color{red}{\ce{[H^+]}} \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-14} \label{eq10}\]
This means that if you know \(\ce{[H^{+}]}\) 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 \(\ce{[H^{+}]}\). 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—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 \(\ce{[H^{+}]}\) 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: \(\ce{[H^{+}]} =1.0 \times 10^{−4}\, M\) 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−]. \[\left [ \ce{OH^{-}} \right ]=\dfrac{1.0\times 10^{-14}}{\left [ H^+ \right ]} \nonumber\] |
|
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.
- Answer
- \(3.1 \times 10^{−11}\, M\)
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 H+ or OH− ions in the formula unit because \(\ce{[H^{+}]}\) or \(\ce{[OH^{−}]}\)) may not be the same as the concentration of the acid or base itself.
Example \(\PageIndex{2}\): Hydronium Concentration
What is \(\ce{[H^{+}]}\) 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: \(\ce{[H^{+}]}\) = ? 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 \(\ce{[H_3O^{+}]}\). \[\left [ H^{+} \right ]=\dfrac{1.0\times 10^{-14}}{\left [ OH^{-} \right ]} \nonumber\] |
|
Calculate. |
Now substitute the known quantities into the equation and solve. \[\left [ H^{+} \right ]=\frac{1.0\times 10^{-14}}{(0.0088)}=1.1\times 10^{-12}M \nonumber\] \(\ce{[H^{+}]}\) 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 \(\ce{[H^{+}]}\) of an aqueous solution if \(\ce{[OH^{−}]}\) is \(1.0 \times 10^{−9}\, M\)?
- Answer
- 1.0 × 10−5 M
In any aqueous solution, the product of \(\ce{[H^{+}]}\) and \(\ce{[OH^{−}]}\) equals \(1.0 \times 10^{−14}\) (at room temperature).
Contributions & Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality:
Henry Agnew (UC Davis)