Water is amphoteric
Water is one of the most common solvents for acid-base reactions. As we discussed in a previous article on Brønsted-Lowry acids and bases, water is also amphoteric, capable of acting as either a Brønsted-Lowry acid or base.
Practice 11: Identifying the role of water in a reaction
In the following reactions, identify if water is playing the role of an acid, a base, or neither.
[Hint 1]
[Hint 2]
Autoionization of water
Since acids and bases react with each other, this implies that water can react with itself! While that might sound strange, it does happen−minuswater molecules exchange protons with one another to a very small extent. We call this process the autoionization, or self-ionization, of water.
The proton exchange can be written as the following balanced equation:
H2O(l)+H2O(l)⇌H3O+(aq)+OH−(aq)start text, space, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, O, H, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis
space filling models to show two water molecules, where each water molecule is represented as a large red sphere (oxygen) stuck to two small grey sphere (hydrogen). The products are hydronium ion, which has 3 hydrogens and a positive charge, and hydroxide, which has one hydrogen and a negative charge.
One water molecule donates a proton (orange sphere) to a neighboring water molecule, which acts as a Bronsted-Lowry base by accepting that proton. The products of the reversible acid-base reaction are hydronium and hydroxide.
One water molecule is donating a proton and acting as a Bronsted-Lowry acid, while another water molecule accepts the proton, acting as a Bronsted-Lowry base. This results in the formation of hydronium and hydroxide ions in a 1:11, colon, 1 molar ratio. For any sample of pure water, the molar concentrations of hydronium, H3O+start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, and hydroxide, OH−start text, O, H, end text, start superscript, minus, end superscript, must be equal:
[H3O+]=[OH−]inpurewateropen bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, equals, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, space, space, start text, i, n, space, p, u, r, e, space, w, a, t, e, r, end text
Note that this process is readily reversible. Because water is a weak acid and a weak base, the hydronium and hydroxide ions exist in very, very small concentrations relative to that of non-ionized water. Just how small are these concentrations? Let's find out by examining the equilibrium constant for this reaction (also called the autoionization constant), which has the special symbol KwK, start subscript, start text, w, end text, end subscript.
The autoionization constant, KwK, start subscript, start text, w, end text, end subscript
The expression for the autoionization constant is
Kw=[H3O+][OH−](Eq.1)K, start subscript, start text, w, end text, end subscript, equals, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, start text, left parenthesis, E, q, point, space, 1, right parenthesis, end text
Remember that when writing equilibrium expressions, the concentrations of solids and liquids are not included. Therefore, our expression for KwK, start subscript, start text, w, end text, end subscript does not include the concentration of water, which is a pure liquid.
We can calculate the value of KwK, start subscript, start text, w, end text, end subscript at 25∘C25, degrees, start text, C, end text using [H3O+]open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, which is related to the pHstart text, p, H, end text of water. At 25∘C25, degrees, start text, C, end text, the pHstart text, p, H, end text of pure water is 77. Therefore, we can calculate the concentration of hydronium ions in pure water:
[H3O+]=10−pH=10−7Mat25∘Copen bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, equals, 10, start superscript, minus, start text, p, H, end text, end superscript, equals, 10, start superscript, minus, 7, end superscript, start text, space, M, end text, space, space, start text, a, t, space, end text, 25, degrees, start text, C, end text
In the last section, we saw that hydronium and hydroxide form in a 1:11, colon, 1 molar ratio during the autoionization of pure water. We can use that relationship to calculate the concentration of hydroxide in pure water at 25∘C25, degrees, start text, C, end text:
[OH−]=[H3O+]=10−7Mat25∘Copen bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, equals, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, equals, 10, start superscript, minus, 7, end superscript, start text, space, M, end text, space, space, start text, a, t, space, end text, 25, degrees, start text, C, end text
This is a little tough to visualize, but 10−710, start superscript, minus, 7, end superscript is an extremely small number! Within a sample of water, only a small fraction of the water molecules will be in the ionized form.
Now that we know [OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket and [H3O+]open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, we can use these values in our equilibrium expression to calculate KwK, start subscript, start text, w, end text, end subscript at 25∘C25, degrees, start text, C, end text:
Kw=(10−7)×(10−7)=10−14at25∘CK, start subscript, start text, w, end text, end subscript, equals, left parenthesis, 10, start superscript, minus, 7, end superscript, right parenthesis, times, left parenthesis, 10, start superscript, minus, 7, end superscript, right parenthesis, equals, 10, start superscript, minus, 14, end superscript, space, space, start text, a, t, space, end text, 25, degrees, start text, C, end text
Concept check: How many hydroxide and hydronium ions are in one liter of water at 25∘C25, degrees, start text, C, end text?
[Show the answer]
Relationship between the autoionization constant, pHstart text, p, H, end text, and pOHstart text, p, O, H, end text
The fact that KwK, start subscript, start text, w, end text, end subscript is equal to 10−1410, start superscript, minus, 14, end superscript at 25∘C25, degrees, start text, C, end text leads to an interesting and useful new equation. If we take the negative logarithm of both sides of Eq.1start text, E, q, point, space, 1, end text in the previous section, we get the following:
−logKw=−log([H3O+][OH−])=−(log[H3O+]+log[OH−])=−log[H3O+]+(−log[OH−])=pH+pOH
[I don't remember how to use logarithms!!!]
We can abbreviate −logKwminus, log, K, start subscript, start text, w, end text, end subscript as pKwstart text, p, end text, K, start subscript, start text, w, end text, end subscript, which is equal to 1414 at 25∘C25, degrees, start text, C, end text:
pKw=pH+pOH=14at25∘C(Eq.2)start text, p, end text, K, start subscript, start text, w, end text, end subscript, equals, start text, p, H, end text, plus, start text, p, O, H, end text, equals, 14, space, space, start text, a, t, space, end text, 25, degrees, start text, C, end text, start text, left parenthesis, E, q, point, space, 2, end text, right parenthesis
Therefore, the sum of pHstart text, p, H, end text and pOHstart text, p, O, H, end text will always be 1414 for any aqueous solution at 25∘C25, degrees, start text, C, end text. Keep in mind that this relationship will not hold true at other temperatures, because KwK, start subscript, start text, w, end text, end subscript is temperature dependent!
Example 11: Calculating [OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket from pHstart text, p, H, end text
An aqueous solution has a pHstart text, p, H, end text of 1010 at 25∘C25, degrees, start text, C, end text.
What is the concentration of hydroxide ions in the solution?
Method 11: Using Eq. 11
One way to solve this problem is to first find [H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket from the pHstart text, p, H, end text:
[H3O+]=10−pH=10−10M
We can then calculate [OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket using Eq. 1:
Kw[OH−]=[H3O+][OH−]Rearrangetosolvefor[OH−]=[H3O+]KwPluginvaluesforKwand[H3O+]=10−1010−14=10−4M
Method 22: Using Eq. 22
Another way to calculate [OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket is to calculate it from the pOHstart text, p, O, H, end text of the solution. We can use Eq. 2 to calculate the pOHstart text, p, O, H, end text of our solution from the pHstart text, p, H, end text. Rearranging Eq. 2 and solving for the pOHstart text, p, O, H, end text, we get:
pOH=14−pH=14−10=4
We can now use the equation for pOHstart text, p, O, H, end text to solve for [OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket.
[OH−]=10−pOH=10−4M
Using either method of solving the problem, the hydroxide concentration is 10−4M10, start superscript, minus, 4, end superscript, start text, space, M, end text for an aqueous solution with a pHstart text, p, H, end text of 1010 at 25∘C25, degrees, start text, C, end text.
Definitions of acidic, basic, and neutral solutions
We have seen that the concentrations of H3O+start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript and OH−start text, O, H, end text, start superscript, minus, end superscript are equal in pure water, and both have a value of 10−7M10, start superscript, minus, 7, end superscript, start text, space, M, end text at 25∘C25, degrees, start text, C, end text. When the concentrations of hydronium and hydroxide are equal, we say that the solution is neutral. Aqueous solutions can also be acidic or basic depending on the relative concentrations of H3O+start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript and OH−start text, O, H, end text, start superscript, minus, end superscript.
In a neutral solution, [H3O+]=[OH−]open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, equals, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket
In an acidic solution, [H3O+]>[OH−]open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, is greater than, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket
In a basic solution, [OH−]>[H3O+]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, is greater than, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket
Autoionization and Le Chatelier's principle
We also know that in pure water, the concentrations of hydroxide and hydronium are equal. Most of the time, however, we are interested in studying aqueous solutions containing other acids and bases. In that case, what happens to [H3O+]open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket and [OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket?
The moment we dissolve other acids or bases in water, we change [H3O+]open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket and/or [OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket such that the product of the concentrations is no longer is equal to KwK, start subscript, start text, w, end text, end subscript. That means the reaction is no longer at equilibrium. In response, Le Chatelier's principle tells us that the reaction will shift to counteract the change in concentration and establish a new equilibrium.
For example, what if we add an acid to pure water? While pure water at 25∘C25, degrees, start text, C, end text has a hydronium ion concentration of 10−7M10, start superscript, minus, 7, end superscript, start text, M, end text, the added acid increases the concentration of H3O+start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript. In order to get back to equilibrium, the reaction will favor the reverse reaction to use up some of the extra H3O+start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript. This causes the concentration of OH−start text, O, H, end text, start superscript, minus, end superscript to decrease until the product of [H3O+]open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket and [OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket is once again equal to 10−1410, start superscript, minus, 14, end superscript.
Once the reaction reaches its new equilibrium state, we know that:
[H+]>[OH−]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket, is greater than, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket because the added acid increased [H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket. Thus, our solution is acidic!
[OH−]<10−7Mopen bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, is less than, 10, start superscript, minus, 7, end superscript, start text, M, end text because favoring the reverse reaction decreased [OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket to get back to equilibrium.
The important thing to remember is that any aqueous acid-base reaction can be described as shifting the equilibrium concentrations for the autoionization of water. This is really useful, because that means we can apply Eq. 1 and Eq. 2 to all aqueous acid-base reactions, not just pure water!
Autoionization matters for very dilute acid and base solutions
The autoionization of water is usually introduced when first learning about acids and bases, and it is used to derive some extremely useful equations that we've discussed in this article. However, we will often calculate [H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket and pHstart text, p, H, end text for aqueous solutions without including the contribution from the autoionization of water. The reason we can do this is because autoionization usually contributes relatively few ions to the overall [H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket or [OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket compared to the ions from additional acid or base.
The only situation when we need to remember the autoionization of water is when the concentration of our acid or base is extremely dilute. In practice, this means that we need to include the contribution from autoionization when the concentration of H+start text, H, end text, start superscript, plus, end superscript or OH−start text, O, H, end text, start superscript, minus, end superscript is within ~22 orders of magnitude (or less than) of 10−7Mstart text, 10, end text, start superscript, minus, 7, end superscript, start text, M, end text. We will now go through an example of how to calculate the pHstart text, p, H, end text of a very dilute acid solution.
Example 22: Calculating the pHstart text, p, H, end text of a very dilute acid solution
Let's calculate the pHstart text, p, H, end text of a 6.3×10−8M6, point, 3, times, 10, start superscript, minus, 8, end superscript, start text, M, end text HClstart text, H, C, l, end text solution. HClstart text, H, C, l, end text completely dissociates in water, so the concentration of hydronium ions in solution due to HClstart text, H, C, l, end text is also 6.3×10−8M6, point, 3, times, 10, start superscript, minus, 8, end superscript, start text, M, end text.
Try 1: Ignoring the autoionization of water
If we ignore the autoionization of water and simply use the formula for pHstart text, p, H, end text, we get:
pH=−log[H+]=−log[6.3×10−8]=7.20
Easy! We have an aqueous acid solution with a pHstart text, p, H, end text that is greater than 77. But, wait, wouldn't that make it a basic solution? That can't be right!
Try 2: Including the contribution from autoionization to [H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket
Since the concentration of this solution is extremely dilute, the concentration of the hydronium from the hydrochloric acid is close to the [H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket contribution from the autoionization of water. That means:
We have to include the contribution from autoionization to [H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket
Since the autoionization of water is an equilibrium reaction, we must solve for the overall [H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket using the expression for KwK, start subscript, start text, w, end text, end subscript:
Kw=[H+][OH−]=1.0×10−14K, start subscript, start text, w, end text, end subscript, equals, open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, equals, 1, point, 0, times, 10, start superscript, minus, 14, end superscript
If we say that xx is the contribution of autoionization to the equilibrium concentration of H+start text, H, end text, start superscript, plus, end superscript and OH−start text, O, H, end text, start superscript, minus, end superscript, the concentrations at equilibrium will be as follows:
[H+]=6.3×10−8M+xopen bracket, start text, H, end text, start superscript, plus, end superscript, close bracket, equals, 6, point, 3, times, 10, start superscript, minus, 8, end superscript, start text, M, end text, plus, x
[OH−]=xopen bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, equals, x
Plugging these concentrations into our equilibrium expression, we get:
Kw=(6.3×10−8M+x)x=1.0×10−14=x2+6.3×10−8x
Rearranging this expression so that everything is equal to 00 gives the following quadratic equation:
0=x2+6.3×10−8x−1.0×10−140, equals, x, squared, plus, 6, point, 3, times, 10, start superscript, minus, 8, end superscript, x, minus, 1, point, 0, times, 10, start superscript, minus, 14, end superscript
We can solve for xx using the quadratic formula, which gives the following solutions:
x=7.3×10−8M,−1.4×10−7Mx, equals, 7, point, 3, times, 10, start superscript, minus, 8, end superscript, start text, M, end text, comma, minus, 1, point, 4, times, 10, start superscript, minus, 7, end superscript, start text, M, end text
Since the concentration of OH−start text, O, H, end text, start superscript, minus, end superscript can't be negative, we can eliminate the second solution. If we plug in the first value of xx to get the equilibrium concentration of H+start text, H, end text, start superscript, plus, end superscript and calculate pHstart text, p, H, end text, we get:
pH=−log[H+]=−log[6.3×10−8+x]=−log[6.3×10−8+7.3×10−8]=−log[1.36×10−7]=6.87
Thus we can see that once we include the autoionization of water, our very dilute HClstart text, H, C, l, end text solution has a pHstart text, p, H, end text that is weakly acidic. Whew!