Water autoionization and Kw (article) | Khan Academy (2023)

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 1111: 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)\qquad\qquad\text{ H}_2\text{O}(l)+\text{H}_2\text{O}(l)\rightleftharpoons\text{H}_3\text{O}^+(aq)+\text{OH}^-(aq)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.

(Video) Autoionization of water | Acids and bases | AP Chemistry | Khan Academy

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 1:11:11:11, colon, 1 molar ratio. For any sample of pure water, the molar concentrations of hydronium, H3O+\text{H}_3\text{O}^+H3O+start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, and hydroxide, OH\text{OH}^-OHstart text, O, H, end text, start superscript, minus, end superscript, must be equal:

[H3O+]=[OH]inpurewater[\text{H}_3\text{O}^+]=[\text{OH}^-]~~\text{in pure water}[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_\text{w}KwK, start subscript, start text, w, end text, end subscript.

The autoionization constant, KwK_\text{w}KwK, start subscript, start text, w, end text, end subscript

The expression for the autoionization constant is

Kw=[H3O+][OH](Eq.1)K_\text{w}=[\text{H}_3\text{O}^+][\text{OH}^-]\quad\quad\text{(Eq. 1)}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_\text{w}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_\text{w}KwK, start subscript, start text, w, end text, end subscript at 25C25\,^\circ\text{C}25C25, degrees, start text, C, end text using [H3O+][\text{H}_3\text{O}^+][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 pH\text{pH}pHstart text, p, H, end text of water. At 25C25\,^\circ\text{C}25C25, degrees, start text, C, end text, the pH\text{pH}pHstart text, p, H, end text of pure water is 7777. Therefore, we can calculate the concentration of hydronium ions in pure water:

[H3O+]=10pH=107Mat25C[\text{H}_3\text{O}^+]=10^{-\text{pH}}=10^{-7}\text{ M}~~\text{at }25\,^\circ\text{C}[H3O+]=10pH=107Mat25Copen 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 1:11:11:11, 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 25C25^\circ\text{C}25C25, degrees, start text, C, end text:

[OH]=[H3O+]=107Mat25C[\text{OH}^-]=[\text{H}_3\text{O}^+]=10^{-7}\text{ M}~~\text{at }25\,^\circ\text{C}[OH]=[H3O+]=107Mat25Copen 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 10710^{-7}10710, 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][\text{OH}^-][OH]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket and [H3O+][\text{H}_3\text{O}^+][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_\text{w}KwK, start subscript, start text, w, end text, end subscript at 25C25^\circ\text{C}25C25, degrees, start text, C, end text:

Kw=(107)×(107)=1014at25CK_\text{w}=(10^{-7})\times(10^{-7})=10^{-14}~~\text{at }25\,^\circ\text{C}Kw=(107)×(107)=1014at25CK, 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 25C25^\circ\text{C}25C25, degrees, start text, C, end text?

[Show the answer]

(Video) Autoionization of water | Chemical processes | MCAT | Khan Academy

Relationship between the autoionization constant, pH\text{pH}pHstart text, p, H, end text, and pOH\text{pOH}pOHstart text, p, O, H, end text

The fact that KwK_\text{w}KwK, start subscript, start text, w, end text, end subscript is equal to 101410^{-14}101410, start superscript, minus, 14, end superscript at 25C25\,^\circ\text{C}25C25, 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.1\text{Eq. 1}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\begin{aligned}-\log{K_\text{w}}&=-\log({[\text{H}_3\text{O}^+}][\text{OH}^-])\\\\&=-\big(\log[\text{H}_3\text{O}^+]+\log[\text{OH}^-]\big)\\\\&=-\log[\text{H}_3\text{O}^+]+(-\log[\text{OH}^-])\\\\&=\text{pH}+\text{pOH}\end{aligned}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 logKw-\log{K_\text{w}}logKwminus, log, K, start subscript, start text, w, end text, end subscript as pKw\text{p}K_\text wpKwstart text, p, end text, K, start subscript, start text, w, end text, end subscript, which is equal to 14141414 at 25C25\,^\circ\text{C}25C25, degrees, start text, C, end text:

pKw=pH+pOH=14at25C(Eq.2)\text{p}K_\text{w}=\text{pH}+\text{pOH}=14~~\text{at }25\,^\circ \text C\quad\quad\text{(Eq. 2})pKw=pH+pOH=14at25C(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 pH\text{pH}pHstart text, p, H, end text and pOH\text{pOH}pOHstart text, p, O, H, end text will always be 14141414 for any aqueous solution at 25C25\,^\circ\text{C}25C25, degrees, start text, C, end text. Keep in mind that this relationship will not hold true at other temperatures, because KwK_\text{w}KwK, start subscript, start text, w, end text, end subscript is temperature dependent!

Example 1111: Calculating [OH][\text{OH}^-][OH]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket from pH\text{pH}pHstart text, p, H, end text

An aqueous solution has a pH\text{pH}pHstart text, p, H, end text of 10101010 at 25C25\,^\circ\text{C}25C25, degrees, start text, C, end text.

What is the concentration of hydroxide ions in the solution?

Method 1111: Using Eq. 1111

One way to solve this problem is to first find [H+][\text{H}^+][H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket from the pH\text{pH}pHstart text, p, H, end text:

[H3O+]=10pH=1010M\begin{aligned}[\text{H}_3\text{O}^+]&=10^{-\text{pH}}\\\\&=10^{-10}\,\text M\\\end{aligned}[H3O+]=10pH=1010M

We can then calculate [OH][\text{OH}^-][OH]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket using Eq. 1:

Kw=[H3O+][OH]Rearrangetosolvefor[OH][OH]=Kw[H3O+]PluginvaluesforKwand[H3O+]=10141010=104M\begin{aligned}K_\text{w}&=[\text{H}_3\text{O}^+][\text{OH}^-]~~~\quad\quad\text{Rearrange to solve for }[\text{OH}^-]\\\\[\text{OH}^-]&=\dfrac{K_\text{w}}{[\text{H}_3\text{O}^+]}\qquad\quad\qquad\text{Plug in values for }K_\text w \,\text{and [H}_3 \text O^+]\\\\&=\dfrac{10^{-14}}{10^{-10}}\\\\&=10^{-4}\text{ M}\end{aligned}Kw[OH]=[H3O+][OH]Rearrangetosolvefor[OH]=[H3O+]KwPluginvaluesforKwand[H3O+]=10101014=104M

Method 2222: Using Eq. 2222

Another way to calculate [OH][\text{OH}^-][OH]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket is to calculate it from the pOH\text{pOH}pOHstart text, p, O, H, end text of the solution. We can use Eq. 2 to calculate the pOH\text{pOH}pOHstart text, p, O, H, end text of our solution from the pH\text{pH}pHstart text, p, H, end text. Rearranging Eq. 2 and solving for the pOH\text{pOH}pOHstart text, p, O, H, end text, we get:

pOH=14pH=1410=4\begin{aligned}\text{pOH}&=14-\text{pH}\\\\&=14-10\\\\&=4\end{aligned}pOH=14pH=1410=4

(Video) Autoionization of Water and Kw

We can now use the equation for pOH\text{pOH}pOHstart text, p, O, H, end text to solve for [OH][\text{OH}^-][OH]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket.

[OH]=10pOH=104M\begin{aligned}[\text{OH}^-]&=10^{-\text{pOH}}\\\\&=10^{-4}\text{ M}\end{aligned}[OH]=10pOH=104M

Using either method of solving the problem, the hydroxide concentration is 104M10^{-4}\text{ M}104M10, start superscript, minus, 4, end superscript, start text, space, M, end text for an aqueous solution with a pH\text{pH}pHstart text, p, H, end text of 10101010 at 25C25\,^\circ\text{C}25C25, degrees, start text, C, end text.

Definitions of acidic, basic, and neutral solutions

We have seen that the concentrations of H3O+\text{H}_3\text{O}^+H3O+start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript and OH\text{OH}^-OHstart text, O, H, end text, start superscript, minus, end superscript are equal in pure water, and both have a value of 107M10^{-7}\text{ M}107M10, start superscript, minus, 7, end superscript, start text, space, M, end text at 25C25\,^\circ\text{C}25C25, 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+\text{H}_3\text{O}^+H3O+start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript and OH\text{OH}^-OHstart text, O, H, end text, start superscript, minus, end superscript.

  • In a neutral solution, [H3O+]=[OH][\text{H}_3\text{O}^+]=[\text{OH}^-][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][\text{H}_3\text{O}^+]>[\text{OH}^-][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+][\text{OH}^-]>[\text{H}_3\text{O}^+][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+][\text{H}_3\text{O}^+][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][\text{OH}^-][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+][\text{H}_3\text{O}^+][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][\text{OH}^-][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_\text{w}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 25C25\,^\circ \text C25C25, degrees, start text, C, end text has a hydronium ion concentration of 107M10^{-7}\,\text M107M10, start superscript, minus, 7, end superscript, start text, M, end text, the added acid increases the concentration of H3O+\text{H}_3\text{O}^+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+\text{H}_3\text{O}^+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\text{OH}^-OHstart text, O, H, end text, start superscript, minus, end superscript to decrease until the product of [H3O+][\text{H}_3\text{O}^+][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][\text{OH}^-][OH]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket is once again equal to 101410^{-14}101410, start superscript, minus, 14, end superscript.

Once the reaction reaches its new equilibrium state, we know that:

  • [H+]>[OH][\text H^+]>[\text{OH}^-][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+][\text H^+][H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket. Thus, our solution is acidic!

  • [OH]<107M[\text{OH}^-]<10^{-7}\,\text M[OH]<107Mopen 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][\text{OH}^-][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+][\text H^+][H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket and pH\text{pH}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+][\text H^+][H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket or [OH][\text{OH}^-][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+\text H^+H+start text, H, end text, start superscript, plus, end superscript or OH\text{OH}^-OHstart text, O, H, end text, start superscript, minus, end superscript is within ~2222 orders of magnitude (or less than) of 107M\text{10}^{-7}\,\text M107Mstart 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 pH\text{pH}pHstart text, p, H, end text of a very dilute acid solution.

Example 2222: Calculating the pH\text{pH}pHstart text, p, H, end text of a very dilute acid solution

Let's calculate the pH\text{pH}pHstart text, p, H, end text of a 6.3×108M6.3 \times 10^{-8}\,\text M6.3×108M6, point, 3, times, 10, start superscript, minus, 8, end superscript, start text, M, end text HCl\text{HCl}HClstart text, H, C, l, end text solution. HCl\text{HCl}HClstart text, H, C, l, end text completely dissociates in water, so the concentration of hydronium ions in solution due to HCl\text{HCl}HClstart text, H, C, l, end text is also 6.3×108M6.3 \times 10^{-8}\,\text M6.3×108M6, point, 3, times, 10, start superscript, minus, 8, end superscript, start text, M, end text.

(Video) Autoionization of water | Water, acids, and bases | Biology | Khan Academy

Try 1: Ignoring the autoionization of water

If we ignore the autoionization of water and simply use the formula for pH\text{pH}pHstart text, p, H, end text, we get:

pH=log[H+]=log[6.3×108]=7.20\begin{aligned}\text{pH}&=-\text{log}[\text H^+]\\\\&=-\text{log}[6.3 \times 10^{-8}]\\\\&=7.20\end{aligned}pH=log[H+]=log[6.3×108]=7.20

Easy! We have an aqueous acid solution with a pH\text{pH}pHstart text, p, H, end text that is greater than 7777. But, wait, wouldn't that make it a basic solution? That can't be right!

Try 2: Including the contribution from autoionization to [H+][\text{H}^+][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+][\text{H}^+][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+][\text{H}^+][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+][\text{H}^+][H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket using the expression for KwK_\text{w}KwK, start subscript, start text, w, end text, end subscript:

Kw=[H+][OH]=1.0×1014K_\text{w} =[\text H^+][\text{OH}^-]=1.0\times10^{-14}Kw=[H+][OH]=1.0×1014K, 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 xxxx is the contribution of autoionization to the equilibrium concentration of H+\text H^+H+start text, H, end text, start superscript, plus, end superscript and OH\text{OH}^-OHstart text, O, H, end text, start superscript, minus, end superscript, the concentrations at equilibrium will be as follows:

[H+]=6.3×108M+x[\text H^+]=6.3 \times 10^{-8}\,\text M+x[H+]=6.3×108M+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]=x[\text {OH}^-]=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=(6.3×108M+x)x=1.0×1014=x2+6.3×108x\begin{aligned}K_\text{w} &=(6.3 \times 10^{-8}\,\text M+x)x=1.0\times10^{-14}\\\\&=x^2+6.3 \times 10^{-8}x\end{aligned}Kw=(6.3×108M+x)x=1.0×1014=x2+6.3×108x

Rearranging this expression so that everything is equal to 0000 gives the following quadratic equation:

0=x2+6.3×108x1.0×10140=x^2+6.3 \times 10^{-8}x-1.0\times10^{-14}0=x2+6.3×108x1.0×10140, 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 xxxx using the quadratic formula, which gives the following solutions:

x=7.3×108M,1.4×107Mx=7.3 \times 10^{-8}\,\text M, -1.4 \times 10^{-7}\,\text Mx=7.3×108M,1.4×107Mx, 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\text{OH}^-OHstart 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 xxxx to get the equilibrium concentration of H+\text H^+H+start text, H, end text, start superscript, plus, end superscript and calculate pH\text{pH}pHstart text, p, H, end text, we get:

pH=log[H+]=log[6.3×108+x]=log[6.3×108+7.3×108]=log[1.36×107]=6.87\begin{aligned}\text{pH}&=-\text{log}[\text H^+]\\\\&=-\text{log}[6.3 \times 10^{-8}+x]\\\\&=-\text{log}[6.3 \times 10^{-8}+7.3 \times 10^{-8}]\\\\&=-\text{log}[1.36 \times 10^{-7}]\\\\&=6.87\end{aligned}pH=log[H+]=log[6.3×108+x]=log[6.3×108+7.3×108]=log[1.36×107]=6.87

Thus we can see that once we include the autoionization of water, our very dilute HCl\text{HCl}HClstart text, H, C, l, end text solution has a pH\text {pH}pHstart text, p, H, end text that is weakly acidic. Whew!

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