in Example 2 , i didn't understand why [H] = 6.3x10...+ x ,why was x added to the concentration of H what does it have to do with that ?

The HCl is so dilute that we must consider both the concentration of H⁺ from the HCl *and* the concentration of H⁺ from the ionization of the water. [H⁺] from HCl = 6.3 × 10⁻⁸ mol/L [H⁺] from H₂O = x mol/L ∴ Total [H⁺] = (6.3 × 10⁻⁸ + x) mol/L

In this article, the value of Kw has been given as 10^-14. But should it not be 10^-14(mol/L)^2? Why have the units been dropped?

That's a really great question! My not so great answer is that it is pretty common in textbooks to drop the units for concentrations when calculating equilibrium constants. This online source has some more detailed explanations (and sources) for why that is the case: http://chemistry.stackexchange.com/questions/1137/why-are-equilibrium-constants-unitless

In exercise 2, the question states that the solution has a hydronium ion conc. of 6.3*10^-8 M. It doesn't specifically say that this measure excludes the number of H+ ions that come from water. Do we always have to assume that?

The article also left me wondering this. If someone states that "a HCl solution with a hydronium ion concentration of 6.3 * 10^-8 M" I would assume that the aqueous solution has the mentioned hydronium ion concentration and use that to calculate the pH. Shouldn't the creator of the article have seperately mentioned that the concentration excludes the H+ coming from autoionization of water. Or maybe I'm missing something? Idk.

How come we multiply the pH value by 2 when the temperature is 0C°?

In pure water, at any temperature, [H3O+] equals [OH-]. At 0 C, pKw = 14.9. pKw = pH + pOH. As [H3O+] equals [OH-], then pH must equal pOH because these are just the negative logs of the respective concentrations, which are equal. Therefore, the equation becomes pKw = pH + pH = 2 x pH. Therefore, pH = pKw/2. So the division by two has nothing to do with temperature. It is just because [H3O+] equals [OH-], which is the case at any temperature for _pure_ water.

I understand how adding an acid such as HCl to water increases the number of H30 ions as the released H+ attaches to H20, but I dont get it why should it decrease the number of OH in a way that keep the product of both constant. Say for the sake of easy math there was 1000 of each ion, the product of both would be 1 million. However, since there is aproximatly 560 million H20 for every 1 OH in neutral water, if I dump over 1 billion HCl molecules there should be about 1 billion more H30 ions and only about 2 fewer OH ions. and 1 billion and a thousand X 998 dosent make a million.

you can use reaction equilibrium to explain this: at 25 degree Celsius, K(w)= 10^(-14) Let [H+] and [OH-] be the initial concentration of H+ and OH- in water Before adding acid: K(w) = [H+]*[OH-] = 10^(-14) After adding acid with x mol/l concentration of [H+]: Q(c) = {[H+] + x}*[OH-] Since {[H+] + x} > [H+] => {[H+] + x}*[OH-] > [H+]*[OH-] => Q(c) > K(w) According to Lechatelier's equilibrium law, the reaction must reach equilibrium by decreasing the new [H+] in the solution. To do this, it must use OH- to react with this newly added H+. As a result, after the reaction reaches equilibrium, both [H+] and [OH-] decreases.

We see that the pKa of water is 14 at 25 degrees Celsius, since the -log of 10^-14 is 14. However, my organic chemistry textbook says the pKa of water is 15.7 at 25 degrees Celsius. Can you please shed some light on this?

There is an article about this via the following link: http://chemwiki.ucdavis.edu/Core/Organic_Chemistry/Fundamentals/What_is_the_pKa_of_water%3F Apparently the organic textbooks are wrong, and the pKa at 25 degrees Celcius is 14. I hope this helps ;)

In Try 2, why is the contribution of [H+} from autoionization not 1 x 10^7?

The autoionization of water is an equilibrium reaction. H₂O ⇌ H⁺ + OH⁻ The addition of H⁺ from the HCl pushes the position of equilibrium to the left, so the [H⁺] from autoionization of water is less than 10⁻⁷ mol/L (it's actually 1.0 × 10⁻⁸ mol/L.

do the ph and the poh change with the change of the temperature

Yes. If the water changes temperature then so does the Kw value, and pH and pOH depend on the kW value. Hope that helps.

When an acid is added to pure water, the resultant solution will have more hydronium ions and fewer hydroxide ions. So, the new equilibrium could be something like [H3O+] = 10^-9 and [OH-] = 10^-5. Is this correct?

If it has more hydronium ions and fewer hydroxide ions, then H3O+ might have a concentration of 10^-5 and OH- could have 10^-9. You have them in the reverse order. But yes, that is possible.

Main content

AP®︎/College Chemistry

Course: AP®︎/College Chemistry > Unit 12

Lesson 1: Acids, bases, and pH

Water autoionization and Kw

Google Classroom

Autoionization of water, the autoionization constant Kw, and the relationship between [H⁺] and [OH⁻] in aqueous solutions.

Key points

Water can undergo autoionization to form start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript and start text, O, H, end text, start superscript, minus, end superscript ions.
The equilibrium constant for the autoionization of water, K, start subscript, start text, w, end text, end subscript, is 10, start superscript, minus, 14, end superscript at 25, degrees, start text, C, end text.
In a neutral solution, 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, 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, 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
For aqueous solutions at 25, degrees, start text, C, end text, the following relationships are always true:

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, equals, 10, start superscript, minus, 14, end superscript

start text, p, H, end text, plus, start text, p, O, H, end text, equals, 14

The contribution of the autoionization of water to 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 open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket becomes significant for extremely dilute acid and base solutions.

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 1: 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 happenminuswater 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:

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

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, colon, 1 molar ratio. For any sample of pure water, the molar concentrations of hydronium, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, and hydroxide, start text, O, H, end text, start superscript, minus, end superscript, must be equal:

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, space, space, start text, i, n, space, p, u, r, e, space, w, a, t, e, r, end text

[What is pure water?]

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 K, start subscript, start text, w, end text, end subscript.

The autoionization constant, K, start subscript, start text, w, end text, end subscript

The expression for the autoionization constant is

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 K, 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 K, start subscript, start text, w, end text, end subscript at 25, degrees, start text, C, end text using 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 start text, p, H, end text of water. At 25, degrees, start text, C, end text, the start text, p, H, end text of pure water is 7. Therefore, we can calculate the concentration of hydronium ions in pure water:

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, 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, 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, degrees, start text, C, end text:

open 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, 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 open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket and 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 K, start subscript, start text, w, end text, end subscript at 25, degrees, start text, C, end text:

K, 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, degrees, start text, C, end text?

[Show the answer]

Relationship between the autoionization constant, start text, p, H, end text, and start text, p, O, H, end text

The fact that K, start subscript, start text, w, end text, end subscript is equal to 10, start superscript, minus, 14, end superscript at 25, degrees, start text, C, end text leads to an interesting and useful new equation. If we take the negative logarithm of both sides of start text, E, q, point, space, 1, end text in the previous section, we get the following:

\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}

[I don't remember how to use logarithms!!!]

We can abbreviate minus, log, K, start subscript, start text, w, end text, end subscript as start text, p, end text, K, start subscript, start text, w, end text, end subscript, which is equal to 14 at 25, degrees, start text, C, end text:

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 start text, p, H, end text and start text, p, O, H, end text will always be 14 for any aqueous solution at 25, degrees, start text, C, end text. Keep in mind that this relationship will not hold true at other temperatures, because K, start subscript, start text, w, end text, end subscript is temperature dependent!

Example 1: Calculating open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket from start text, p, H, end text

An aqueous solution has a start text, p, H, end text of 10 at 25, degrees, start text, C, end text.

What is the concentration of hydroxide ions in the solution?

Method 1: Using Eq. 1

One way to solve this problem is to first find open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket from the start text, p, H, end text:

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

We can then calculate open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket using Eq. 1:

\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}

Method 2: Using Eq. 2

Another way to calculate open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket is to calculate it from the start text, p, O, H, end text of the solution. We can use Eq. 2 to calculate the start text, p, O, H, end text of our solution from the start text, p, H, end text. Rearranging Eq. 2 and solving for the start text, p, O, H, end text, we get:

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

We can now use the equation for start text, p, O, H, end text to solve for open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket.

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

Using either method of solving the problem, the hydroxide concentration is 10, start superscript, minus, 4, end superscript, start text, space, M, end text for an aqueous solution with a start text, p, H, end text of 10 at 25, degrees, start text, C, end text.

Definitions of acidic, basic, and neutral solutions

We have seen that the concentrations of start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript and start text, O, H, end text, start superscript, minus, end superscript are equal in pure water, and both have a value of 10, start superscript, minus, 7, end superscript, start text, space, M, end text at 25, 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 start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript and start text, O, H, end text, start superscript, minus, end superscript.

In a neutral solution, 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, 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, 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

Practice 2: Calculating start text, p, H, end text of water at 0, degrees, start text, C, end text

If the start text, p, end text, K, start subscript, start text, w, end text, end subscript of a sample of pure water at 0, degrees, start text, C, end text is 14, point, 9, what is the start text, p, H, end text of pure water at this temperature?

[Hint 1]

[Hint 2]

Practice 3: Calculating start text, p, end text, K, start subscript, start text, w, end text, end subscript at 40, degrees, start text, C, end text

The start text, p, H, end text of pure water at 40, degrees, start text, C, end text is measured to be 6, point, 75.

Based on this information, what is the start text, p, end text, K, start subscript, start text, w, end text, end subscript of water at 40, degrees, start text, C, end text?

[Hint 1]

[Hint 2]

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 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 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 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 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 K, 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, degrees, start text, C, end text has a hydronium ion concentration of 10, start superscript, minus, 7, end superscript, start text, M, end text, the added acid increases the concentration of 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 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 start text, O, H, end text, start superscript, minus, end superscript to decrease until the product of 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 open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket is once again equal to 10, start superscript, minus, 14, end superscript.

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

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 open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket. Thus, our solution is acidic!
open 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 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 open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket and start 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 open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket or 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 start text, H, end text, start superscript, plus, end superscript or start text, O, H, end text, start superscript, minus, end superscript is within ~2 orders of magnitude (or less than) of start 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 start text, p, H, end text of a very dilute acid solution.

Example 2: Calculating the start text, p, H, end text of a very dilute acid solution

Let's calculate the start text, p, H, end text of a 6, point, 3, times, 10, start superscript, minus, 8, end superscript, start text, M, end text start text, H, C, l, end text solution. start text, H, C, l, end text completely dissociates in water, so the concentration of hydronium ions in solution due to start text, H, C, l, end text is also 6, 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 start text, p, H, end text, we get:

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

Easy! We have an aqueous acid solution with a start text, p, H, end text that is greater than 7. But, wait, wouldn't that make it a basic solution? That can't be right!

Try 2: Including the contribution from autoionization to 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 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 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 open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket using the expression for K, start subscript, start text, w, end text, end subscript:

K, 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 x is the contribution of autoionization to the equilibrium concentration of start text, H, end text, start superscript, plus, end superscript and start text, O, H, end text, start superscript, minus, end superscript, the concentrations at equilibrium will be as follows:

open 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

open 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:

\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}

Rearranging this expression so that everything is equal to 0 gives the following quadratic equation:

0, 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 x using the quadratic formula, which gives the following solutions:

x, 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 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 x to get the equilibrium concentration of start text, H, end text, start superscript, plus, end superscript and calculate start text, p, H, end text, we get:

\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}

Thus we can see that once we include the autoionization of water, our very dilute start text, H, C, l, end text solution has a start text, p, H, end text that is weakly acidic. Whew!

Summary

Water can undergo autoionization to form start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript and start text, O, H, end text, start superscript, minus, end superscript ions.
The equilibrium constant for the autoionization of water, K, start subscript, start text, w, end text, end subscript, is 10, start superscript, minus, 14, end superscript at 25, degrees, start text, C, end text.
In a neutral solution, 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, 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, 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
For aqueous solutions at 25, degrees, start text, C, end text, the following relationships are always true:

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, equals, 10, start superscript, minus, 14, end superscript

start text, p, H, end text, plus, start text, p, O, H, end text, equals, 14

The contribution of the autoionization of water to 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 open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket becomes significant for extremely dilute acid and base solutions.

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sarvin9574
Posted 7 years ago. Direct link to sarvin9574's post “in Example 2 , i didn't ...”
in Example 2 , i didn't understand why [H] = 6.3x10...+ x ,why was x added to the concentration of H what does it have to do with that ?
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(17 votes)
Answer
- Ernest Zinck
  Posted 7 years ago. Direct link to Ernest Zinck's post “The HCl is so dilute that...”
  The HCl is so dilute that we must consider both the concentration of H⁺ from the HCl and the concentration of H⁺ from the ionization of the water.
  [H⁺] from HCl = 6.3 × 10⁻⁸ mol/L
  [H⁺] from H₂O = x mol/L
  ∴ Total [H⁺] = (6.3 × 10⁻⁸ + x) mol/L
  Comment on Ernest Zinck's post “The HCl is so dilute that...”
  (32 votes)
Amit Mukherjee
Posted 7 years ago. Direct link to Amit Mukherjee's post “In this article, the valu...”
In this article, the value of Kw has been given as 10^-14. But should it not be 10^-14(mol/L)^2? Why have the units been dropped?
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(11 votes)
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- yuki
  Posted 7 years ago. Direct link to yuki's post “That's a really great que...”
  That's a really great question! My not so great answer is that it is pretty common in textbooks to drop the units for concentrations when calculating equilibrium constants. This online source has some more detailed explanations (and sources) for why that is the case: http://chemistry.stackexchange.com/questions/1137/why-are-equilibrium-constants-unitless
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just_miku.py
Posted 6 years ago. Direct link to just_miku.py's post “How come we multiply the ...”
How come we multiply the pH value by 2 when the temperature is 0C°?
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(4 votes)
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- RogerP
  Posted 6 years ago. Direct link to RogerP's post “In pure water, at any tem...”
  In pure water, at any temperature, [H3O+] equals [OH-].
  
  At 0 C, pKw = 14.9.
  
  pKw = pH + pOH. As [H3O+] equals [OH-], then pH must equal pOH because these are just the negative logs of the respective concentrations, which are equal.
  
  Therefore, the equation becomes pKw = pH + pH = 2 x pH. Therefore, pH = pKw/2.
  
  So the division by two has nothing to do with temperature. It is just because [H3O+] equals [OH-], which is the case at any temperature for pure water.
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Amit Mukherjee
Posted 7 years ago. Direct link to Amit Mukherjee's post “When an acid is added to ...”
When an acid is added to pure water, the resultant solution will have more hydronium ions and fewer hydroxide ions. So, the new equilibrium could be something like [H3O+] = 10^-9 and [OH-] = 10^-5. Is this correct?
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(0 votes)
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- Matt B
  Posted 7 years ago. Direct link to Matt B's post “If it has more hydronium ...”
  If it has more hydronium ions and fewer hydroxide ions, then H3O+ might have a concentration of 10^-5 and OH- could have 10^-9. You have them in the reverse order. But yes, that is possible.
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Nasrullah Sami
Posted 5 years ago. Direct link to Nasrullah Sami's post “In exercise 2, the questi...”
In exercise 2, the question states that the solution has a hydronium ion conc. of 6.3*10^-8 M. It doesn't specifically say that this measure excludes the number of H+ ions that come from water. Do we always have to assume that?
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(5 votes)
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- RolandKan3
  Posted 5 years ago. Direct link to RolandKan3's post “The article also left me ...”
  The article also left me wondering this. If someone states that "a HCl solution with a hydronium ion concentration of 6.3 * 10^-8 M" I would assume that the aqueous solution has the mentioned hydronium ion concentration and use that to calculate the pH. Shouldn't the creator of the article have seperately mentioned that the concentration excludes the H+ coming from autoionization of water. Or maybe I'm missing something? Idk.
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johnny
Posted 7 years ago. Direct link to johnny's post “I understand how adding a...”
I understand how adding an acid such as HCl to water increases the number of H30 ions as the released H+ attaches to H20, but I dont get it why should it decrease the number of OH in a way that keep the product of both constant. Say for the sake of easy math there was 1000 of each ion, the product of both would be 1 million. However, since there is aproximatly 560 million H20 for every 1 OH in neutral water, if I dump over 1 billion HCl molecules there should be about 1 billion more H30 ions and only about 2 fewer OH ions. and 1 billion and a thousand X 998 dosent make a million.
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(2 votes)
Answer
- Nhi Tran
  Posted 7 years ago. Direct link to Nhi Tran's post “you can use reaction equi...”
  you can use reaction equilibrium to explain this:
  at 25 degree Celsius, K(w)= 10^(-14)
  Let [H+] and [OH-] be the initial concentration of H+ and OH- in water
  Before adding acid: K(w) = [H+]*[OH-] = 10^(-14)
  After adding acid with x mol/l concentration of [H+]: Q(c) = {[H+] + x}*[OH-]
  Since {[H+] + x} > [H+] => {[H+] + x}*[OH-] > [H+]*[OH-] => Q(c) > K(w)
  According to Lechatelier's equilibrium law, the reaction must reach equilibrium by decreasing the new [H+] in the solution. To do this, it must use OH- to react with this newly added H+.
  As a result, after the reaction reaches equilibrium, both [H+] and [OH-] decreases.
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  (4 votes)
Eyasu Kebede
Posted 4 years ago. Direct link to Eyasu Kebede's post “In this question: "Suppos...”
In this question: "Suppose a nanotechnological innovation allows every single charged ion to be precisely identified and removed from a small volume of water. Which of the following describes K_a for the water at the end of the process, assuming that the filtered water is given adequate time to re-equilibrate?"

The answer was Ka = Kw = 10^-14. But, I thought Ka would equal 10^-7. Could you explain this?
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(3 votes)
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Elliot Natanov
Posted 7 years ago. Direct link to Elliot Natanov's post “We see that the pKa of wa...”
We see that the pKa of water is 14 at 25 degrees Celsius, since the -log of 10^-14 is 14. However, my organic chemistry textbook says the pKa of water is 15.7 at 25 degrees Celsius. Can you please shed some light on this?
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(2 votes)
Answer
- Sander Sloof
  Posted 7 years ago. Direct link to Sander Sloof's post “There is an article about...”
  There is an article about this via the following link: http://chemwiki.ucdavis.edu/Core/Organic_Chemistry/Fundamentals/What_is_the_pKa_of_water%3F
  Apparently the organic textbooks are wrong, and the pKa at 25 degrees Celcius is 14. I hope this helps ;)
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  (3 votes)
nickboreham
Posted 7 years ago. Direct link to nickboreham's post “In Try 2, why is the cont...”
In Try 2, why is the contribution of [H+} from autoionization not 1 x 10^7?
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(1 vote)
Answer
- Ernest Zinck
  Posted 7 years ago. Direct link to Ernest Zinck's post “The autoionization of wat...”
  The autoionization of water is an equilibrium reaction.
  H₂O ⇌ H⁺ + OH⁻
  The addition of H⁺ from the HCl pushes the position of equilibrium to the left, so the [H⁺] from autoionization of water is less than 10⁻⁷ mol/L (it's actually 1.0 × 10⁻⁸ mol/L.
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  (4 votes)
shem
Posted a year ago. Direct link to shem's post “do the ph and the poh cha...”
do the ph and the poh change with the change of the temperature
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(1 vote)
Answer
- Richard
  Posted a year ago. Direct link to Richard's post “Yes. If the water changes...”
  Yes. If the water changes temperature then so does the Kw value, and pH and pOH depend on the kW value.
  
  Hope that helps.
  Comment on Richard's post “Yes. If the water changes...”
  (3 votes)

AP®︎/College Chemistry

Course: AP®︎/College Chemistry > Unit 12

Key points

Water is amphoteric

Practice 1111: Identifying the role of water in a reaction

Autoionization of water

The autoionization constant, KwK_\text{w}Kw​K, start subscript, start text, w, end text, end subscript

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

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

Method 1111: Using Eq. 1111

Method 2222: Using Eq. 2222

Definitions of acidic, basic, and neutral solutions

Practice 2222: Calculating pH\text{pH}pHstart text, p, H, end text of water at 0 ∘C0\,^\circ \text C0∘C0, degrees, start text, C, end text

Practice 3333: Calculating pKw\text pK_{\text w}pKw​start text, p, end text, K, start subscript, start text, w, end text, end subscript at 40 ∘C40\,^\circ\text{C}40∘C40, degrees, start text, C, end text

Autoionization and Le Chatelier's principle

Autoionization matters for very dilute acid and base solutions

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

Try 1: Ignoring the autoionization of water

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

Summary

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Practice 1: Identifying the role of water in a reaction

The autoionization constant, K, start subscript, start text, w, end text, end subscript

Relationship between the autoionization constant, start text, p, H, end text, and start text, p, O, H, end text

Example 1: Calculating open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket from start text, p, H, end text

Method 1: Using Eq. 1

Method 2: Using Eq. 2

Practice 2: Calculating start text, p, H, end text of water at 0, degrees, start text, C, end text

Practice 3: Calculating start text, p, end text, K, start subscript, start text, w, end text, end subscript at 40, degrees, start text, C, end text

Example 2: Calculating the start text, p, H, end text of a very dilute acid solution

Try 2: Including the contribution from autoionization to open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket