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 ;)

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.

"Since the concentration of OH- can't be negative, we can eliminate the second solution. Doesn't it sound too loose and unlogical? We can just casually remove a certain value from calculation?

If you're solving a quadratic equation, you can receive a maximum of two real solutions. In this case we get that with one being positive and one being negative. So it makes mathematical sense that we get two answers, but physically only the positive answer makes sense. A negative concentration means you have less than 0 moles of hydroxide, or less than nothing. We can have no hydroxide possible, 0 moles of hydroxide in the solution, but not less than that. It's similar to problems in geometry where we can only have positive areas or volumes since we're considering real physical objects even if the algebra sometimes allows a negative answer. Hope that helps.

Main content

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 $H_{3} O^{+}$ ‍ and ${OH}^{-}$ ‍ ions.
The equilibrium constant for the autoionization of water, $K_{w}$ ‍, is $10^{- 14}$ ‍ at $25^{\circ} C$ ‍.
In a neutral solution, $[H_{3} O^{+}] = [{OH}^{-}]$ ‍
In an acidic solution, $[H_{3} O^{+}] > [{OH}^{-}]$ ‍
In a basic solution, $[{OH}^{-}] > [H_{3} O^{+}]$ ‍
For aqueous solutions at $25^{\circ} C$ ‍, the following relationships are always true:

$K_{w} = [H_{3} O^{+}] [{OH}^{-}] = 10^{- 14}$ ‍

$pH + pOH = 14$ ‍

The contribution of the autoionization of water to $[H_{3} O^{+}]$ ‍ and $[{OH}^{-}]$ ‍ 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.

1

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

-

water 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:

H_{2} O (l) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {OH}^{-} (a q)

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 : 1

molar ratio. For any sample of pure water, the molar concentrations of hydronium,

H_{3} O^{+}

, and hydroxide,

{OH}^{-}

, must be equal:

[H_{3} O^{+}] = [{OH}^{-}] in 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_{w}

The autoionization constant, $K_{w}$ ‍

The expression for the autoionization constant is

K_{w} = [H_{3} O^{+}] [{OH}^{-}] (Eq. 1)

Remember that when writing equilibrium expressions, the concentrations of solids and liquids are not included. Therefore, our expression for

K_{w}

does not include the concentration of water, which is a pure liquid.

We can calculate the value of

K_{w}

25^{\circ} C

using

[H_{3} O^{+}]

, which is related to the

pH

of water. At

25^{\circ} C

, the

pH

of pure water is

7

. Therefore, we can calculate the concentration of hydronium ions in pure water:

[H_{3} O^{+}] = 10^{- pH} = 10^{- 7} M at 25^{\circ} C

In the last section, we saw that hydronium and hydroxide form in a

1 : 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^{\circ} C

[{OH}^{-}] = [H_{3} O^{+}] = 10^{- 7} M at 25^{\circ} C

This is a little tough to visualize, but

10^{- 7}

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}^{-}]

and

[H_{3} O^{+}]

, we can use these values in our equilibrium expression to calculate

K_{w}

25^{\circ} C

K_{w} = (10^{- 7}) \times (10^{- 7}) = 10^{- 14} at 25^{\circ} C

Concept check: How many hydroxide and hydronium ions are in one liter of water at $25^{\circ} C$ ‍?

We know that at

25^{\circ} C

, the following relationship is true:

[{OH}^{-}] = [H_{3} O^{+}] = 10^{- 7} \frac{mol}{L}

That means one liter of water contains

10^{- 7} mol

each of hydronium ions and hydroxide ions. We can convert from moles to the number of ions using Avogadro's number:

10^{- 7} mol \times \frac{6.022 \times 10^{23} ions}{1 mol} = 6.022 \times 10^{16} ions

This sounds like a huge number! To put this in perspective, we can compare the number of hydronium and hydroxide to the number of water molecules in the same volume. There are

3.34 \times 10^{25}

molecules of water in one liter of water. That is nearly

10^{9}

larger than the number of

H_{3} O^{+}

and

{OH}^{-}

ions in solution!

Relationship between the autoionization constant, $pH$ ‍, and $pOH$ ‍

The fact that

K_{w}

is equal to

10^{- 14}

25^{\circ} C

leads to an interesting and useful new equation. If we take the negative logarithm of both sides of

Eq. 1

in the previous section, we get the following:

\begin{aligned} - \log K_{w} & = - \log ([H_{3} O^{+}] [{OH}^{-}]) \\ = - (\log [H_{3} O^{+}] + \log [{OH}^{-}]) \\ = - \log [H_{3} O^{+}] + (- \log [{OH}^{-}]) \\ = pH + pOH \end{aligned}

We can abbreviate

- \log K_{w}

p K_{w}

, which is equal to

14

25^{\circ} C

p K_{w} = pH + pOH = 14 at 25^{\circ} C (Eq. 2)

Therefore, the sum of

pH

and

pOH

will always be

14

for any aqueous solution at

25^{\circ} C

. Keep in mind that this relationship will not hold true at other temperatures, because

K_{w}

is temperature dependent!

Example $1$ ‍: Calculating $[{OH}^{-}]$ ‍ from $pH$ ‍

An aqueous solution has a

pH

10

25^{\circ} C

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

[H^{+}]

from the

pH

\begin{aligned} [H_{3} O^{+}] & = 10^{- pH} \\ = 10^{- 10} M \end{aligned}

We can then calculate

[{OH}^{-}]

using Eq. 1:

\begin{aligned} K_{w} & = [H_{3} O^{+}] [{OH}^{-}] Rearrange to solve for [{OH}^{-}] \\ [{OH}^{-}] & = \frac{K_{w}}{[H_{3} O^{+}]} Plug in values for K_{w} {and [H}_{3} O^{+}] \\ = \frac{10^{- 14}}{10^{- 10}} \\ = 10^{- 4} M \end{aligned}

Method $2$ ‍: Using Eq. $2$ ‍

Another way to calculate

[{OH}^{-}]

is to calculate it from the

pOH

of the solution. We can use Eq. 2 to calculate the

pOH

of our solution from the

pH

. Rearranging Eq. 2 and solving for the

pOH

, we get:

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

We can now use the equation for

pOH

to solve for

[{OH}^{-}]

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

Using either method of solving the problem, the hydroxide concentration is

10^{- 4} M

for an aqueous solution with a

pH

10

25^{\circ} C

Definitions of acidic, basic, and neutral solutions

We have seen that the concentrations of

H_{3} O^{+}

and

{OH}^{-}

are equal in pure water, and both have a value of

10^{- 7} M

25^{\circ} C

. 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

H_{3} O^{+}

and

{OH}^{-}

In a neutral solution, $[H_{3} O^{+}] = [{OH}^{-}]$ ‍
In an acidic solution, $[H_{3} O^{+}] > [{OH}^{-}]$ ‍
In a basic solution, $[{OH}^{-}] > [H_{3} O^{+}]$ ‍

Practice $2$ ‍: Calculating $pH$ ‍ of water at $0^{\circ} C$ ‍

If the $p K_{w}$ ‍ of a sample of pure water at $0^{\circ} C$ ‍ is $14.9$ ‍, what is the $pH$ ‍ of pure water at this temperature?

Practice $3$ ‍: Calculating $p K_{w}$ ‍ at $40^{\circ} C$ ‍

The

pH

of pure water at

40^{\circ} C

is measured to be

6.75

Based on this information, what is the $p K_{w}$ ‍ of water at $40^{\circ} C$ ‍?

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

[H_{3} O^{+}]

and

[{OH}^{-}]

The moment we dissolve other acids or bases in water, we change

[H_{3} O^{+}]

and/or

[{OH}^{-}]

such that the product of the concentrations is no longer is equal to

K_{w}

. 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^{\circ} C

has a hydronium ion concentration of

10^{- 7} M

, the added acid increases the concentration of

H_{3} O^{+}

. In order to get back to equilibrium, the reaction will favor the reverse reaction to use up some of the extra

H_{3} O^{+}

. This causes the concentration of

{OH}^{-}

to decrease until the product of

[H_{3} O^{+}]

and

[{OH}^{-}]

is once again equal to

10^{- 14}

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

$[H^{+}] > [{OH}^{-}]$ ‍ because the added acid increased $[H^{+}]$ ‍. Thus, our solution is acidic!
$[{OH}^{-}] < 10^{- 7} M$ ‍ because favoring the reverse reaction decreased $[{OH}^{-}]$ ‍ 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^{+}]

and

pH

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^{+}]

[{OH}^{-}]

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^{+}

{OH}^{-}

is within ~

2

orders of magnitude (or less than) of

10^{- 7} M

. We will now go through an example of how to calculate the

pH

of a very dilute acid solution.

Example $2$ ‍: Calculating the $pH$ ‍ of a very dilute acid solution

Let's calculate the

pH

of a

6.3 \times 10^{- 8} M

HCl

solution.

HCl

completely dissociates in water, so the concentration of hydronium ions in solution due to

HCl

is also

6.3 \times 10^{- 8} M

Try 1: Ignoring the autoionization of water

If we ignore the autoionization of water and simply use the formula for

pH

, we get:

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

Easy! We have an aqueous acid solution with a

pH

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 $[H^{+}]$ ‍

Since the concentration of this solution is extremely dilute, the concentration of the hydronium from the hydrochloric acid is close to the

[H^{+}]

contribution from the autoionization of water. That means:

We have to include the contribution from autoionization to $[H^{+}]$ ‍
Since the autoionization of water is an equilibrium reaction, we must solve for the overall $[H^{+}]$ ‍ using the expression for $K_{w}$ ‍:

K_{w} = [H^{+}] [{OH}^{-}] = 1.0 \times 10^{- 14}

If we say that

x

is the contribution of autoionization to the equilibrium concentration of

H^{+}

and

{OH}^{-}

, the concentrations at equilibrium will be as follows:

[H^{+}] = 6.3 \times 10^{- 8} M + x

[{OH}^{-}] = x

Plugging these concentrations into our equilibrium expression, we get:

\begin{aligned} K_{w} & = (6.3 \times 10^{- 8} M + x) x = 1.0 \times 10^{- 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 = x^{2} + 6.3 \times 10^{- 8} x - 1.0 \times 10^{- 14}

We can solve for

x

using the quadratic formula, which gives the following solutions:

x = 7.3 \times 10^{- 8} M, - 1.4 \times 10^{- 7} M

Since the concentration of

{OH}^{-}

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

H^{+}

and calculate

pH

, we get:

\begin{aligned} pH & = - log [H^{+}] \\ = - log [6.3 \times 10^{- 8} + x] \\ = - log [6.3 \times 10^{- 8} + 7.3 \times 10^{- 8}] \\ = - 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

HCl

solution has a

pH

that is weakly acidic. Whew!

Summary

Water can undergo autoionization to form $H_{3} O^{+}$ ‍ and ${OH}^{-}$ ‍ ions.
The equilibrium constant for the autoionization of water, $K_{w}$ ‍, is $10^{- 14}$ ‍ at $25^{\circ} C$ ‍.
In a neutral solution, $[H_{3} O^{+}] = [{OH}^{-}]$ ‍
In an acidic solution, $[H_{3} O^{+}] > [{OH}^{-}]$ ‍
In a basic solution, $[{OH}^{-}] > [H_{3} O^{+}]$ ‍
For aqueous solutions at $25^{\circ} C$ ‍, the following relationships are always true:

$K_{w} = [H_{3} O^{+}] [{OH}^{-}] = 10^{- 14}$ ‍

$pH + pOH = 14$ ‍

The contribution of the autoionization of water to $[H_{3} O^{+}]$ ‍ and $[{OH}^{-}]$ ‍ becomes significant for extremely dilute acid and base solutions.

Attributions

This article was adapted from the following articles:

"Water Autoionization" from UC Davis ChemWiki, CC BY-NC-SA 3.0
“Temperature dependence of the pH of pure water” from UC Davis ChemWiki, CC BY-NC-SA 3.0

The modified article is licensed under a CC-BY-NC-SA 4.0 license.

Additional References

Zumdahl, S.S., and Zumdahl S.A. (2003). Atomic Structure and Periodicity. In Chemistry (6th ed., pp. 290-94), Boston, MA: Houghton Mifflin Company.

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sarvin9574
Posted 8 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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(18 votes)
Answer
- Ernest Zinck
  Posted 8 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...”
  (33 votes)
Amit Mukherjee
Posted 8 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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(12 votes)
Answer
- yuki
  Posted 8 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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  (11 votes)
just_miku.py
Posted 7 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)
Answer
- RogerP
  Posted 7 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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  (10 votes)
Amit Mukherjee
Posted 8 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)
Answer
- Matt B
  Posted 8 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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  (16 votes)
Nasrullah Sami
Posted 6 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?
Button navigates to signup pageComment on Nasrullah Sami's post “In exercise 2, the questi...”
(5 votes)
Answer
- RolandKan3
  Posted 6 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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  (3 votes)
johnny
Posted 8 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 8 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 5 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)
Answer
wanglx123456789
Posted 3 years ago. Direct link to wanglx123456789's post “"Since the concentration ...”
"Since the concentration of OH- can't be negative, we can eliminate the second solution.
Doesn't it sound too loose and unlogical? We can just casually remove a certain value from calculation?
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(0 votes)
Answer
- Richard
  Posted 3 years ago. Direct link to Richard's post “If you're solving a quadr...”
  If you're solving a quadratic equation, you can receive a maximum of two real solutions. In this case we get that with one being positive and one being negative. So it makes mathematical sense that we get two answers, but physically only the positive answer makes sense.
  
  A negative concentration means you have less than 0 moles of hydroxide, or less than nothing. We can have no hydroxide possible, 0 moles of hydroxide in the solution, but not less than that.
  
  It's similar to problems in geometry where we can only have positive areas or volumes since we're considering real physical objects even if the algebra sometimes allows a negative answer.
  
  Hope that helps.
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  (7 votes)
Elliot Natanov
Posted 8 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 8 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)
vitti
Posted 4 years ago. Direct link to vitti's post “Why don't we include soli...”
Why don't we include solids/liquids in Kw calculations? Are we only including gasses then?
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(3 votes)
Answer

Chemistry library

Course: Chemistry library > Unit 11

Key points

Water is amphoteric

Practice 1‍ : Identifying the role of water in a reaction

Autoionization of water

The autoionization constant, Kw‍

Relationship between the autoionization constant, pH‍ , and pOH‍

Example 1‍ : Calculating [OH−]‍ from pH‍

Method 1‍ : Using Eq. 1‍

Method 2‍ : Using Eq. 2‍

Definitions of acidic, basic, and neutral solutions

Practice 2‍ : Calculating pH‍ of water at 0∘C‍

Practice 3‍ : Calculating pKw‍ at 40∘C‍

Autoionization and Le Chatelier's principle

Autoionization matters for very dilute acid and base solutions

Example 2‍ : Calculating the pH‍ of a very dilute acid solution

Try 1: Ignoring the autoionization of water

Try 2: Including the contribution from autoionization to [H+]‍

Summary

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Practice $1$ ‍: Identifying the role of water in a reaction

The autoionization constant, $K_{w}$ ‍

Relationship between the autoionization constant, $pH$ ‍, and $pOH$ ‍

Example $1$ ‍: Calculating $[{OH}^{-}]$ ‍ from $pH$ ‍

Method $1$ ‍: Using Eq. $1$ ‍

Method $2$ ‍: Using Eq. $2$ ‍

Practice $2$ ‍: Calculating $pH$ ‍ of water at $0^{\circ} C$ ‍

Practice $3$ ‍: Calculating $p K_{w}$ ‍ at $40^{\circ} C$ ‍

Example $2$ ‍: Calculating the $pH$ ‍ of a very dilute acid solution

Try 2: Including the contribution from autoionization to $[H^{+}]$ ‍