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Autoionization of water (Kw)

 

AUTOIONIZATION of WATER

--->Auto Ionization of Water Tutorial<---

notes fromhttp://dbhs.wvusd.k12.ca.us/webdocs/AcidBase/Kw.html

H2O(l) + H2O(l) <==> H3O+(aq) + OH(aq)

Kw = [H3O+] [OH]

Kw = water autoionization constant=1.0 x 10-14 @ 25 C

From the chemical equation just above, it can be seen that H3O+ and OH concentrations are in the molar ratio of one-to-one. This means that [H3O+] = [OH].

Therefore the values of [H3O+] and [OH] can be determined by taking the square root of Kw. Hence, both [H3O+] and [OH] equal 1.00 x 107 M in pure water.

Kw = [H3O+] [OH]

1.0 x 10-14= [x][x]  

x=1.00 x 107 M

Result #1: The pH of pure water is 7

By definition, pH = -log [H3O+]

The pH of pure water then equals -log 107, which is 7.

At other temperatures neutrality changes:

T (C) Kw pH
0 1.14 x 10-15 7.47
10 2.93 x 10-15 7.27
20 6.81 x 10-15 7.08
25 1.00 x 10-14 7.00
30 1.47 x 10-14 6.92
40 2.92 x 10-14 6.77
50 5.48 x 10-14 6.63
100 5.13 x 10-13 6.14

Result #2: If the pH or the pOH is known, the other can be found.

Take the negative logarithm of each side of the Kw equation as follows:

- log Kw = -log [H3O+] + -log [OH]

-log 1.00 x 1014 = -log [H3O+] + -log [OH]

Note the use of the add sign on the right side of the equation. The result is usually written as:

pKw = pH + pOH = 14

This is an extremely important equation. Learn it well.


Result #3: If the [H3O+] or the [OH] is known, the other can be found.

Simply divide Kw by the known value to get the other.

Suppose [H3O+] is known, then:

[OH] = Kw / [H3O+]

Suppose [OH] is known, then:

[H3O+] = Kw / [OH]


Result #4: If one variable ( [H3O+] or [OH] ) changes value (either up or down), the other variable will change in the opposite direction.

The change in values will still preserve this fundamental equality:

Kw = [H3O+] [OH]

Suppose [H3O+] became larger, therefore the [OH] becomes smaller.
Suppose [OH] became larger, therefore the [H3O+] becomes smaller.

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