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 Autoionization of water (Kw)

AUTOIONIZATION of WATER

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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 10¯7 M in pure water.

Kw = [H3O+] [OH¯]

1.0 x 10-14= [x][x]

x=1.00 x 10¯7 M

Result #1: The pH of pure water is 7

By definition, pH = -log [H3O+]

The pH of pure water then equals -log 10¯7, 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 10¯14 = -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.