AUTOIONIZATION of WATER

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

__notes fromhttp://dbhs.wvusd.k12.ca.us/webdocs/AcidBase/Kw.html__

H_{2}O_{(l)} + H_{2}O_{(l)}
<==> H_{3}O^{+}_{(aq)} + OH¯_{(aq)}

K_{w} = [H_{3}O^{+}] [OH¯]

K_{w} = water autoionization constant=1.0 x 10^{-14
}@ 25 °C

From the chemical equation just above, it can be seen that H_{3}O^{+}
and OH¯ concentrations are in the molar ratio of one-to-one. This means that [H_{3}O^{+}]
= [OH¯].

Therefore the values of [H_{3}O^{+}] and [OH¯]
can be determined by taking the square root of K_{w}. Hence, both [H_{3}O^{+}]
and [OH¯] equal 1.00 x 10¯^{7} M in pure water.

K_{w} = [H_{3}O^{+}] [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 [H_{3}O^{+}]

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

Take the negative logarithm of each side of the K_{w} equation as
follows:

- log K_{w} = -log [H_{3}O^{+}] + -log [OH¯]

-log 1.00 x 10¯^{14} = -log [H_{3}O^{+}] + -log [OH¯]

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

pK_{w} = pH + pOH = 14

This is an extremely important equation. Learn it well.

Result #3: If the [H_{3}O^{+}] or the [OH¯] is
known, the other can be found.

Simply divide K_{w} by the known value to get the other.

Suppose [H_{3}O^{+}] is known, then:

[OH¯] = K_{w} / [H_{3}O^{+}]

Suppose [OH¯] is known, then:

[H_{3}O^{+}] = K_{w} / [OH¯]

Result #4: If one variable ( [H_{3}O^{+}] 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:

K_{w} = [H_{3}O^{+}] [OH¯]

Suppose [H_{3}O^{+}] became larger, therefore the [OH¯]
becomes smaller.

Suppose [OH¯] became larger, therefore the [H_{3}O^{+}] becomes
smaller.

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