SAT Subject Test Chemistry,10 edition

(Marvins-Underground-K-12) #1
H 2 O(l)        H+  (aq)    +   OH− (aq)

This dissociation is an equilibrium reaction and is therefore described by an equilibrium constant,
Kw, known as the water dissociation constant:


Kw  =   [H+][OH-]   =   10−14   (at 25°C)

One can take the logarithm of both sides and manipulate the equation by using the properties of
logarithms:


log ([H+][OH−]) =   log (10−14)
log [H+] + log [OH−] = log (10−14), since log (xy) = log x + log y
= −14, since log (10p) = p
−log [H+] − log [OH−] = 14, where we have taken the negative of both sides
pH + pOH = 14

In pure H 2 O, [H+] is equal to [OH−], since equimolar amounts of H+ and of OH− are formed from the
dissociation process. The pH and pOH would therefore also be equal, both having a value of 7. A
solution with equal concentrations of H+ and OH− is neutral. A pH below 7 indicates a relative excess
of H+ ions, and therefore an acidic solution; a pH above 7 indicates a relative excess of OH− ions, and
therefore a basic solution.


It is important to realize that even when the pH deviates from the value of 7, the water dissociation
equilibrium still holds. If an acid, for example, HCl, dissociates in water at 25 °C and causes an
increase in proton concentration such that the pH falls below 7, the hydroxide ion concentration will
have to decrease so as to maintain the relation Kw = [H+][OH−] = 10−14. Despite the higher
concentration of H+ relative to OH−, the solution does not acquire a net positive charge because the
conjugate base of the dissociated acid will be negatively charged (for example, Cl−) and thus will
maintain charge neutrality.

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