What is the equation to find pH and pOH?

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Answer 1 · 2016-06-02T18:21:08

$p H + p O H = 14$ $pH + pOH = 14$

$2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}$ $2H_2O rightleftharpoons H_3O^+ + OH^-$

The auto-ionization constant of water is given by:

$K_{W} = [H_{3} O^{+}] [O H^{-}]$ $K_W = [ H_3O^+] [ OH^-]$

Apply the $log$ $log$ function to both sides of the equation:

$log (K_{W}) = log ([H_{3} O^{+}] [O H^{-}])$ $log(K_W )=log( [ H_3O^+] [ OH^-])$

$log (K_{W}) = log [H_{3} O^{+}] + log [O H^{-}]$ $log(K_W )= log[ H_3O^+] + log[ OH^-]$

Multiply both sides of the equation by a negative sign

$-{log(K_W )}= -{\ log[ H_3O^+] + log[ OH^-]\ }$

$-log(K_W )= -log[ H_3O^+] - log[ OH^-]$

$-log(1.0xx10^-14)= -log[ H_3O^+] - log[ OH^-]$

$14 = -log[ H_3O^+] - log[ OH^-]$

$pH = -log[ H_3O^+] " and " pOH= - log[ OH^-]$

$14 = pH + pOH =>$

$pH + pOH = 14$