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Ionization of water, [H3O+] and [OH-}.

Submitted by Tara14 on Tue, 12/11/2012 - 16:25

Water ionizes to a very small degree into hydronium ions, \rm H_3O^+, and hydroxide ions, \rm OH^-:

\rm 2H_2O(l) \rightleftharpoons H_3O^+(aq)+OH^-(aq)
K_{\rm w}=\rm [H_3O^+][OH^-]=1.00 \times 10^{-14}

The very small value of the equilibrium constant, K_{\rm w}, should give you an appreciation for how few water molecules actually ionize in pure water.

  • In neutral solutions, \rm [H_3O^+]=[OH^-].
  • In basic solutions, \rm [H_3O^+]<[OH^-].
  • In acidic solutions, \rm [H_3O^+]>[OH^-].

But in all aqueous solutions, the product of the hydronium and hydroxide concentrations is equal to K_{\rm w}. Thus,  K_{\rm w} allows you to calculate \rm [H_3O^+] from \rm [OH^-], or vice versa.

5.00×10−3mol of \rm HBr are dissolved in water to make 12.0L of solution. What is the concentration of hydroxide ions, \rm [OH^-], in this solution?

so 5.00×10−3mol of \rm HBr are dissolved in water to make 12.0L so [H3O+] = 5.00×10−3/12 moles per litre. This is a strong acid which fully ionises.

[OH-] = Kw/[H3O+]