## anonymous one year ago When [OH-] changes from 10-9 to 10-11 in a solution: Kw = [H+][OH-] What happens to [H+]? A. Increases B.decreases C.remains constant

Kw refers to the ionic product of water, which is essentially the equilibrium constant for the following equilibrium reaction (known as the autoionisation of water): $H _{2}O _{(l)} \leftarrow \rightarrow H ^{+}_{(aq)} + OH ^{-}_{(aq)}$ This occurs because water molecules can act as both acids, in which they lose a H+ to form OH-, or as bases, in which they gain a H+ to form H3O+. The H3O+ ion (called an hydroxonium ion) is equivalent to the H+ ion in the above equation. This is what is happening in the forward direction (left to right) for this reaction These ions are relatively unstable however (i.e. highly reactive), meaning that they will quickly react back together to form H2O in the opposite direction (right to left) once more. This process is constantly happening in any solution containing water molecules. So, at any given time there will be a very small number of H+ ions and OH- ions floating about in an aqueous solution, with the majority of the time the left-hand side of the equilibrium favored. This leads to a very small value for Kw (1 x 10^-14 mol^2 dm-6 at 25 degrees Celsius). However, the nature of this reaction, in which a water molecule is able to 'split itself up' to form two ionic species, means that one H+ ion and one OH- ion is always formed. We can't have more OH- ions than H+ ions or vice versa, they are like a 'pair' for which there will always be one to match up with. A drop in concentration of the OH- ions, represented as [OH-] in the formula for Kw from 10-9 to 10-11 (mol dm-3) must mean a corresponding drop in the H+ concentration in solution to the same value also, as for the autoionisation of water: $[H ^{+}] = [OH ^{-}]$ Hope that helps! :)