Confusion on ion product of water The ion product of water is defined as $$ K_w = [\ce{H+}][\ce{OH-}] $$ and in pure water $ [\ce{H+}] = [\ce{OH-}]= 1.0 \times 10^{-7}$ and $ K_w = 1.0 \times 10^{-14}$. It is said that if we dissolve some acidic substance in water then $[\ce{H+}]$ will increase and $[\ce{OH-}]$ will decrease and the value of $K_w$ remains constant. I understand that $[\ce{H+}]$ increases because acids donate $\ce{H+}$ (according to Brønsted definition of acids) but how $[\ce{OH-}]$ decreases. I mean a water molecule i.e. $\ce{H2O}$ will certainly break into $\ce{H+}$ and $\ce{OH-}$ and to this $\ce{H+}$ our acid has added it's part and hence it's concentration has increased but why the value of $[\ce{OH-}]$ gets lowered from $1.0 \times 10^{-14}$ , if water molecule were to ionize then whenever $\ce{H+}$ gets formed simultaneously we would get $\ce{OH-}$.  

I can cite a common example problem  


  The concentration of $\ce{OH-}$ ions in a certain household ammonia cleansing solution is $0.0025$. Calculate the concentration of $\ce{H+}$ ions. 


We can solve this problem by using the equation $$ K_w = 1.0 \times 10^{-14}$$
$$ [\ce{H+}] [\ce{OH-}] = 1.0 \times 10^{-14}$$
and if put the value of $[\ce{OH-}]$ in the above equation and solve $[\ce{H+}]$, then we would get $ 4.0 \times 10^{-12}$. Here we observe that the value of $[\ce{H+}]$ has decreased from the it's original value in pure water, well this is understandable because ammonia being a base would consume $\ce{H+}$ but how the value of $\ce{OH-}$ has increased.  

I want to know that how the chemical reaction can cause the increase or decrease of $\ce{OH-}$ when each molecule of $\ce{H2O}$ always going to yield one $\ce{H+}$ and one $\ce{OH-}$ always.  

Thank you. Any help will be much appreciated.
It is not a chemical problem, but a trivial mathematical problem.

If you have an equation $x \cdot y = c$, where $x$, $y$ are variables and $c$ is constant, then if $x$ increases, $y$ must decrease, otherwise $c$ is not a constant.

The key part is to understand that  chemical equilibrium means existence of 2 ongoing opposite chemical reactions of the same rate. Like

$$\ce{H2O &lt;=&gt; H+ + OH-}$$

where the rate of ion creation equals the rate of ion recombination.

If there is an excess or deficit  of either of  $\ce{H+}$   or $\ce{OH-}$ ions, the rate of their recombination changes, while the rate of their creation remains the same. As consequence, the product of their concentrations converges quickly towards $K_\mathrm{w}$ to be in the equilibrium again.

The ammonia reacts :

$$\ce{NH3 + H+ &lt;=&gt;&gt; NH4+}$$

That creates deficit of $\ce{H+}$. As consequence, dissociation

$$\ce{H2O -&gt; H+ + OH-}$$

Is faster then recombination

$$\ce{ H+ + OH- -&gt; H2O}$$

An alternative reaction mechanism is 
ammonia reacting with water:

$$\ce{NH3 + H2O &lt;=&gt; NH4+ + OH-}$$

what directly produces the excess of $\ce{OH-}$ that recombines with the most of the present $\ce{H+}$.

The production of $\ce{OH-}$ and elimination of $\ce{H+}$ continues, until their concentrations satisfy both the basicity constant of $\ce{NH3}$ ( or equivalently acidity constant of $\ce{NH4+}$ ) and the ion product of water  $K_\mathrm{w}$ 

$$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$$ 

$$K_\mathrm{b, \ce{NH3}}=\frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}=\frac{[\ce{NH4+}]K_\mathrm{w}}{[\ce{NH3}] [\ce{H+}] }=\frac{K_\mathrm{w}}{K_\mathrm{a, \ce{NH4+}}}$$