acid base - 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.

Answer

Another way to look at the problem:

If we add $\ce{H+}$ to the water through adding acid, then some of the $\ce{H+}$ would just remain in the solution as is, and some of them would react with $\ce{OH-}$ in the reaction $\ce{H+ + OH- -> H_2O}$. How much of the added $\ce{H+}$ reacts? To calculate this, you need to know that $\ce{[H+][OH-] = K_w}$ remains constant. So while $\ce{H+}$ is increased, $\ce{OH-}$ is decreased through $\ce{H+ + OH- -> H_2O}$.

If we were to add $\ce{OH-}$ to the water through bases, the same thing would happen: a portion would react with $\ce{H+}$ and a portion would remain in the solution, and it will happen in such a way that is dictated by the constant value of $K_w$.