Calculating the ph of a strong acid the concentration of h


In calculating the pH of a strong acid the concentration of H+ ions is taken to be the same as the initial concentration of the reactant. According to the equations I've seen [H+][OH-] = 10^(-14) the ion product constant of water @ 25 C. So, if the acid is added to water the [OH-] must decrease. What exactly happens to the [OH-] ions? I've given it some thought and I'm at a loss. If they were consumed by the addition of the H+ ions the equilibrium concentration of the H+ ions would not be the same as the concentration added to the solution from the acid. The whole concept is becoming very convoluted for me and I would greatly appreciate an explanation of what exactly is happening and why exactly it is happening. Thanks in advance.

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Chemistry: Calculating the ph of a strong acid the concentration of h
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