Is it necessary to consider common ion effect when suppose a strong acid like HCl is added to water?

For example in this question: 10 mL of $10^{-6 } \ce{mol/L \ HCl}$ solution is mixed with 90 mL of $\ce{H2O}$. pH will change approximately? Normally by using $SV=constant$ I get pH will change by 1 unit (without taking common ion effect).

How will common ion effect affect the answer? Help.

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Yes, you will definitely need to account for common ion effect.

However, use Kw to compute the pH. Assume that there is a complete dissociation by $\ce{HCl}$. However, you would need to solve the according quadratic for Kw.

as such, $\ce{H+= concentration~~ H+ of~~ HCl + x ~~and ~~OH- = x}$.

You should get a result around 6.79=pH.

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