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.


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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