A solution of `HCI` has p`H = 5`. If `1mL` of it is diluted to `1L` what will be the `pH` of resulting solution?

Explanation

`HCI_(I)=10^(-5)M` since `pH=5`
Meq. Of `HCI_(I)` in mL `=10^(-5)xx1`
Meq. Of `HCI_(II)` in `1000mL =Nxx1000`
Since II is prepared by diluting I and Meq. does not change on dilution.
i.e., Meq. Of `HCI(conc.)= Meq. Of HCI(dil.)`
`:. 10^(-5)xx1=Nxx1000`
`:. N_(HCI_(II))=10^(-8)`
Now proceed as Problem 17 part (e ).
`:. pH=6.9788`