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

Meq. of `HCl_1` in mL `=10^(-5) xx 1`
Meq. of `HCl_(II)` in 1000 mL `= N xx 1000`
Since II is prepared by diluting I and Meq. does not change on dilution
i.e, Meq of HCl (conc.) =Meq of HCl (dil.)
`therefore 10^(-5) xx 1=N xx 1000`
`therefore N_(HCl_(II))=10^(-8)`
Now proceed as Problem 50 part (e)
`therefore pH=6.9788`