Calculate pH of -
`10^(2) M HCl`

Explanation

Strong acids ionize completely at normal dilutions
`10^(2) M HCl`
` HCl to H^(+) +Cl^(-)`
Conc. before dissociation `10^(2)M" 0 0"`
Conc. after dissociation `"0 "10^(2)" "10^(2)`
`therefore [H^(+)]=10^2M`
`therefore pH=-2`
But this is not true. This may be explained as follows :
Sorenson’s originally intended pH to be related to `[H^+],` but his fundamental method of measurement – the hydrogen electrode – is now known to de- pend on thermodynamics activities rather than `[H^+],` i.e., on log `a_(H^(+))=[H^(+)]f_(H)`. In dilute solutions `f_(H^(+))` is near enough to unity and thus, `-log [H^+]`. Thus, pH defined by `-log [H^+]` is not only of little theo- retical significance, but in fact cannot be measured directly. It has therefore, came to be accepted that pH `=-log_(10+), i.e.," pH of "10^2" M HCl"` cannot be calculated and it practically lies near to zero.