Question

What is the resultant `pH` of the solution.....?

What is the resultant `pH` of the solution prepared by taking a `10*mL` volume of `HCl(aq)` of `0.10*mol*L^-1` concentration, and diluting this to a `1*L` volume?

Answer 1

`pH=3`

Explanation:

`pH=-log_(10)[H_3O^+]`

`[H_3O^+]=(10xx10^-3*Lxx0.1*mol*L^-1)/(1*L)`

`10^-3*mol*L^-1`

And thus `pH=-log_(10)[H_3O^+]`

`-log_(10)(10^-3)=3...........`

And I did this all without a calculator. Note that normally, we would always add acid to water, and not water to acid. However, here the acid concentration is sufficiently dilute to get away with reverse addition.