Question

What would be the final pH if I add 25 ml of water to 25 ml of a 0.1N HCl?

Answer 1

`pH_"final"=1.30`

Explanation:

By definition, `pH=-log_10[H_3O^+]`

Initially, we gots `0.1*mol*L^-1` `HCl`...

...and `-log_10(0.1)=-log_10(10^-1)=-(-1)=1`...

We dilute the solution by half....

`[H_3O^+]=(25xx10^-3*Lxx0.10*mol*L^-1)/(50xx10^-3*L)=0.05*mol*L^-1`...

`pH=-log_10(0.05)=1.30`