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

1 Answer
anor277
Feb 1, 2018

#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#

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