Question

Assuming the volumes are additive, what is the `pH` of a solution prepared from `10*mL` of `0.010*mol*L^-1` `HCl`, and `990*mL` of `0.0001*mol*L^-1` `HCl`?

Answer 1

`pH=3.70`

Explanation:

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

WE gots `10xx10^-3*Lxx10^-2*mol*L^-1=1.0xx10^-4*mol` with respect to `H_3O^+`...from the first solution....

And `990xx10^-3*Lxx10^-4*mol*L^-1=9.90xx10^-5*mol` with respect to `H_3O^+`...from the second solution....

And so in the final LITRE volume we realize a concentration of....

`(1.0xx10^-4*mol+9.90xx10^-5*mol)/(1*L)=1.99xx10^-4*mol*L^-1` with respect to `H_3O^+`...

And `pH=-log_10(1.99xx10^-4)=-(3.70)=3.70`