Question

What is the `pH` of a solution comprising `100*mL` volumes EACH of `HCl(aq)` at `0.20*mol*L^-1` and `NaOH(aq)` at `0.18*mol*L^-1`?

Answer 1

`pH-=2`

Explanation:

We interrogate the chemical reaction....

`HCl(aq) + NaOH(aq) rarr NaCl(aq) + H_2O(l)`.

`"Moles of HCl(aq)"=100xx10^-3*Lxx0.2*mol*L^-1=2.0xx10^-2*mol`.

`"Moles of NaOH(aq)"=100xx10^-3*Lxx0.18*mol*L^-1=1.8xx10^-2*mol`.

There is thus an excess of hydronium ions, to the tune of `2.0xx10^-2*mol-1.8xx10^-2*mol=2.0xx10^-3*mol`

But this is dissolved in a solution whose volume is now `200xx10^-3*L`

And so `[HCl]=(2.0xx10^-3*mol)/(200xx10^-3*L)=0.01*mol*L^-1`

But `pH-=-log_10{[H_3O^+]}=-log_10(0.01)=2.0`