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

Question

1626 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-25T16:44:28

#pH-=2#

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#

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#?