How many mL of 0.20 M hydrochloric acid is required to neutralize 100 mL of 0.80 M potassium hydroxide?

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Answer 1 · 2016-08-11T15:44:30

#"Volume of HCl"=400# #mL#.

#HCl(aq) + KOH(aq) rarr KCl(aq) + H_2O(l)#

Now the reaction is trivial here. It is nevertheless good practice to write the stoichiometric equation to show the #1:1# equivalence.

#"Moles of KOH"# #=# #0.80*mol*L^-1xx0.100*L# #=# #0.080*mol#. Note that here I normalized the volumes, #1*mL# #=# #10^-3L#.

Thus #"volume of HCl"# #=# #(0.080*mol)/(0.20*mol*L^-1)# #=# #0.40*L#

PS Sorry for shouting in the top answer; I still haven't got the hang of this editor.